problem solving with linear function

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• A linear equation represents a straight line on a coordinate plane. It can be written in the form: y = mx + b where m is the slope of the line and b is the y-intercept.
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• To find the linear equation you need to know the slope and the y-intercept of the line. To find the slope use the formula m = (y2 - y1) / (x2 - x1) where (x1, y1) and (x2, y2) are two points on the line. The y-intercept is the point at which x=0.
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• The most basic linear equation is a first-degree equation with one variable, usually written in the form of y = mx + b, where m is the slope of the line and b is the y-intercept.

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4.2 Modeling with Linear Functions

Learning objectives.

In this section, you will:

• Build linear models from verbal descriptions.
• Model a set of data with a linear function.

Elan is a college student who plans to spend a summer in Seattle. Elan has saved $3,500 for their trip and anticipates spending $400 each week on rent, food, and activities. How can we write a linear model to represent this situation? What would be the x -intercept, and what can Elan learn from it? To answer these and related questions, we can create a model using a linear function. Models such as this one can be extremely useful for analyzing relationships and making predictions based on those relationships. In this section, we will explore examples of linear function models.

Building Linear Models from Verbal Descriptions

When building linear models to solve problems involving quantities with a constant rate of change, we typically follow the same problem strategies that we would use for any type of function. Let’s briefly review them:

• Identify changing quantities, and then define descriptive variables to represent those quantities. When appropriate, sketch a picture or define a coordinate system.
• Carefully read the problem to identify important information. Look for information that provides values for the variables or values for parts of the functional model, such as slope and initial value.
• Carefully read the problem to determine what we are trying to find, identify, solve, or interpret.
• Identify a solution pathway from the provided information to what we are trying to find. Often this will involve checking and tracking units, building a table, or even finding a formula for the function being used to model the problem.
• When needed, write a formula for the function.
• Solve or evaluate the function using the formula.
• Reflect on whether your answer is reasonable for the given situation and whether it makes sense mathematically.
• Clearly convey your result using appropriate units, and answer in full sentences when necessary.

Now let’s take a look at the student in Seattle. In Elan’s situation, there are two changing quantities: time and money. The amount of money they have remaining while on vacation depends on how long they stay. We can use this information to define our variables, including units.

So, the amount of money remaining depends on the number of weeks: M ( t ) M ( t ) .

Notice that the unit of dollars per week matches the unit of our output variable divided by our input variable. Also, because the slope is negative, the linear function is decreasing. This should make sense because she is spending money each week.

The rate of change is constant, so we can start with the linear model M ( t ) = m t + b . M ( t ) = m t + b . Then we can substitute the intercept and slope provided.

To find the t- intercept (horizontal axis intercept), we set the output to zero, and solve for the input.

The t -intercept (horizontal axis intercept) is 8.75 weeks. Because this represents the input value when the output will be zero, we could say that Elan will have no money left after 8.75 weeks.

When modeling any real-life scenario with functions, there is typically a limited domain over which that model will be valid—almost no trend continues indefinitely. Here the domain refers to the number of weeks. In this case, it doesn’t make sense to talk about input values less than zero. A negative input value could refer to a number of weeks before Elan saved $3,500, but the scenario discussed poses the question once they saved $3,500 because this is when the trip and subsequent spending starts. It is also likely that this model is not valid after the t -intercept (horizontal axis intercept), unless Elan uses a credit card and goes into debt. The domain represents the set of input values, so the reasonable domain for this function is 0 ≤ t ≤ 8.75. 0 ≤ t ≤ 8.75.

In this example, we were given a written description of the situation. We followed the steps of modeling a problem to analyze the information. However, the information provided may not always be the same. Sometimes we might be provided with an intercept. Other times we might be provided with an output value. We must be careful to analyze the information we are given, and use it appropriately to build a linear model.

Using a Given Intercept to Build a Model

Some real-world problems provide the vertical axis intercept, which is the constant or initial value. Once the vertical axis intercept is known, the t -intercept (horizontal axis intercept) can be calculated. Suppose, for example, that Hannah plans to pay off a no-interest loan from her parents. Her loan balance is $1,000. She plans to pay $250 per month until her balance is $0. The y -intercept is the initial amount of her debt, or $1,000. The rate of change, or slope, is -$250 per month. We can then use the slope-intercept form and the given information to develop a linear model.

Now we can set the function equal to 0, and solve for x x to find the x -intercept.

The x -intercept is the number of months it takes her to reach a balance of $0. The x -intercept is 4 months, so it will take Hannah four months to pay off her loan.

Using a Given Input and Output to Build a Model

Many real-world applications are not as direct as the ones we just considered. Instead they require us to identify some aspect of a linear function. We might sometimes instead be asked to evaluate the linear model at a given input or set the equation of the linear model equal to a specified output.

Given a word problem that includes two pairs of input and output values, use the linear function to solve a problem.

• Identify the input and output values.
• Convert the data to two coordinate pairs.
• Find the slope.
• Write the linear model.
• Use the model to make a prediction by evaluating the function at a given x -value.
• Use the model to identify an x -value that results in a given y -value.
• Answer the question posed.

Using a Linear Model to Investigate a Town’s Population

A town’s population has been growing linearly. In 2004, the population was 6,200. By 2009, the population had grown to 8,100. Assume this trend continues.

• ⓐ Predict the population in 2013.
• ⓑ Identify the year in which the population will reach 15,000.

The two changing quantities are the population size and time. While we could use the actual year value as the input quantity, doing so tends to lead to very cumbersome equations because the y -intercept would correspond to the year 0, more than 2000 years ago!

To make computation a little nicer, we will define our input as the number of years since 2004.

To predict the population in 2013 ( t = 9 t = 9 ), we would first need an equation for the population. Likewise, to find when the population would reach 15,000, we would need to solve for the input that would provide an output of 15,000. To write an equation, we need the initial value and the rate of change, or slope.

To determine the rate of change, we will use the change in output per change in input.

The problem gives us two input-output pairs. Converting them to match our defined variables, the year 2004 would correspond to t = 0 , t = 0 , giving the point ( 0 , 6200 ) . ( 0 , 6200 ) . Notice that through our clever choice of variable definition, we have “given” ourselves the y -intercept of the function. The year 2009 would correspond to t = 5, t = 5, giving the point ( 5 , 8100 ) . ( 5 , 8100 ) .

The two coordinate pairs are ( 0 , 6200 ) ( 0 , 6200 ) and ( 5 , 8100 ) . ( 5 , 8100 ) . Recall that we encountered examples in which we were provided two points earlier in the chapter. We can use these values to calculate the slope.

We already know the y -intercept of the line, so we can immediately write the equation:

To predict the population in 2013, we evaluate our function at t = 9. t = 9.

If the trend continues, our model predicts a population of 9,620 in 2013.

To find when the population will reach 15,000, we can set P ( t ) = 15000 P ( t ) = 15000 and solve for t . t .

Our model predicts the population will reach 15,000 in a little more than 23 years after 2004, or somewhere around the year 2027.

A company sells doughnuts. They incur a fixed cost of $25,000 for rent, insurance, and other expenses. It costs $0.25 to produce each doughnut.

ⓐ Write a linear model to represent the cost C C of the company as a function of x , x , the number of doughnuts produced. ⓑ Find and interpret the y -intercept.

A city’s population has been growing linearly. In 2008, the population was 28,200. By 2012, the population was 36,800. Assume this trend continues.

• ⓐ Predict the population in 2014.
• ⓑ Identify the year in which the population will reach 54,000.

Using a Diagram to Build a Model

It is useful for many real-world applications to draw a picture to gain a sense of how the variables representing the input and output may be used to answer a question. To draw the picture, first consider what the problem is asking for. Then, determine the input and the output. The diagram should relate the variables. Often, geometrical shapes or figures are drawn. Distances are often traced out. If a right triangle is sketched, the Pythagorean Theorem relates the sides. If a rectangle is sketched, labeling width and height is helpful.

Using a Diagram to Model Distance Walked

Anna and Emanuel start at the same intersection. Anna walks east at 4 miles per hour while Emanuel walks south at 3 miles per hour. They are communicating with a two-way radio that has a range of 2 miles. How long after they start walking will they fall out of radio contact?

In essence, we can partially answer this question by saying they will fall out of radio contact when they are 2 miles apart, which leads us to ask a new question:

"How long will it take them to be 2 miles apart"?

In this problem, our changing quantities are time and position, but ultimately we need to know how long will it take for them to be 2 miles apart. We can see that time will be our input variable, so we’ll define our input and output variables.

Because it is not obvious how to define our output variable, we’ll start by drawing a picture such as Figure 2 .

Initial Value: They both start at the same intersection so when t = 0 , t = 0 , the distance traveled by each person should also be 0. Thus the initial value for each is 0.

Rate of Change: Anna is walking 4 miles per hour and Emanuel is walking 3 miles per hour, which are both rates of change. The slope for A A is 4 and the slope for E E is 3.

Using those values, we can write formulas for the distance each person has walked.

For this problem, the distances from the starting point are important. To notate these, we can define a coordinate system, identifying the “starting point” at the intersection where they both started. Then we can use the variable, A , A , which we introduced above, to represent Anna’s position, and define it to be a measurement from the starting point in the eastward direction. Likewise, can use the variable, E , E , to represent Emanuel’s position, measured from the starting point in the southward direction. Note that in defining the coordinate system, we specified both the starting point of the measurement and the direction of measure.

We can then define a third variable, D , D , to be the measurement of the distance between Anna and Emanuel. Showing the variables on the diagram is often helpful, as we can see from Figure 3 .

Recall that we need to know how long it takes for D , D , the distance between them, to equal 2 miles. Notice that for any given input t , t , the outputs A ( t ) , E ( t ) , A ( t ) , E ( t ) , and D ( t ) D ( t ) represent distances.

Figure 2 shows us that we can use the Pythagorean Theorem because we have drawn a right angle.

Using the Pythagorean Theorem, we get:

In this scenario we are considering only positive values of t , t , so our distance D ( t ) D ( t ) will always be positive. We can simplify this answer to D ( t ) = 5 t . D ( t ) = 5 t . This means that the distance between Anna and Emanuel is also a linear function. Because D D is a linear function, we can now answer the question of when the distance between them will reach 2 miles. We will set the output D ( t ) = 2 D ( t ) = 2 and solve for t . t .

They will fall out of radio contact in 0.4 hour, or 24 minutes.

Should I draw diagrams when given information based on a geometric shape?

Yes. Sketch the figure and label the quantities and unknowns on the sketch.

Using a Diagram to Model Distance Between Cities

There is a straight road leading from the town of Westborough to Agritown 30 miles east and 10 miles north. Partway down this road, it junctions with a second road, perpendicular to the first, leading to the town of Eastborough. If the town of Eastborough is located 20 miles directly east of the town of Westborough, how far is the road junction from Westborough?

It might help here to draw a picture of the situation. See Figure 4 . It would then be helpful to introduce a coordinate system. While we could place the origin anywhere, placing it at Westborough seems convenient. This puts Agritown at coordinates ( 3 0 , 1 0 ) , ( 3 0 , 1 0 ) , and Eastborough at ( 2 0 , 0 ) . ( 2 0 , 0 ) .

Using this point along with the origin, we can find the slope of the line from Westborough to Agritown.

Now we can write an equation to describe the road from Westborough to Agritown.

From this, we can determine the perpendicular road to Eastborough will have slope m = – 3. m = – 3. Because the town of Eastborough is at the point (20, 0), we can find the equation.

We can now find the coordinates of the junction of the roads by finding the intersection of these lines. Setting them equal,

The roads intersect at the point (18, 6). Using the distance formula, we can now find the distance from Westborough to the junction.

One nice use of linear models is to take advantage of the fact that the graphs of these functions are lines. This means real-world applications discussing maps need linear functions to model the distances between reference points.

There is a straight road leading from the town of Timpson to Ashburn 60 miles east and 12 miles north. Partway down the road, it junctions with a second road, perpendicular to the first, leading to the town of Garrison. If the town of Garrison is located 22 miles directly east of the town of Timpson, how far is the road junction from Timpson?

Modeling a Set of Data with Linear Functions

Real-world situations including two or more linear functions may be modeled with a system of linear equations . Remember, when solving a system of linear equations, we are looking for points the two lines have in common. Typically, there are three types of answers possible, as shown in Figure 5 .

Given a situation that represents a system of linear equations, write the system of equations and identify the solution.

• Identify the input and output of each linear model.
• Identify the slope and y -intercept of each linear model.
• Find the solution by setting the two linear functions equal to another and solving for x , x , or find the point of intersection on a graph.

Building a System of Linear Models to Choose a Truck Rental Company

Jamal is choosing between two truck-rental companies. The first, Keep on Trucking, Inc., charges an up-front fee of $20, then 59 cents a mile. The second, Move It Your Way, charges an up-front fee of $16, then 63 cents a mile 4 . When will Keep on Trucking, Inc. be the better choice for Jamal?

The two important quantities in this problem are the cost and the number of miles driven. Because we have two companies to consider, we will define two functions in Table 1 .

A linear function is of the form f ( x ) = m x + b . f ( x ) = m x + b . Using the rates of change and initial charges, we can write the equations

Using these equations, we can determine when Keep on Trucking, Inc., will be the better choice. Because all we have to make that decision from is the costs, we are looking for when Move It Your Way, will cost less, or when K ( d ) < M ( d ) . K ( d ) < M ( d ) . The solution pathway will lead us to find the equations for the two functions, find the intersection, and then see where the K ( d ) K ( d ) function is smaller.

These graphs are sketched in Figure 6 , with K ( d ) K ( d ) in blue.

To find the intersection, we set the equations equal and solve:

This tells us that the cost from the two companies will be the same if 100 miles are driven. Either by looking at the graph, or noting that K ( d ) K ( d ) is growing at a slower rate, we can conclude that Keep on Trucking, Inc. will be the cheaper price when more than 100 miles are driven, that is d > 100 d > 100 .

Access this online resource for additional instruction and practice with linear function models.

• Interpreting a Linear Function

4.2 Section Exercises

Explain how to find the input variable in a word problem that uses a linear function.

Explain how to find the output variable in a word problem that uses a linear function.

Explain how to interpret the initial value in a word problem that uses a linear function.

Explain how to determine the slope in a word problem that uses a linear function.

Find the area of a parallelogram bounded by the y -axis, the line x = 3 , x = 3 , the line f ( x ) = 1 + 2 x , f ( x ) = 1 + 2 x , and the line parallel to f ( x ) f ( x ) passing through ( 2 , 7 ) . ( 2 , 7 ) .

Find the area of a triangle bounded by the x -axis, the line f ( x ) = 12 – 1 3 x , f ( x ) = 12 – 1 3 x , and the line perpendicular to f ( x ) f ( x ) that passes through the origin.

Find the area of a triangle bounded by the y -axis, the line f ( x ) = 9 – 6 7 x , f ( x ) = 9 – 6 7 x , and the line perpendicular to f ( x ) f ( x ) that passes through the origin.

Find the area of a parallelogram bounded by the x -axis, the line g ( x ) = 2 , g ( x ) = 2 , the line f ( x ) = 3 x , f ( x ) = 3 x , and the line parallel to f ( x ) f ( x ) passing through ( 6 , 1 ) . ( 6 , 1 ) .

For the following exercises, consider this scenario: A town’s population has been decreasing at a constant rate. In 2010 the population was 5,900. By 2012 the population had dropped to 4,700. Assume this trend continues.

Predict the population in 2016.

Identify the year in which the population will reach 0.

For the following exercises, consider this scenario: A town’s population has been increased at a constant rate. In 2010 the population was 46,020. By 2012 the population had increased to 52,070. Assume this trend continues.

Identify the year in which the population will reach 75,000.

For the following exercises, consider this scenario: A town has an initial population of 75,000. It grows at a constant rate of 2,500 per year for 5 years.

Find the linear function that models the town’s population P P as a function of the year, t , t , where t t is the number of years since the model began.

Find a reasonable domain and range for the function P . P .

If the function P P is graphed, find and interpret the x - and y -intercepts.

If the function P P is graphed, find and interpret the slope of the function.

When will the population reach 100,000?

What is the population in the year 12 years from the onset of the model?

For the following exercises, consider this scenario: The weight of a newborn is 7.5 pounds. The baby gained one-half pound a month for its first year.

Find the linear function that models the baby’s weight W W as a function of the age of the baby, in months, t . t .

Find a reasonable domain and range for the function W . W .

If the function W W is graphed, find and interpret the x - and y -intercepts.

If the function W is graphed, find and interpret the slope of the function.

When did the baby weight 10.4 pounds?

What is the output when the input is 6.2?

For the following exercises, consider this scenario: The number of people afflicted with the common cold in the winter months steadily decreased by 205 each year from 2005 until 2010. In 2005, 12,025 people were inflicted.

Find the linear function that models the number of people inflicted with the common cold C C as a function of the year, t . t .

Find a reasonable domain and range for the function C . C .

If the function C C is graphed, find and interpret the x - and y -intercepts.

If the function C C is graphed, find and interpret the slope of the function.

When will the output reach 0?

In what year will the number of people be 9,700?

For the following exercises, use the graph in Figure 7 , which shows the profit, y , y , in thousands of dollars, of a company in a given year, t , t , where t t represents the number of years since 1980.

Find the linear function y , y , where y y depends on t , t , the number of years since 1980.

Find and interpret the y -intercept.

Find and interpret the x -intercept.

Find and interpret the slope.

For the following exercises, use the graph in Figure 8 , which shows the profit, y , y , in thousands of dollars, of a company in a given year, t , t , where t t represents the number of years since 1980.

For the following exercises, use the median home values in Mississippi and Hawaii (adjusted for inflation) shown in Table 2 . Assume that the house values are changing linearly.

In which state have home values increased at a higher rate?

If these trends were to continue, what would be the median home value in Mississippi in 2010?

If we assume the linear trend existed before 1950 and continues after 2000, the two states’ median house values will be (or were) equal in what year? (The answer might be absurd.)

For the following exercises, use the median home values in Indiana and Alabama (adjusted for inflation) shown in Table 3 . Assume that the house values are changing linearly.

If these trends were to continue, what would be the median home value in Indiana in 2010?

Real-World Applications

In 2004, a school population was 1001. By 2008 the population had grown to 1697. Assume the population is changing linearly.

• ⓐ How much did the population grow between the year 2004 and 2008?
• ⓑ How long did it take the population to grow from 1001 students to 1697 students?
• ⓒ What is the average population growth per year?
• ⓓ What was the population in the year 2000?
• ⓔ Find an equation for the population, P , P , of the school t years after 2000.
• ⓕ Using your equation, predict the population of the school in 2011.

In 2003, a town’s population was 1431. By 2007 the population had grown to 2134. Assume the population is changing linearly.

• ⓐ How much did the population grow between the year 2003 and 2007?
• ⓑ How long did it take the population to grow from 1431 people to 2134 people?
• ⓔ Find an equation for the population, P , P , of the town t t years after 2000.
• ⓕ Using your equation, predict the population of the town in 2014.

A phone company has a monthly cellular plan where a customer pays a flat monthly fee and then a certain amount of money per minute used for voice or video calling. If a customer uses 410 minutes, the monthly cost will be $71.50. If the customer uses 720 minutes, the monthly cost will be $118.

• ⓐ Find a linear equation for the monthly cost of the cell plan as a function of x , the number of monthly minutes used.
• ⓑ Interpret the slope and y -intercept of the equation.
• ⓒ Use your equation to find the total monthly cost if 687 minutes are used.

A phone company has a monthly cellular data plan where a customer pays a flat monthly fee of $10 and then a certain amount of money per megabyte (MB) of data used on the phone. If a customer uses 20 MB, the monthly cost will be $11.20. If the customer uses 130 MB, the monthly cost will be $17.80.

• ⓐ Find a linear equation for the monthly cost of the data plan as a function of x , x , the number of MB used.
• ⓒ Use your equation to find the total monthly cost if 250 MB are used.

In 1991, the moose population in a park was measured to be 4,360. By 1999, the population was measured again to be 5,880. Assume the population continues to change linearly.

• ⓐ Find a formula for the moose population, P since 1990.
• ⓑ What does your model predict the moose population to be in 2003?

In 2003, the owl population in a park was measured to be 340. By 2007, the population was measured again to be 285. The population changes linearly. Let the input be years since 2003.

• ⓐ Find a formula for the owl population, P . P . Let the input be years since 2003.
• ⓑ What does your model predict the owl population to be in 2012?

The Federal Helium Reserve held about 16 billion cubic feet of helium in 2010 and is being depleted by about 2.1 billion cubic feet each year.

• ⓐ Give a linear equation for the remaining federal helium reserves, R , R , in terms of t , t , the number of years since 2010.
• ⓑ In 2015, what will the helium reserves be?
• ⓒ If the rate of depletion doesn’t change, in what year will the Federal Helium Reserve be depleted?

Suppose the world’s oil reserves in 2014 are 1,820 billion barrels. If, on average, the total reserves are decreasing by 25 billion barrels of oil each year:

• ⓐ Give a linear equation for the remaining oil reserves, R , R , in terms of t , t , the number of years since now.
• ⓑ Seven years from now, what will the oil reserves be?
• ⓒ If the rate at which the reserves are decreasing is constant, when will the world’s oil reserves be depleted?

You are choosing between two different prepaid cell phone plans. The first plan charges a rate of 26 cents per minute. The second plan charges a monthly fee of $19.95 plus 11 cents per minute. How many minutes would you have to use in a month in order for the second plan to be preferable?

You are choosing between two different window washing companies. The first charges $5 per window. The second charges a base fee of $40 plus $3 per window. How many windows would you need to have for the second company to be preferable?

When hired at a new job selling jewelry, you are given two pay options:

Option A: Base salary of $17,000 a year with a commission of 12% of your sales

Option B: Base salary of $20,000 a year with a commission of 5% of your sales

How much jewelry would you need to sell for option A to produce a larger income?

When hired at a new job selling electronics, you are given two pay options:

Option A: Base salary of $14,000 a year with a commission of 10% of your sales

Option B: Base salary of $19,000 a year with a commission of 4% of your sales

How much electronics would you need to sell for option A to produce a larger income?

Option A: Base salary of $20,000 a year with a commission of 12% of your sales

Option B: Base salary of $26,000 a year with a commission of 3% of your sales

Option A: Base salary of $10,000 a year with a commission of 9% of your sales

Option B: Base salary of $20,000 a year with a commission of 4% of your sales

• 4 Rates retrieved Aug 2, 2010 from http://www.budgettruck.com and http://www.uhaul.com/

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Solving Linear Equations

Solving linear equations means finding the value of the variable(s) given in the linear equations. A linear equation is a combination of an algebraic expression and an equal to (=) symbol. It has a degree of 1 or it can be called a first-degree equation. For example, x + y = 4 is a linear equation. Sometimes, we may have to find the values of variables involved in a linear equation. When we are given two or more such linear equations, we can find the values of each variable by solving linear equations. There are a few methods to solve linear equations. Let us discuss each of these methods in detail.

Solving Linear Equations in One Variable

A linear equation in one variable is an equation of degree one and has only one variable term. It is of the form 'ax+b = 0', where 'a' is a non zero number and 'x' is a variable. By solving linear equations in one variable, we get only one solution for the given variable. An example for this is 3x - 6 = 0. The variable 'x' has only one solution, which is calculated as 3x - 6 = 0 3x = 6 x = 6/3 x = 2

For solving linear equations with one variable, simplify the equation such that all the variable terms are brought to one side and the constant value is brought to the other side. If there are any fractional terms then find the LCM ( Least Common Multiple ) and simplify them such that the variable terms are on one side and the constant terms are on the other side. Let us work out a small example to understand this.

4x + 8 = 8x - 10. To find the value of 'x', let us simplify and bring the 'x' terms to one side and the constant terms to another side.

4x - 8x = -10 - 8 -4x = -18 4x = 18 x = 18/4 On simplifying, we get x = 9/2.

Solving Linear Equations by Substitution Method

The substitution method is one of the methods of solving linear equations. In the substitution method , we rearrange the equation such that one of the values is substituted in the second equation. Now that we are left with an equation that has only one variable, we can solve it and find the value of that variable. In the two given equations, any equation can be taken and the value of a variable can be found and substituted in another equation. For solving linear equations using the substitution method, follow the steps mentioned below. Let us understand this with an example of solving the following system of linear equations. x + y = 6 --------------(1) 2x + 4y = 20 -----------(2)

Step 1: Find the value of one of the variables using any one of the equations. In this case, let us find the value of 'x' from equation (1). x + y = 6 ---------(1) x = 6 - y Step 2: Substitute the value of the variable found in step 1 in the second linear equation. Now, let us substitute the value of 'x' in the second equation 2x + 4y = 20.

x = 6 - y Substituting the value of 'x' in 2x + 4y = 20, we get,

2(6 - y) + 4y = 20 12 - 2y + 4y = 20 12 + 2y = 20 2y = 20 - 12 2y = 8 y = 8/2 y = 4 Step 3: Now substitute the value of 'y' in either equation (1) or (2). Let us substitute the value of 'y' in equation (1).

x + y = 6 x + 4 = 6 x = 6 - 4 x = 2 Therefore, by substitution method, the linear equations are solved, and the value of x is 2 and y is 4.

Solving Linear Equations by Elimination Method

The elimination method is another way to solve a system of linear equations. Here we make an attempt to multiply either the 'x' variable term or the 'y' variable term with a constant value such that either the 'x' variable terms or the 'y' variable terms cancel out and gives us the value of the other variable. Let us understand the steps of solving linear equations by elimination method . Consider the given linear equations: 2x + y = 11 ----------- (1) x + 3y = 18 ---------- (2) Step 1: Check whether the terms are arranged in a way such that the 'x' term is followed by a 'y' term and an equal to sign and after the equal to sign the constant term should be present. The given set of linear equations are already arranged in the correct way which is ax+by=c or ax+by-c=0.

Step 2: The next step is to multiply either one or both the equations by a constant value such that it will make either the 'x' terms or the 'y' terms cancel out which would help us find the value of the other variable. Now in equation (2), let us multiply every term by the number 2 to make the coefficients of x the same in both the equations. x + 3y = 18 ---------- (2) Multiplying all the terms in equation (2) by 2, we get,

2(x) + 2(3y) = 2(18). Now equation (2) becomes, 2x + 6y = 36 -----------(2)

Therefore, y = 5. Step 4: Using the value obtained in step 3, find out the value of another variable by substituting the value in any of the equations. Let us substitute the value of 'y' in equation (1). We get, 2x + y = 11 2x + 5 = 11 2x = 11 - 5 2x = 6 x = 6/2 x = 3

Therefore, by solving linear equations, we get the value of x = 3 and y = 5.

Graphical Method of Solving Linear Equations

Another method for solving linear equations is by using the graph. When we are given a system of linear equations, we graph both the equations by finding values for 'y' for different values of 'x' in the coordinate system. Once it is done, we find the point of intersection of these two lines. The (x,y) values at the point of intersection give the solution for these linear equations. Let us take two linear equations and solve them using the graphical method.

x + y = 8 -------(1)

y = x + 2 --------(2)

Let us take some values for 'x' and find the values for 'y' for the equation x + y = 8. This can also be rewritten as y = 8 - x.

Let us take some values for 'x' and find the values for 'y' in the equation y = x + 2.

Plotting these points on the coordinate plane, we get a graph like this.

Now, we find the point of intersection of these lines to find the values of 'x' and 'y'. The two lines intersect at the point (3,5). Therefore, x = 3 and y = 5 by using the graphical method of solving linear equations .

This method is also used to find the optimal solution of linear programming problems. Let us look at one more method of solving linear equations, which is the cross multiplication method.

Cross Multiplication Method of Solving Linear Equations

The cross multiplication method enables us to solve linear equations by picking the coefficients of all the terms ('x' , 'y' and the constant terms) in the format shown below and apply the formula for finding the values of 'x' and 'y'.

Topics Related to Solving Linear Equations

Check the given articles related to solving linear equations.

• Linear Equations
• Application of Linear Equations
• Two-Variable Linear Equations
• Linear Equations and Half Planes
• One Variable Linear Equations and Inequations

Solving Linear Equations Examples

Example 1: Solve the following linear equations by the substitution method.

3x + y = 13 --------- (1) 2x + 3y = 18 -------- (2)

By using the substitution method of solving linear equations, let us take the first equation and find the value of 'y' and substitute it in the second equation.

From equation (1), y = 13-3x. Now, substituting the value of 'y' in equation (2), we get, 2x + 3 (13 - 3x) = 18 2x + 39 - 9x = 18 -7x + 39 = 18 -7x = 18 - 39 -7x = -21 x = -21/-7 x = 3 Now, let us substitute the value of 'x = 3' in equation (1) and find the value of 'y'. 3x + y = 13 ------- (1) 3(3) + y = 13 9 + y = 13 y = 13 - 9 y = 4

Therefore, by the substitution method, the value of x is 3 and y is 4.

Example 2: Using the elimination method of solving linear equations find the values of 'x' and 'y'.

3x + y = 21 ------ (1) 2x + 3y = 28 -------- (2)

By using the elimination method, let us make the 'y' variable to be the same in both the equations (1) and (2). To do this let us multiply all the terms of the first equation by 3. Therefore equation (1) becomes,

3(3x) + 3(y) = 63 9x + 3y = 63 ---------- (3) The second equation is, 2x + 3y = 28 Now let us cancel the 'y' terms and find the value of 'x' by subtracting equation (2) from equation (3). This is done by changing the signs of all the terms in equation (2).

Example 3: Using the cross multiplication method of solving linear equations, solve the following equations.

x + 2y - 16 = 0 --------- (1) 4x - y - 10 = 0 ---------- (2)

Compare the given equation with \(a_{1}\)x + \(b_{1}\)y + \(c_{1}\) = 0, and \(a_{2}\)x+\(b_{2}\)y+\(c_{2}\) = 0. From the given equations,

\(a_{1}\) = 1, \(a_{2}\) = 4, \(b_{1}\) = 2, \(b_{2}\) = -1, \(c_{1}\) = -16, and \(c_{2}\) = -10.

By cross multiplication method,

x = \(b_{1}\)\(c_{2}\) - \(b_{2}\)\(c_{1}\)/\(a_{1}\)\(b_{2}\) - \(a_{2}\)\(b_{1}\) y = \(c_{1}\)\(a_{2}\) - \(c_{2}\)\(a_{1}\) / \(a_{1}\)\(b_{2}\) - \(a_{2}\)\(b_{1}\)

Substituting the values in the formula we get,

x = ((2)(-10)) - ((-1)(-16)) / ((1)(-1)) - ((4)(2)) x = (-20-16)/(-1-8) x = -36/-9 x = 36/9 x = 4 y = ((-16)(4)) - ((-10)(1)) / ((1)(-1)) - ((4)(2)) y = (-64 + 10) / (-1 - 8) y = -54 / -9 y = 54/9 y = 6 Therefore, by the cross multiplication method, the value of x is 4 and y is 6.

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Practice Questions on Solving Linear Equations

Faqs on solving linear equations, what does it mean by solving linear equations.

An equation that has a degree of 1 is called a linear equation. We can have one variable linear equations , two-variable linear equations , linear equations with three variables, and more depending on the number of variables in it. Solving linear equations means finding the values of all the variables present in the equation. This can be done by substitution method, elimination method, graphical method, and the cross multiplication method . All these methods are different ways of finding the values of the variables.

How to Use the Substitution Method for Solving Linear Equations?

The substitution method of solving equations states that for a given system of linear equations, find the value of either 'x' or 'y' from any of the given equations and then substitute the value found of 'x' or 'y' in another equation so that the other unknown value can be found.

How to Use the Elimination Method for Solving Linear Equations?

In the elimination method of solving linear equations, we multiply a constant or a number with one equation or both the equations such that either the 'x' terms or the 'y' terms are the same. Then we cancel out the same term in both the equations by either adding or subtracting them and find the value of one variable (either 'x' or 'y'). After finding one of the values, we substitute the value in one of the equations and find the other unknown value.

What is the Graphical Method of Solving Linear Equations?

In the graphical method of solving linear equations, we find the value of 'y' from the given equations by putting the values of x as 0, 1, 2, 3, and so on, and plot a graph in the coordinate system for the line for various values of 'x' for both the system of linear equations. We will see that these two lines intersect at a point. This point is the solution for the given system of linear equations. If there is no intersection point between two lines, then we consider them as parallel lines , and if we found that both the lines lie on each other, those are known as coincident lines and have infinitely many solutions.

What are the Steps of Solving Linear Equations that has One Variable?

A linear equation is an equation with degree 1. To solve a linear equation that has one variable we bring the variable to one side and the constant value to the other side. Then, a non-zero number may be added, subtracted, multiplied, or divided on both sides of the equation. For example, a linear equation with one variable will be of the form 'x - 4 = 2'. To find the value of 'x', we add the constant value '4' to both sides of the equation. Therefore, the value of 'x = 6'.

What are the Steps of Solving Linear Equations having Three Variables?

To solve a system of linear equations that has three variables, we take any two equations and variables. We then take another pair of linear equations and also solve for the same variable. Now that, we have two linear equations with two variables, we can use the substitution method or elimination method, or any other method to solve the values of two unknown variables. After finding these two variables, we substitute them in any of the three equations to find the third unknown variable.

What are the 4 Methods of Solving Linear Equations?

The methods for solving linear equations are given below:

• Substitution method
• Elimination method
• Cross multiplication method
• Graphical method

• Mathematicians
• Math Lessons
• Square Roots
• Math Calculators

How to Solve a Linear Function – A Step-by-Step Guide

To solve a linear function, I always begin by identifying its standard form, which is typically expressed as $y = mx + b$ . In this equation, (m) represents the slope of the line , and (b) denotes the y-intercept , where the line crosses the y-axis.

By knowing these components, I can graph the function or work with it algebraically. It’s crucial to remember that in a linear equation , the variables (x) and (y) are to the first power, indicating a constant rate of change and resulting in a straight line when graphed.

Understanding how to manipulate this equation to isolate the variable of interest is the key to finding solutions.

Whether I’m substituting values to find points on the line or reformatting the equation to point-slope or standard form depends on the context of the problem.

Encountering various linear equation word problems sharpens my skills, as they often require a tailored approach for each unique scenario.

Stay tuned, as I’m about to unveil the step-by-step process to confidently tackle any linear function that comes my way.

Steps for Solving Linear Functions

When I approach linear functions , I often think of them as a puzzle where my goal is to find the value of the variable that makes the function true. A linear function takes the form of $f(x) = mx + b$, where $m$ is the slope or rate of change , and $b$ is the y-intercept . To make things easier, I’ve broken down the process into clear steps:

Understanding Function Notation : A linear function is typically written as $f(x)$, which is equivalent to $y$. It represents the output for a given input $x$.

Identifying Key Components :

• Slope ($m$) : Determines how steep the line is on a graph.
• Y-intercept ($b$) : Where the line crosses the y-axis.

Writing the Equation : Start by placing the known values of the slope and y-intercept into their proper places in the equation $y = mx + b$.

Plotting the Graph : On a coordinate plane , plot the y-intercept and use the slope to find a second point. Connect these points to visualize the function as a straight line .

Solving for a Specific Value :

• If you need $y$ for a particular $x$, substitute the $x$ value into the equation and solve for $y$.
• To find an $x$ value for a given $y$, you can rearrange the equation to isolate $x$ and then substitute the $y$ value.

Remember, linear equations involving two lines can be parallel or perpendicular based on their slopes:

• Parallel lines share the same slope ($m$).
• Perpendicular lines have slopes that are negative reciprocals of each other.

The process of solving for a linear function is straightforward if you take it step by step. Just remember that you’re working with constants and a variable to find points that lie on a line, and that’s the heart of linear equations .

Applications in Real-World Contexts

In my day-to-day life, I often encounter situations where linear functions are incredibly useful.

 These functions, which can be written in the form $f(x) = mx + b$, where $m$ is the slope and $b$ is the y-intercept, model relationships with a constant rate of change. Here are a few examples of how I’ve seen linear functions shine in real-world contexts:

Budgeting and Finance: For instance, if I wanted to understand how my savings account grows over time, I could use a linear function . If I start with $200 and save $50 each week, the amount of money, $M$, after $t$ weeks is represented by $M(t) = 50t + 200$.

Cooking and Recipes: When I’m cooking and need to adjust the recipe according to the number of guests, I use linear functions to scale the ingredients. If a recipe calls for 2 cups of flour for 4 cookies, the equation would look like $c = \frac{1}{2}n$, where $c$ is cups of flour and $n$ is the number of cookies.

Travel and Distance: If I go for a jog and keep a steady pace, the distance I cover can be predicted with a linear function . The distance $d$ in miles, at a constant speed $s$, after jogging for $t$ hours, would be $d(t) = st$.

These are simple applications, but they show how understanding linear functions can apply to anything from managing finances to making dinner or exercising, making this concept a valuable tool in my everyday life.

Solving linear functions can be a satisfying experience, as it sharpens my problem-solving skills and enhances my understanding of algebra.

I’ve learned that by following a methodical approach, such as using the slope-intercept form, which is denoted as $f(x) = mx + b$ , handling linear functions becomes much more straightforward.

It’s important to remember the role of $m$ , which represents the slope, and $b$, the y-intercept, in graphing these linear functions .

When I embark on solving linear equations , it’s crucial to apply systematic methods—graphing, substitution, elimination, or matrices—based on the context of the problem. Each method has its own merits, and knowing when to use each can save me time and effort.

Moreover, understanding how to interpret the solutions in real-life scenarios is an integral part of the learning process.

Whether I am predicting outcomes or determining the relationship between variables, linear functions serve as a fundamental tool in various fields, from economics to engineering.

I encourage fellow learners to practice regularly, as this will certainly fortify their abilities to solve and apply linear functions effectively. The more I engage with these concepts, the more intuitive and rewarding they become.

• Pre Calculus
• Probability
• Sets & Set Theory
• Trigonometry

Linear Equations

A linear equation is an equation for a straight line

These are all linear equations:

Let us look more closely at one example:

Example: y = 2x + 1 is a linear equation:

The graph of y = 2x+1 is a straight line

• When x increases, y increases twice as fast , so we need 2x
• When x is 0, y is already 1. So +1 is also needed
• And so: y = 2x + 1

Here are some example values:

Check for yourself that those points are part of the line above!

Different Forms

There are many ways of writing linear equations, but they usually have constants (like "2" or "c") and must have simple variables (like "x" or "y").

Examples: These are linear equations:

But the variables (like "x" or "y") in Linear Equations do NOT have:

• Exponents (like the 2 in x 2 )
• Square roots , cube roots , etc

Examples: These are NOT linear equations:

Slope-intercept form.

The most common form is the slope-intercept equation of a straight line :

Example: y = 2x + 1

• Slope: m = 2
• Intercept: b = 1

Point-Slope Form

Another common one is the Point-Slope Form of the equation of a straight line:

Example: y − 3 = (¼)(x − 2)

It is in the form y − y 1 = m(x − x 1 ) where:

General Form

And there is also the General Form of the equation of a straight line:

Example: 3x + 2y − 4 = 0

It is in the form Ax + By + C = 0 where:

There are other, less common forms as well.

As a Function

Sometimes a linear equation is written as a function , with f(x) instead of y :

And functions are not always written using f(x):

The Identity Function

There is a special linear function called the "Identity Function":

And here is its graph:

It is called "Identity" because what comes out is identical to what goes in:

Constant Functions

Another special type of linear function is the Constant Function ... it is a horizontal line:

No matter what value of "x", f(x) is always equal to some constant value.

Using Linear Equations

You may like to read some of the things you can do with lines:

• Finding the Midpoint of a Line Segment
• Finding Parallel and Perpendicular Lines
• Finding the Equation of a Line from 2 Points

Grade 8 Mathematics Module: “Solving Problems Involving Linear Functions”

This Self-Learning Module (SLM) is prepared so that you, our dear learners, can continue your studies and learn while at home. Activities, questions, directions, exercises, and discussions are carefully stated for you to understand each lesson.

Each SLM is composed of different parts. Each part shall guide you step-by-step as you discover and understand the lesson prepared for you.

Pre-tests are provided to measure your prior knowledge on lessons in each SLM. This will tell you if you need to proceed on completing this module or if you need to ask your facilitator or your teacher’s assistance for better understanding of the lesson. At the end of each module, you need to answer the post-test to self-check your learning. Answer keys are provided for each activity and test. We trust that you will be honest in using these.

Please use this module with care. Do not put unnecessary marks on any part of this SLM. Use a separate sheet of paper in answering the exercises and tests. And read the instructions carefully before performing each task.

If you have any questions in using this SLM or any difficulty in answering the tasks in this module, do not hesitate to consult your teacher or facilitator.

In this module, you will be acquainted with solving problems involving linear functions. The scope of this module enables you to use it in many different learning situations. The lesson is arranged to follow the standard sequence of the course. But the order in which you read them can be changed to correspond with the textbook you are now using.

This module contains:

• Lesson 1: Solving Problems Involving Linear Functions

After going through this module, you are expected to:

1. identify steps in modeling and solving word problems involving linear functions;

2. create linear functions that represent relation between quantities; and

3. apply the concepts of linear function in solving real-life problems.

Grade 8 Mathematics Quarter 2 Self-Learning Module: “Solving Problems Involving Linear Functions”

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• Published: 10 April 2024

A hybrid particle swarm optimization algorithm for solving engineering problem

• Jinwei Qiao 1 , 2 ,
• Guangyuan Wang 1 , 2 ,
• Zhi Yang 1 , 2 ,
• Xiaochuan Luo 3 ,
• Jun Chen 1 , 2 ,
• Kan Li 4 &
• Pengbo Liu 1 , 2  

Scientific Reports volume  14 , Article number:  8357 ( 2024 ) Cite this article

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• Computational science
• Mechanical engineering

To overcome the disadvantages of premature convergence and easy trapping into local optimum solutions, this paper proposes an improved particle swarm optimization algorithm (named NDWPSO algorithm) based on multiple hybrid strategies. Firstly, the elite opposition-based learning method is utilized to initialize the particle position matrix. Secondly, the dynamic inertial weight parameters are given to improve the global search speed in the early iterative phase. Thirdly, a new local optimal jump-out strategy is proposed to overcome the "premature" problem. Finally, the algorithm applies the spiral shrinkage search strategy from the whale optimization algorithm (WOA) and the Differential Evolution (DE) mutation strategy in the later iteration to accelerate the convergence speed. The NDWPSO is further compared with other 8 well-known nature-inspired algorithms (3 PSO variants and 5 other intelligent algorithms) on 23 benchmark test functions and three practical engineering problems. Simulation results prove that the NDWPSO algorithm obtains better results for all 49 sets of data than the other 3 PSO variants. Compared with 5 other intelligent algorithms, the NDWPSO obtains 69.2%, 84.6%, and 84.6% of the best results for the benchmark function ( \({f}_{1}-{f}_{13}\) ) with 3 kinds of dimensional spaces (Dim = 30,50,100) and 80% of the best optimal solutions for 10 fixed-multimodal benchmark functions. Also, the best design solutions are obtained by NDWPSO for all 3 classical practical engineering problems.

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Introduction.

In the ever-changing society, new optimization problems arise every moment, and they are distributed in various fields, such as automation control 1 , statistical physics 2 , security prevention and temperature prediction 3 , artificial intelligence 4 , and telecommunication technology 5 . Faced with a constant stream of practical engineering optimization problems, traditional solution methods gradually lose their efficiency and convenience, making it more and more expensive to solve the problems. Therefore, researchers have developed many metaheuristic algorithms and successfully applied them to the solution of optimization problems. Among them, Particle swarm optimization (PSO) algorithm 6 is one of the most widely used swarm intelligence algorithms.

However, the basic PSO has a simple operating principle and solves problems with high efficiency and good computational performance, but it suffers from the disadvantages of easily trapping in local optima and premature convergence. To improve the overall performance of the particle swarm algorithm, an improved particle swarm optimization algorithm is proposed by the multiple hybrid strategy in this paper. The improved PSO incorporates the search ideas of other intelligent algorithms (DE, WOA), so the improved algorithm proposed in this paper is named NDWPSO. The main improvement schemes are divided into the following 4 points: Firstly, a strategy of elite opposition-based learning is introduced into the particle population position initialization. A high-quality initialization matrix of population position can improve the convergence speed of the algorithm. Secondly, a dynamic weight methodology is adopted for the acceleration coefficients by combining the iterative map and linearly transformed method. This method utilizes the chaotic nature of the mapping function, the fast convergence capability of the dynamic weighting scheme, and the time-varying property of the acceleration coefficients. Thus, the global search and local search of the algorithm are balanced and the global search speed of the population is improved. Thirdly, a determination mechanism is set up to detect whether the algorithm falls into a local optimum. When the algorithm is “premature”, the population resets 40% of the position information to overcome the local optimum. Finally, the spiral shrinking mechanism combined with the DE/best/2 position mutation is used in the later iteration, which further improves the solution accuracy.

The structure of the paper is given as follows: Sect. “ Particle swarm optimization (PSO) ” describes the principle of the particle swarm algorithm. Section “ Improved particle swarm optimization algorithm ” shows the detailed improvement strategy and a comparison experiment of inertia weight is set up for the proposed NDWPSO. Section “ Experiment and discussion ” includes the experimental and result discussion sections on the performance of the improved algorithm. Section “ Conclusions and future works ” summarizes the main findings of this study.

Literature review

This section reviews some metaheuristic algorithms and other improved PSO algorithms. A simple discussion about recently proposed research studies is given.

Metaheuristic algorithms

A series of metaheuristic algorithms have been proposed in recent years by using various innovative approaches. For instance, Lin et al. 7 proposed a novel artificial bee colony algorithm (ABCLGII) in 2018 and compared ABCLGII with other outstanding ABC variants on 52 frequently used test functions. Abed-alguni et al. 8 proposed an exploratory cuckoo search (ECS) algorithm in 2021 and carried out several experiments to investigate the performance of ECS by 14 benchmark functions. Brajević 9 presented a novel shuffle-based artificial bee colony (SB-ABC) algorithm for solving integer programming and minimax problems in 2021. The experiments are tested on 7 integer programming problems and 10 minimax problems. In 2022, Khan et al. 10 proposed a non-deterministic meta-heuristic algorithm called Non-linear Activated Beetle Antennae Search (NABAS) for a non-convex tax-aware portfolio selection problem. Brajević et al. 11 proposed a hybridization of the sine cosine algorithm (HSCA) in 2022 to solve 15 complex structural and mechanical engineering design optimization problems. Abed-Alguni et al. 12 proposed an improved Salp Swarm Algorithm (ISSA) in 2022 for single-objective continuous optimization problems. A set of 14 standard benchmark functions was used to evaluate the performance of ISSA. In 2023, Nadimi et al. 13 proposed a binary starling murmuration optimization (BSMO) to select the effective features from different important diseases. In the same year, Nadimi et al. 14 systematically reviewed the last 5 years' developments of WOA and made a critical analysis of those WOA variants. In 2024, Fatahi et al. 15 proposed an Improved Binary Quantum-based Avian Navigation Optimizer Algorithm (IBQANA) for the Feature Subset Selection problem in the medical area. Experimental evaluation on 12 medical datasets demonstrates that IBQANA outperforms 7 established algorithms. Abed-alguni et al. 16 proposed an Improved Binary DJaya Algorithm (IBJA) to solve the Feature Selection problem in 2024. The IBJA’s performance was compared against 4 ML classifiers and 10 efficient optimization algorithms.

Improved PSO algorithms

Many researchers have constantly proposed some improved PSO algorithms to solve engineering problems in different fields. For instance, Yeh 17 proposed an improved particle swarm algorithm, which combines a new self-boundary search and a bivariate update mechanism, to solve the reliability redundancy allocation problem (RRAP) problem. Solomon et al. 18 designed a collaborative multi-group particle swarm algorithm with high parallelism that was used to test the adaptability of Graphics Processing Units (GPUs) in distributed computing environments. Mukhopadhyay and Banerjee 19 proposed a chaotic multi-group particle swarm optimization (CMS-PSO) to estimate the unknown parameters of an autonomous chaotic laser system. Duan et al. 20 designed an improved particle swarm algorithm with nonlinear adjustment of inertia weights to improve the coupling accuracy between laser diodes and single-mode fibers. Sun et al. 21 proposed a particle swarm optimization algorithm combined with non-Gaussian stochastic distribution for the optimal design of wind turbine blades. Based on a multiple swarm scheme, Liu et al. 22 proposed an improved particle swarm optimization algorithm to predict the temperatures of steel billets for the reheating furnace. In 2022, Gad 23 analyzed the existing 2140 papers on Swarm Intelligence between 2017 and 2019 and pointed out that the PSO algorithm still needs further research. In general, the improved methods can be classified into four categories:

Adjusting the distribution of algorithm parameters. Feng et al. 24 used a nonlinear adaptive method on inertia weights to balance local and global search and introduced asynchronously varying acceleration coefficients.

Changing the updating formula of the particle swarm position. Both papers 25 and 26 used chaotic mapping functions to update the inertia weight parameters and combined them with a dynamic weighting strategy to update the particle swarm positions. This improved approach enables the particle swarm algorithm to be equipped with fast convergence of performance.

The initialization of the swarm. Alsaidy and Abbood proposed 27 a hybrid task scheduling algorithm that replaced the random initialization of the meta-heuristic algorithm with the heuristic algorithms MCT-PSO and LJFP-PSO.

Combining with other intelligent algorithms: Liu et al. 28 introduced the differential evolution (DE) algorithm into PSO to increase the particle swarm as diversity and reduce the probability of the population falling into local optimum.

Particle swarm optimization (PSO)

The particle swarm optimization algorithm is a population intelligence algorithm for solving continuous and discrete optimization problems. It originated from the social behavior of individuals in bird and fish flocks 6 . The core of the PSO algorithm is that an individual particle identifies potential solutions by flight in a defined constraint space adjusts its exploration direction to approach the global optimal solution based on the shared information among the group, and finally solves the optimization problem. Each particle \(i\) includes two attributes: velocity vector \({V}_{i}=\left[{v}_{i1},{v}_{i2},{v}_{i3},{...,v}_{ij},{...,v}_{iD},\right]\) and position vector \({X}_{i}=[{x}_{i1},{x}_{i2},{x}_{i3},...,{x}_{ij},...,{x}_{iD}]\) . The velocity vector is used to modify the motion path of the swarm; the position vector represents a potential solution for the optimization problem. Here, \(j=\mathrm{1,2},\dots ,D\) , \(D\) represents the dimension of the constraint space. The equations for updating the velocity and position of the particle swarm are shown in Eqs. ( 1 ) and ( 2 ).

Here \({Pbest}_{i}^{k}\) represents the previous optimal position of the particle \(i\) , and \({Gbest}\) is the optimal position discovered by the whole population. \(i=\mathrm{1,2},\dots ,n\) , \(n\) denotes the size of the particle swarm. \({c}_{1}\) and \({c}_{2}\) are the acceleration constants, which are used to adjust the search step of the particle 29 . \({r}_{1}\) and \({r}_{2}\) are two random uniform values distributed in the range \([\mathrm{0,1}]\) , which are used to improve the randomness of the particle search. \(\omega\) inertia weight parameter, which is used to adjust the scale of the search range of the particle swarm 30 . The basic PSO sets the inertia weight parameter as a time-varying parameter to balance global exploration and local seeking. The updated equation of the inertia weight parameter is given as follows:

where \({\omega }_{max}\) and \({\omega }_{min}\) represent the upper and lower limits of the range of inertia weight parameter. \(k\) and \(Mk\) are the current iteration and maximum iteration.

Improved particle swarm optimization algorithm

According to the no free lunch theory 31 , it is known that no algorithm can solve every practical problem with high quality and efficiency for increasingly complex and diverse optimization problems. In this section, several improvement strategies are proposed to improve the search efficiency and overcome this shortcoming of the basic PSO algorithm.

Improvement strategies

The optimization strategies of the improved PSO algorithm are shown as follows:

The inertia weight parameter is updated by an improved chaotic variables method instead of a linear decreasing strategy. Chaotic mapping performs the whole search at a higher speed and is more resistant to falling into local optimal than the probability-dependent random search 32 . However, the population may result in that particles can easily fly out of the global optimum boundary. To ensure that the population can converge to the global optimum, an improved Iterative mapping is adopted and shown as follows:

Here \({\omega }_{k}\) is the inertia weight parameter in the iteration \(k\) , \(b\) is the control parameter in the range \([\mathrm{0,1}]\) .

The acceleration coefficients are updated by the linear transformation. \({c}_{1}\) and \({c}_{2}\) represent the influential coefficients of the particles by their own and population information, respectively. To improve the search performance of the population, \({c}_{1}\) and \({c}_{2}\) are changed from fixed values to time-varying parameter parameters, that are updated by linear transformation with the number of iterations:

where \({c}_{max}\) and \({c}_{min}\) are the maximum and minimum values of acceleration coefficients, respectively.

The initialization scheme is determined by elite opposition-based learning . The high-quality initial population will accelerate the solution speed of the algorithm and improve the accuracy of the optimal solution. Thus, the elite backward learning strategy 33 is introduced to generate the position matrix of the initial population. Suppose the elite individual of the population is \({X}=[{x}_{1},{x}_{2},{x}_{3},...,{x}_{j},...,{x}_{D}]\) , and the elite opposition-based solution of \(X\) is \({X}_{o}=[{x}_{{\text{o}}1},{x}_{{\text{o}}2},{x}_{{\text{o}}3},...,{x}_{oj},...,{x}_{oD}]\) . The formula for the elite opposition-based solution is as follows:

where \({k}_{r}\) is the random value in the range \((\mathrm{0,1})\) . \({ux}_{oij}\) and \({lx}_{oij}\) are dynamic boundaries of the elite opposition-based solution in \(j\) dimensional variables. The advantage of dynamic boundary is to reduce the exploration space of particles, which is beneficial to the convergence of the algorithm. When the elite opposition-based solution is out of bounds, the out-of-bounds processing is performed. The equation is given as follows:

After calculating the fitness function values of the elite solution and the elite opposition-based solution, respectively, \(n\) high quality solutions were selected to form a new initial population position matrix.

The position updating Eq. ( 2 ) is modified based on the strategy of dynamic weight. To improve the speed of the global search of the population, the strategy of dynamic weight from the artificial bee colony algorithm 34 is introduced to enhance the computational performance. The new position updating equation is shown as follows:

Here \(\rho\) is the random value in the range \((\mathrm{0,1})\) . \(\psi\) represents the acceleration coefficient and \({\omega }{\prime}\) is the dynamic weight coefficient. The updated equations of the above parameters are as follows:

where \(f(i)\) denotes the fitness function value of individual particle \(i\) and u is the average of the population fitness function values in the current iteration. The Eqs. ( 11 , 12 ) are introduced into the position updating equation. And they can attract the particle towards positions of the best-so-far solution in the search space.

New local optimal jump-out strategy is added for escaping from the local optimal. When the value of the fitness function for the population optimal particles does not change in M iterations, the algorithm determines that the population falls into a local optimal. The scheme in which the population jumps out of the local optimum is to reset the position information of the 40% of individuals within the population, in other words, to randomly generate the position vector in the search space. M is set to 5% of the maximum number of iterations.

New spiral update search strategy is added after the local optimal jump-out strategy. Since the whale optimization algorithm (WOA) was good at exploring the local search space 35 , the spiral update search strategy in the WOA 36 is introduced to update the position of the particles after the swarm jumps out of local optimal. The equation for the spiral update is as follows:

Here \(D=\left|{x}_{i}\left(k\right)-Gbest\right|\) denotes the distance between the particle itself and the global optimal solution so far. \(B\) is the constant that defines the shape of the logarithmic spiral. \(l\) is the random value in \([-\mathrm{1,1}]\) . \(l\) represents the distance between the newly generated particle and the global optimal position, \(l=-1\) means the closest distance, while \(l=1\) means the farthest distance, and the meaning of this parameter can be directly observed by Fig.  1 .

Spiral updating position.

The DE/best/2 mutation strategy is introduced to form the mutant particle. 4 individuals in the population are randomly selected that differ from the current particle, then the vector difference between them is rescaled, and the difference vector is combined with the global optimal position to form the mutant particle. The equation for mutation of particle position is shown as follows:

where \({x}^{*}\) is the mutated particle, \(F\) is the scale factor of mutation, \({r}_{1}\) , \({r}_{2}\) , \({r}_{3}\) , \({r}_{4}\) are random integer values in \((0,n]\) and not equal to \(i\) , respectively. Specific particles are selected for mutation with the screening conditions as follows:

where \(Cr\) represents the probability of mutation, \(rand\left(\mathrm{0,1}\right)\) is a random number in \(\left(\mathrm{0,1}\right)\) , and \({i}_{rand}\) is a random integer value in \((0,n]\) .

The improved PSO incorporates the search ideas of other intelligent algorithms (DE, WOA), so the improved algorithm proposed in this paper is named NDWPSO. The pseudo-code for the NDWPSO algorithm is given as follows:

The main procedure of NDWPSO.

Comparing the distribution of inertia weight parameters

There are several improved PSO algorithms (such as CDWPSO 25 , and SDWPSO 26 ) that adopt the dynamic weighted particle position update strategy as their improvement strategy. The updated equations of the CDWPSO and the SDWPSO algorithm for the inertia weight parameters are given as follows:

where \({\text{A}}\) is a value in \((\mathrm{0,1}]\) . \({r}_{max}\) and \({r}_{min}\) are the upper and lower limits of the fluctuation range of the inertia weight parameters, \(k\) is the current number of algorithm iterations, and \(Mk\) denotes the maximum number of iterations.

Considering that the update method of inertia weight parameters by our proposed NDWPSO is comparable to the CDWPSO, and SDWPSO, a comparison experiment for the distribution of inertia weight parameters is set up in this section. The maximum number of iterations in the experiment is \(Mk=500\) . The distributions of CDWPSO, SDWPSO, and NDWPSO inertia weights are shown sequentially in Fig.  2 .

The inertial weight distribution of CDWPSO, SDWPSO, and NDWPSO.

In Fig.  2 , the inertia weight value of CDWPSO is a random value in (0,1]. It may make individual particles fly out of the range in the late iteration of the algorithm. Similarly, the inertia weight value of SDWPSO is a value that tends to zero infinitely, so that the swarm no longer can fly in the search space, making the algorithm extremely easy to fall into the local optimal value. On the other hand, the distribution of the inertia weights of the NDWPSO forms a gentle slope by two curves. Thus, the swarm can faster lock the global optimum range in the early iterations and locate the global optimal more precisely in the late iterations. The reason is that the inertia weight values between two adjacent iterations are inversely proportional to each other. Besides, the time-varying part of the inertial weight within NDWPSO is designed to reduce the chaos characteristic of the parameters. The inertia weight value of NDWPSO avoids the disadvantages of the above two schemes, so its design is more reasonable.

Experiment and discussion

In this section, three experiments are set up to evaluate the performance of NDWPSO: (1) the experiment of 23 classical functions 37 between NDWPSO and three particle swarm algorithms (PSO 6 , CDWPSO 25 , SDWPSO 26 ); (2) the experiment of benchmark test functions between NDWPSO and other intelligent algorithms (Whale Optimization Algorithm (WOA) 36 , Harris Hawk Algorithm (HHO) 38 , Gray Wolf Optimization Algorithm (GWO) 39 , Archimedes Algorithm (AOA) 40 , Equilibrium Optimizer (EO) 41 and Differential Evolution (DE) 42 ); (3) the experiment for solving three real engineering problems (welded beam design 43 , pressure vessel design 44 , and three-bar truss design 38 ). All experiments are run on a computer with Intel i5-11400F GPU, 2.60 GHz, 16 GB RAM, and the code is written with MATLAB R2017b.

The benchmark test functions are 23 classical functions, which consist of indefinite unimodal (F1–F7), indefinite dimensional multimodal functions (F8–F13), and fixed-dimensional multimodal functions (F14–F23). The unimodal benchmark function is used to evaluate the global search performance of different algorithms, while the multimodal benchmark function reflects the ability of the algorithm to escape from the local optimal. The mathematical equations of the benchmark functions are shown and found as Supplementary Tables S1 – S3 online.

Experiments on benchmark functions between NDWPSO, and other PSO variants

The purpose of the experiment is to show the performance advantages of the NDWPSO algorithm. Here, the dimensions and corresponding population sizes of 13 benchmark functions (7 unimodal and 6 multimodal) are set to (30, 40), (50, 70), and (100, 130). The population size of 10 fixed multimodal functions is set to 40. Each algorithm is repeated 30 times independently, and the maximum number of iterations is 200. The performance of the algorithm is measured by the mean and the standard deviation (SD) of the results for different benchmark functions. The parameters of the NDWPSO are set as: \({[{\omega }_{min},\omega }_{max}]=[\mathrm{0.4,0.9}]\) , \(\left[{c}_{max},{c}_{min}\right]=\left[\mathrm{2.5,1.5}\right],{V}_{max}=0.1,b={e}^{-50}, M=0.05\times Mk, B=1,F=0.7, Cr=0.9.\) And, \(A={\omega }_{max}\) for CDWPSO; \({[r}_{max},{r}_{min}]=[\mathrm{4,0}]\) for SDWPSO.

Besides, the experimental data are retained to two decimal places, but some experimental data will increase the number of retained data to pursue more accuracy in comparison. The best results in each group of experiments will be displayed in bold font. The experimental data is set to 0 if the value is below 10 –323 . The experimental parameter settings in this paper are different from the references (PSO 6 , CDWPSO 25 , SDWPSO 26 , so the final experimental data differ from the ones within the reference.

As shown in Tables 1 and 2 , the NDWPSO algorithm obtains better results for all 49 sets of data than other PSO variants, which include not only 13 indefinite-dimensional benchmark functions and 10 fixed-multimodal benchmark functions. Remarkably, the SDWPSO algorithm obtains the same accuracy of calculation as NDWPSO for both unimodal functions f 1 –f 4 and multimodal functions f 9 –f 11 . The solution accuracy of NDWPSO is higher than that of other PSO variants for fixed-multimodal benchmark functions f 14 -f 23 . The conclusion can be drawn that the NDWPSO has excellent global search capability, local search capability, and the capability for escaping the local optimal.

In addition, the convergence curves of the 23 benchmark functions are shown in Figs. 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 , 16 , 17 , 18 and 19 . The NDWPSO algorithm has a faster convergence speed in the early stage of the search for processing functions f1-f6, f8-f14, f16, f17, and finds the global optimal solution with a smaller number of iterations. In the remaining benchmark function experiments, the NDWPSO algorithm shows no outstanding performance for convergence speed in the early iterations. There are two reasons of no outstanding performance in the early iterations. On one hand, the fixed-multimodal benchmark function has many disturbances and local optimal solutions in the whole search space. on the other hand, the initialization scheme based on elite opposition-based learning is still stochastic, which leads to the initial position far from the global optimal solution. The inertia weight based on chaotic mapping and the strategy of spiral updating can significantly improve the convergence speed and computational accuracy of the algorithm in the late search stage. Finally, the NDWPSO algorithm can find better solutions than other algorithms in the middle and late stages of the search.

Evolution curve of NDWPSO and other PSO algorithms for f1 (Dim = 30,50,100).

Evolution curve of NDWPSO and other PSO algorithms for f2 (Dim = 30,50,100).

Evolution curve of NDWPSO and other PSO algorithms for f3 (Dim = 30,50,100).

Evolution curve of NDWPSO and other PSO algorithms for f4 (Dim = 30,50,100).

Evolution curve of NDWPSO and other PSO algorithms for f5 (Dim = 30,50,100).

Evolution curve of NDWPSO and other PSO algorithms for f6 (Dim = 30,50,100).

Evolution curve of NDWPSO and other PSO algorithms for f7 (Dim = 30,50,100).

Evolution curve of NDWPSO and other PSO algorithms for f8 (Dim = 30,50,100).

Evolution curve of NDWPSO and other PSO algorithms for f9 (Dim = 30,50,100).

Evolution curve of NDWPSO and other PSO algorithms for f10 (Dim = 30,50,100).

Evolution curve of NDWPSO and other PSO algorithms for f11(Dim = 30,50,100).

Evolution curve of NDWPSO and other PSO algorithms for f12 (Dim = 30,50,100).

Evolution curve of NDWPSO and other PSO algorithms for f13 (Dim = 30,50,100).

Evolution curve of NDWPSO and other PSO algorithms for f14, f15, f16.

Evolution curve of NDWPSO and other PSO algorithms for f17, f18, f19.

Evolution curve of NDWPSO and other PSO algorithms for f20, f21, f22.

Evolution curve of NDWPSO and other PSO algorithms for f23.

To evaluate the performance of different PSO algorithms, a statistical test is conducted. Due to the stochastic nature of the meta-heuristics, it is not enough to compare algorithms based on only the mean and standard deviation values. The optimization results cannot be assumed to obey the normal distribution; thus, it is necessary to judge whether the results of the algorithms differ from each other in a statistically significant way. Here, the Wilcoxon non-parametric statistical test 45 is used to obtain a parameter called p -value to verify whether two sets of solutions are different to a statistically significant extent or not. Generally, it is considered that p  ≤ 0.5 can be considered as a statistically significant superiority of the results. The p -values calculated in Wilcoxon’s rank-sum test comparing NDWPSO and other PSO algorithms are listed in Table  3 for all benchmark functions. The p -values in Table  3 additionally present the superiority of the NDWPSO because all of the p -values are much smaller than 0.5.

In general, the NDWPSO has the fastest convergence rate when finding the global optimum from Figs. 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 , 16 , 17 , 18 and 19 , and thus we can conclude that the NDWPSO is superior to the other PSO variants during the process of optimization.

Comparison experiments between NDWPSO and other intelligent algorithms

Experiments are conducted to compare NDWPSO with several other intelligent algorithms (WOA, HHO, GWO, AOA, EO and DE). The experimental object is 23 benchmark functions, and the experimental parameters of the NDWPSO algorithm are set the same as in Experiment 4.1. The maximum number of iterations of the experiment is increased to 2000 to fully demonstrate the performance of each algorithm. Each algorithm is repeated 30 times individually. The parameters of the relevant intelligent algorithms in the experiments are set as shown in Table 4 . To ensure the fairness of the algorithm comparison, all parameters are concerning the original parameters in the relevant algorithm literature. The experimental results are shown in Tables 5 , 6 , 7 and 8 and Figs. 20 , 21 , 22 , 23 , 24 , 25 , 26 , 27 , 28 , 29 , 30 , 31 , 32 , 33 , 34 , 35 and 36 .

Evolution curve of NDWPSO and other algorithms for f1 (Dim = 30,50,100).

Evolution curve of NDWPSO and other algorithms for f2 (Dim = 30,50,100).

Evolution curve of NDWPSO and other algorithms for f3(Dim = 30,50,100).

Evolution curve of NDWPSO and other algorithms for f4 (Dim = 30,50,100).

Evolution curve of NDWPSO and other algorithms for f5 (Dim = 30,50,100).

Evolution curve of NDWPSO and other algorithms for f6 (Dim = 30,50,100).

Evolution curve of NDWPSO and other algorithms for f7 (Dim = 30,50,100).

Evolution curve of NDWPSO and other algorithms for f8 (Dim = 30,50,100).

Evolution curve of NDWPSO and other algorithms for f9(Dim = 30,50,100).

Evolution curve of NDWPSO and other algorithms for f10 (Dim = 30,50,100).

Evolution curve of NDWPSO and other algorithms for f11 (Dim = 30,50,100).

Evolution curve of NDWPSO and other algorithms for f12 (Dim = 30,50,100).

Evolution curve of NDWPSO and other algorithms for f13 (Dim = 30,50,100).

Evolution curve of NDWPSO and other algorithms for f14, f15, f16.

Evolution curve of NDWPSO and other algorithms for f17, f18, f19.

Evolution curve of NDWPSO and other algorithms for f20, f21, f22.

Evolution curve of NDWPSO and other algorithms for f23.

The experimental data of NDWPSO and other intelligent algorithms for handling 30, 50, and 100-dimensional benchmark functions ( \({f}_{1}-{f}_{13}\) ) are recorded in Tables 8 , 9 and 10 , respectively. The comparison data of fixed-multimodal benchmark tests ( \({f}_{14}-{f}_{23}\) ) are recorded in Table 11 . According to the data in Tables 5 , 6 and 7 , the NDWPSO algorithm obtains 69.2%, 84.6%, and 84.6% of the best results for the benchmark function ( \({f}_{1}-{f}_{13}\) ) in the search space of three dimensions (Dim = 30, 50, 100), respectively. In Table 8 , the NDWPSO algorithm obtains 80% of the optimal solutions in 10 fixed-multimodal benchmark functions.

The convergence curves of each algorithm are shown in Figs. 20 , 21 , 22 , 23 , 24 , 25 , 26 , 27 , 28 , 29 , 30 , 31 , 32 , 33 , 34 , 35 and 36 . The NDWPSO algorithm demonstrates two convergence behaviors when calculating the benchmark functions in 30, 50, and 100-dimensional search spaces. The first behavior is the fast convergence of NDWPSO with a small number of iterations at the beginning of the search. The reason is that the Iterative-mapping strategy and the position update scheme of dynamic weighting are used in the NDWPSO algorithm. This scheme can quickly target the region in the search space where the global optimum is located, and then precisely lock the optimal solution. When NDWPSO processes the functions \({f}_{1}-{f}_{4}\) , and \({f}_{9}-{f}_{11}\) , the behavior can be reflected in the convergence trend of their corresponding curves. The second behavior is that NDWPSO gradually improves the convergence accuracy and rapidly approaches the global optimal in the middle and late stages of the iteration. The NDWPSO algorithm fails to converge quickly in the early iterations, which is possible to prevent the swarm from falling into a local optimal. The behavior can be demonstrated by the convergence trend of the curves when NDWPSO handles the functions \({f}_{6}\) , \({f}_{12}\) , and \({f}_{13}\) , and it also shows that the NDWPSO algorithm has an excellent ability of local search.

Combining the experimental data with the convergence curves, it is concluded that the NDWPSO algorithm has a faster convergence speed, so the effectiveness and global convergence of the NDWPSO algorithm are more outstanding than other intelligent algorithms.

Experiments on classical engineering problems

Three constrained classical engineering design problems (welded beam design, pressure vessel design 43 , and three-bar truss design 38 ) are used to evaluate the NDWPSO algorithm. The experiments are the NDWPSO algorithm and 5 other intelligent algorithms (WOA 36 , HHO, GWO, AOA, EO 41 ). Each algorithm is provided with the maximum number of iterations and population size ( \({\text{Mk}}=500,\mathrm{ n}=40\) ), and then repeats 30 times, independently. The parameters of the algorithms are set the same as in Table 4 . The experimental results of three engineering design problems are recorded in Tables 9 , 10 and 11 in turn. The result data is the average value of the solved data.

Welded beam design

The target of the welded beam design problem is to find the optimal manufacturing cost for the welded beam with the constraints, as shown in Fig.  37 . The constraints are the thickness of the weld seam ( \({\text{h}}\) ), the length of the clamped bar ( \({\text{l}}\) ), the height of the bar ( \({\text{t}}\) ) and the thickness of the bar ( \({\text{b}}\) ). The mathematical formulation of the optimization problem is given as follows:

Welded beam design.

In Table 9 , the NDWPSO, GWO, and EO algorithms obtain the best optimal cost. Besides, the standard deviation (SD) of t NDWPSO is the lowest, which means it has very good results in solving the welded beam design problem.

Pressure vessel design

Kannan and Kramer 43 proposed the pressure vessel design problem as shown in Fig.  38 to minimize the total cost, including the cost of material, forming, and welding. There are four design optimized objects: the thickness of the shell \({T}_{s}\) ; the thickness of the head \({T}_{h}\) ; the inner radius \({\text{R}}\) ; the length of the cylindrical section without considering the head \({\text{L}}\) . The problem includes the objective function and constraints as follows:

Pressure vessel design.

The results in Table 10 show that the NDWPSO algorithm obtains the lowest optimal cost with the same constraints and has the lowest standard deviation compared with other algorithms, which again proves the good performance of NDWPSO in terms of solution accuracy.

Three-bar truss design

This structural design problem 44 is one of the most widely-used case studies as shown in Fig.  39 . There are two main design parameters: the area of the bar1 and 3 ( \({A}_{1}={A}_{3}\) ) and area of bar 2 ( \({A}_{2}\) ). The objective is to minimize the weight of the truss. This problem is subject to several constraints as well: stress, deflection, and buckling constraints. The problem is formulated as follows:

Three-bar truss design.

From Table 11 , NDWPSO obtains the best design solution in this engineering problem and has the smallest standard deviation of the result data. In summary, the NDWPSO can reveal very competitive results compared to other intelligent algorithms.

Conclusions and future works

An improved algorithm named NDWPSO is proposed to enhance the solving speed and improve the computational accuracy at the same time. The improved NDWPSO algorithm incorporates the search ideas of other intelligent algorithms (DE, WOA). Besides, we also proposed some new hybrid strategies to adjust the distribution of algorithm parameters (such as the inertia weight parameter, the acceleration coefficients, the initialization scheme, the position updating equation, and so on).

23 classical benchmark functions: indefinite unimodal (f1-f7), indefinite multimodal (f8-f13), and fixed-dimensional multimodal(f14-f23) are applied to evaluate the effective line and feasibility of the NDWPSO algorithm. Firstly, NDWPSO is compared with PSO, CDWPSO, and SDWPSO. The simulation results can prove the exploitative, exploratory, and local optima avoidance of NDWPSO. Secondly, the NDWPSO algorithm is compared with 5 other intelligent algorithms (WOA, HHO, GWO, AOA, EO). The NDWPSO algorithm also has better performance than other intelligent algorithms. Finally, 3 classical engineering problems are applied to prove that the NDWPSO algorithm shows superior results compared to other algorithms for the constrained engineering optimization problems.

Although the proposed NDWPSO is superior in many computation aspects, there are still some limitations and further improvements are needed. The NDWPSO performs a limit initialize on each particle by the strategy of “elite opposition-based learning”, it takes more computation time before speed update. Besides, the” local optimal jump-out” strategy also brings some random process. How to reduce the random process and how to improve the limit initialize efficiency are the issues that need to be further discussed. In addition, in future work, researchers will try to apply the NDWPSO algorithm to wider fields to solve more complex and diverse optimization problems.

Data availability

The datasets used and/or analyzed during the current study available from the corresponding author on reasonable request.

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Acknowledgements

This work was supported by Key R&D plan of Shandong Province, China (2021CXGC010207, 2023CXGC01020); First batch of talent research projects of Qilu University of Technology in 2023 (2023RCKY116); Introduction of urgently needed talent projects in Key Supported Regions of Shandong Province; Key Projects of Natural Science Foundation of Shandong Province (ZR2020ME116); the Innovation Ability Improvement Project for Technology-based Small- and Medium-sized Enterprises of Shandong Province (2022TSGC2051, 2023TSGC0024, 2023TSGC0931); National Key R&D Program of China (2019YFB1705002), LiaoNing Revitalization Talents Program (XLYC2002041) and Young Innovative Talents Introduction & Cultivation Program for Colleges and Universities of Shandong Province (Granted by Department of Education of Shandong Province, Sub-Title: Innovative Research Team of High Performance Integrated Device).

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On the Construction of an Optimal Network of Observation Points when Solving Inverse Linear Problems of Gravimetry and Magnetometry

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• I. E. Stepanova 1 ,
• D. V. Lukyanenko 2 ,
• I. I. Kolotov 2 ,
• A. V. Shchepetilov 2 ,
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• A. N. Levashov 2  

Unique solvability of systems of linear algebraic equations is studied to which many inverse problems of geophysics are reduced as a result of discretization after applying the method of integral equations or integral representations. Examples of singular and nonsingular systems of various dimensions that arise when processing magnetometric and gravimetric data from experimental observations are discussed. Conclusions are drawn about methods for constructing an optimal network of experimental observation points.

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ACKNOWLEDGMENTS

We are grateful to A.S. Leonov for useful remarks and interest in this work.

This work was supported and carried out within the state assignment at the Schmidt Institute of Physics of the Earth of the Russian Academy of Sciences.

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Stepanova, I.E., Lukyanenko, D.V., Kolotov, I.I. et al. On the Construction of an Optimal Network of Observation Points when Solving Inverse Linear Problems of Gravimetry and Magnetometry. Comput. Math. and Math. Phys. 64 , 381–391 (2024). https://doi.org/10.1134/S0965542524030151

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2.5: Applications of Linear Equations

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Learning Objectives

• Identify key words and phrases, translate sentences to mathematical equations, and develop strategies to solve problems.
• Solve word problems involving relationships between numbers.
• Solve geometry problems involving perimeter.
• Solve percent and money problems including simple interest.
• Set up and solve uniform motion problems.

Key Words, Translation, and Strategy

Algebra simplifies the process of solving real-world problems. This is done by using letters to represent unknowns, restating problems in the form of equations, and offering systematic techniques for solving those equations. To solve problems using algebra, first translate the wording of the problem into mathematical statements that describe the relationships between the given information and the unknowns. Usually, this translation to mathematical statements is the difficult step in the process. The key to the translation is to carefully read the problem and identify certain key words and phrases.

Here are some examples of translated key phrases.

When translating sentences into mathematical statements, be sure to read the sentence several times and identify the key words and phrases.

Example \(\PageIndex{1}\)

Four less than twice some number is \(16\).

First, choose a variable for the unknown number and identify the key words and phrases. Let \(x\) represent the unknown indicated by “some number.”

Figure \(\PageIndex{1}\)

Remember that subtraction is not commutative. For this reason, take care when setting up differences. In this example, \(4−2x=16\) is an incorrect translation.

\(2x−4=16\)

It is important to first identify the variable— let x represent… —and state in words what the unknown quantity is. This step not only makes your work more readable but also forces you to think about what you are looking for. Usually, if you know what you are asked to find, then the task of finding it is achievable.

Example \(\PageIndex{2}\)

When \(7\) is subtracted from \(3\) times the sum of a number and \(12\), the result is \(20\).

Let \(n\) represent the unknown number.

Figure \(\PageIndex{2}\)

\(3(n+12)−7=20\)

To understand why parentheses are needed, study the structures of the following two sentences and their translations:

The key is to focus on the phrase “ 3 times the sum .” This prompts us to group the sum within parentheses and then multiply by 3. Once an application is translated into an algebraic equation, solve it using the techniques you have learned.

Guidelines for Setting Up and Solving Word Problems

• Step 1 : Read the problem several times, identify the key words and phrases, and organize the given information.
• Step 2 : Identify the variables by assigning a letter or expression to the unknown quantities.
• Step 3 : Translate and set up an algebraic equation that models the problem.
• Step 4 : Solve the resulting algebraic equation.
• Step 5 : Finally, answer the question in sentence form and make sure it makes sense (check it).

For now, set up all of your equations using only one variable. Avoid two variables by looking for a relationship between the unknowns.

Problems Involving Relationships between Real Numbers

We classify applications involving relationships between real numbers broadly as number problems. These problems can sometimes be solved using some creative arithmetic, guessing, and checking. Solving in this manner is not a good practice and should be avoided. Begin by working through the basic steps outlined in the general guidelines for solving word problems.

Example \(\PageIndex{3}\)

A larger integer is \(2\) less than \(3\) times a smaller integer. The sum of the two integers is \(18\). Find the integers.

Identify variables : Begin by assigning a variable to the smaller integer.

Let \(x\) represent the smaller integer.

Use the first sentence to identify the larger integer in terms of the variable \(x\): “A larger integer is 2 less than 3 times a smaller.”

Let \(3x-2\) represent the larger integer.

Set up an equation : Add the expressions that represent the two integers, and set the resulting expression equal to \(18\) as indicated in the second sentence: “The sum of the two integers is \(18\).”

\(x+(3x-2)=18\)

Solve : Solve the equation to obtain the smaller integer \(x\).

\(\begin{aligned} x+(3x-2)&=18 \\ x+3x-2&=18 \\ 4x-2&=18 \\ 4x-2\color{Cerulean}{+2}&=18\color{Cerulean}{+2} \\ 4x&=20 \\ \frac{4x}{\color{Cerulean}{4}}&=\frac{20}{\color{Cerulean}{4}} \\ x&=5 \end{aligned}\)

Back substitute : Use the expression \(3x−2\) to find the larger integer—this is called back substituting.

\(3x-2=3(\color{OliveGreen}{5}\color{black}{)-2=15-2=13}\)

Answer the question : The two integers are \(5\) and \(13\).

Check : \(5 + 13 = 18\). The answer makes sense.

Example \(\PageIndex{4}\)

The difference between two integers is \(2\). The larger integer is \(6\) less than twice the smaller. Find the integers.

Use the relationship between the two integers in the second sentence, “The larger integer is 6 less than twice the smaller ,” to identify the unknowns in terms of one variable.

Let \(2x-6\) represent the larger integer.

Since the difference is positive, subtract the smaller integer from the larger.

\((2x-6)-x=2\)

\(\begin{aligned} \color{OliveGreen}{2x}\color{black}{-6}\color{OliveGreen}{-x}&=2 \\ x-6&=2 \\ x-6\color{Cerulean}{+6}&=2\color{Cerulean}{+6} \\ x&=8 \end{aligned}\)

Use \(2x − 6\) to find the larger integer.

\(2x-6=2(\color{Cerulean}{8}\color{black}{)-6=16-6=10}\)

The two integers are \(8\) and \(10\). These integers clearly solve the problem.

It is worth mentioning again that you can often find solutions to simple problems by guessing and checking. This is so because the numbers are chosen to simplify the process of solving, so that the algebraic steps are not too tedious. You learn how to set up algebraic equations with easier problems, so that you can use these ideas to solve more difficult problems later.

Example \(\PageIndex{5}\)

• The sum of two consecutive even integers is \(46\). Find the integers.

The key phrase to focus on is “consecutive even integers.”

Let \(x\) represent the first even integer.

Let \(x+2\) represent the next even integer.

Add the even integers and set them equal to \(46\).

\(x+(x+2)=46\)

\(\begin{aligned}\color{OliveGreen}{x+x}\color{black}{+2}&=46\\2x+2&=46\\2x+2\color{Cerulean}{-2}&=46\color{Cerulean}{-2}\\2x&=44\\x&=22 \end{aligned}\)

Use \(x + 2\) to find the next even integer.

\(x+2=\color{Cerulean}{22}\color{black}{+2=24}\)

The consecutive even integers are \(22\) and \(24\).

It should be clear that consecutive even integers are separated by two units. However, it may not be so clear that odd integers are as well.

Figure \(\PageIndex{3}\)

Example \(\PageIndex{6}\)

The sum of two consecutive odd integers is \(36\). Find the integers.

The key phrase to focus on is “consecutive odd integers.”

Let \(x\) represent the first odd integer.

Let \(x+2\) represent the next odd integer.

Add the two odd integers and set the expression equal to \(36\).

\(x+(x+2)=36\)

\(\begin{aligned} \color{OliveGreen}{x+x}\color{black}{+2}&=36 \\ 2x+2&=36 \\ 2x+2\color{Cerulean}{-2}&=36\color{Cerulean}{-2} \\ 2x&=34 \\ \frac{2x}{\color{Cerulean}{2}}&=\frac{34}{\color{Cerulean}{2}} \\ x&=17 \end{aligned}\)

Use \(x + 2\) to find the next odd integer.

\(x+2=\color{OliveGreen}{17}\color{black}{+2=19}\)

The consecutive odd integers are \(17\) and \(19\).

The algebraic setup for even and odd integer problems is the same. A common mistake is to use \(x\) and \(x + 3\) when identifying the variables for consecutive odd integers. This is incorrect because adding 3 to an odd number yields an even number: for example, \(5 + 3 = 8\). An incorrect setup is very likely to lead to a decimal answer, which may be an indication that the problem was set up incorrectly.

Example \(\PageIndex{7}\)

The sum of three consecutive integers is \(24\). Find the integers.

Consecutive integers are separated by one unit.

Let \(x\) represent the first integer.

Let \(x+1\) represent the next integer.

Let \(x+2\) represent the third integer.

Add the integers and set the sum equal to \(24\).

\(x+(x+1)+(x+2)=24\)

\(\begin{aligned} \color{OliveGreen}{x+x}\color{black}{+1}\color{OliveGreen}{+x}\color{black}{+2}&=24\\ 3x+3&=24 \\ 3x+3\color{Cerulean}{-3}&=24\color{Cerulean}{-3} \\ 3x&=21 \\ x&=7 \end{aligned}\)

Back substitute to find the other two integers.

\(x+1=\color{OliveGreen}{7}\color{black}{+1=8}\)

\(x+2=\color{OliveGreen}{7}\color{black}{+2=9}\)

The three consecutive integers are \(7, 8\) and \(9\), where \(7 + 8 + 9 = 24\).

Exercise \(\PageIndex{1}\)

The sum of three consecutive odd integers is \(87\). Find the integers.

The integers are \(27, 29\), and \(31\).

Geometry Problems (Perimeter)

Recall that the perimeter of a polygon is the sum of the lengths of all the outside edges. In addition, it is helpful to review the following perimeter formulas \((π≈3.14159)\).

Keep in mind that you are looking for a relationship between the unknowns so that you can set up algebraic equations using only one variable. When working with geometry problems, it is often helpful to draw a picture.

Example \(\PageIndex{8}\)

A rectangle has a perimeter measuring \(64\) feet. The length is \(4\) feet more than \(3\) times the width. Find the dimensions of the rectangle.

The sentence “The length is 4 feet more than 3 times the width ” gives the relationship between the two variables.

Figure \(\PageIndex{4}\)

Let \(w\) represent the width of the rectangle.

Let \(3w+4\) represent the length.

The sentence “A rectangle has a perimeter measuring \(64\) feet” suggests an algebraic setup. Substitute \(64\) for the perimeter and the expression for the length into the appropriate formula as follows:

\(\begin{aligned} P&=\:\:\:\:\quad 2l + 2w \\ \color{Cerulean}{\downarrow}&\:\:\:\qquad\quad\color{Cerulean}{\downarrow} \\ \color{OliveGreen}{64}&=2(\color{OliveGreen}{3w+4}\color{black}{)+2w} \end{aligned}\)

Once you have set up an algebraic equation with one variable, solve for the width, \(w\).

\(\begin{aligned} 64&=\color{OliveGreen}{6w}\color{black}{+8+}\color{OliveGreen}{2w} \\ 64&=8w+8 \\ 64\color{Cerulean}{-8}&=8w+8\color{Cerulean}{-8} \\ 56&=8w \\ \frac{56}{\color{Cerulean}{8}}&=\frac{8w}{\color{Cerulean}{8}} \\ 7&=w \end{aligned}\)

Use \(3w + 4\) to find the length.

\(l=3w+4=3(\color{OliveGreen}{7}\color{black}{)+4=21+4=25}\)

The rectangle measures \(7\) feet by \(25\) feet. To check, add all of the sides:

\(P=7\text{ ft+}7\text{ ft+}25\text{ ft+}25\text{ ft}=64\text{ ft}\)

Example \(\PageIndex{9}\)

Two sides of a triangle are \(5\) and \(7\) inches longer than the third side. If the perimeter measures \(21\) inches, find the length of each side.

Figure \(\PageIndex{5}\)

The first sentence describes the relationships between the unknowns.

Let \(x\) represent the length of the third side.

Let \(x+5\) and \(x+7\) represent the lengths of the other two sides.

Substitute these expressions into the appropriate formula and use \(21\) for the perimeter \(P\).

\(\begin{aligned} P&=a+b+c \\ \color{OliveGreen}{21}&=\color{OliveGreen}{x}\color{black}{+}\color{OliveGreen}{(x+5)}\color{black}{+}\color{OliveGreen}{(x+7)} \end{aligned}\)

You now have an equation with one variable to solve.

\(\begin{aligned} 21&=x+x+5+x+7 \\ 21&=3x+12 \\ 21\color{Cerulean}{-12}&=3x+12\color{Cerulean}{-12} \\ 9&=3x\\ \frac{9}{\color{Cerulean}{3}}&=\frac{3x}{\color{Cerulean}{3}} \\ 3&=x \end{aligned}\)

Back substitute.

\(x+5=\color{OliveGreen}{3}\color{black}{+5=8}\)

\(x+5=\color{OliveGreen}{3}\color{black}{+7=10}\)

The three sides of the triangle measure \(3\) inches, \(8\) inches, and \(10\) inches. The check is left to the reader.

Exercise \(\PageIndex{2}\)

The length of a rectangle is \(1\) foot less than twice its width. If the perimeter is \(46\) feet, find the dimensions.

Width: \(8\) feet; length: \(15\) feet

Problems Involving Money and Percents

Whenever setting up an equation involving a percentage, we usually need to convert the percentage to a decimal or fraction. If the question asks for a percentage, then do not forget to convert your answer to a percent at the end. Also, when money is involved, be sure to round off to two decimal places.

Example \(\PageIndex{10}\)

If a pair of shoes costs $\(52.50\) including a \(7\frac{1}{4}\)% tax, what is the original cost of the item before taxes are added?

Begin by converting \(7\frac{1}{4}\)% to a decimal.

\(7\frac{1}{4}%=7.25%=0.0725\)

The amount of tax is this rate times the original cost of the item. The original cost of the item is what you are asked to find.

Let \(c\) represent the cost of the item \(\underline{\text{before taxes}}\) are added.

\(\color{Cerulean}{amount\:of\:tax\:=\:tax\:rate\:\cdot\:cost\:of\:item}\)

\(=0.0725\cdot c\)

\(\color{Cerulean}{total\:cost\:=\:cost\:of\:item\:+\:amount\:of\:tax}\)

\(52.50=c+0.0725c\)

Use this equation to solve for \(c\), the original cost of the item.

\(\begin{aligned} 52.50&=\color{OliveGreen}{1c+0.0725c} \\ 52.50&=1.0725c \\ \frac{52.50}{\color{Cerulean}{1.0725}}&=\frac{1.0725c}{\color{Cerulean}{1.0725}} \\ 48.95& \approx c \end{aligned}\)

The cost of the item before taxes is $\(48.95\). Check this by multiplying $\(48.95\) by \(0.0725\) to obtain the tax and add it to this cost.

Example \(\PageIndex{11}\)

Given a \(5\frac{1}{8}\)% annual interest rate, how long will it take $\(1,200\) to yield $\(307.50\) in simple interest?

Let \(t\) represent the time needed to earn $\(307.50\) at \(5.125\)%.

Organize the data needed to use the simple interest formula \(I=prt\).

Next, substitute all of the known quantities into the formula and then solve for the only unknown, \(t\).

\(\begin{aligned} I&=prt \\ \color{OliveGreen}{307.50}&=\color{OliveGreen}{1200}\color{black}{(}\color{OliveGreen}{0.05125}\color{black}{)t} \\ 307.50&=61.5t \\ \frac{307.50}{\color{Cerulean}{61.5}}&=\frac{61.5t}{\color{Cerulean}{61.5}} \\ 5&=t \end{aligned}\)

It takes \(5\) years for $\(1,200\) invested at \(5\frac{1}{8}\)% to earn $\(307.50\) in simple interest.

Example \(\PageIndex{12}\)

Mary invested her total savings of $\(3,400\) in two accounts. Her mutual fund account earned \(8\)% last year and her CD earned \(5\)%. If her total interest for the year was $\(245\), how much was in each account?

The relationship between the two unknowns is that they total $3,400. When a total is involved, a common technique used to avoid two variables is to represent the second unknown as the difference of the total and the first unknown.

Let \(x\) represent the amount invested in the mutual fund at \(8\)%\(=0.08\).

Let \(3,400-x\) represent the remaining amount invested in the CD at \(5\)%\(=0.05\).

The total interest is the sum of the interest earned from each account.

\(\color{Cerulean}{mutual\:fund\:interest\:+\:CD\:interest\:=\:total\:interest}\)

\(0.08x+0.05(3,400-x)=245\)

This equation models the problem with one variable. Solve for \(x\).

\(\begin{aligned} 0.08x+0.05(3,400-x)&=245 \\ \color{OliveGreen}{0.08x}\color{black}{+170}\color{OliveGreen}{-0.05x}&=245 \\ 0.03x+170\color{Cerulean}{-170} &=245\color{Cerulean}{-170} \\ 0.03x&=75 \\ \frac{0.03x}{\color{Cerulean}{0.03}}&=\frac{75}{\color{Cerulean}{0.03}} \\ x&=2,500 \end{aligned}\)

\(3,400-x=3,400-\color{OliveGreen}{2,500}\color{black}{=900}\)

Mary invested $\(2,500\) at \(8\)% in a mutual fund and $\(900\) at \(5\)% in a CD.

Example \(\PageIndex{13}\)

Joe has a handful of dimes and quarters that values $\(5.30\). He has one fewer than twice as many dimes than quarters. How many of each coin does he have?

Begin by identifying the variables.

Let \(q\) represent the number of quarters Joe is holding.

Let \(2q-1\) represent the number of dimes.

To determine the total value of a number of coins, multiply the number of coins by the value of each coin. For example, \(5\) quarters have a value $\(0.25 ⋅ 5 =\) $\(1.25\).

\(\color{Cerulean}{value\:in\:quarters\:+\:value\:in\:dimes\:=\:total\:value\:of\:coins}\)

\(0.25q+0.10(2q-1)=5.30\)

Solve for the number of quarters, \(q\).

\(\begin{aligned} 0.25q+0.10(2q-1)&=5.30 \\ \color{OliveGreen}{0.25q+0.20q}\color{black}{-0.10}&=5.30 \\ 0.45q-0.10&=5.30 \\ 0.45q-0.10\color{Cerulean}{+0.10}&=5.30\color{Cerulean}{+0.10} \\ 0.45q&=5.40 \\ \frac{0.45q}{\color{Cerulean}{0.45}}&=\frac{5.40}{\color{Cerulean}{0.45}} \\ q&=12 \end{aligned}\)

Back substitute into \(2q − 1\) to find the number of dimes.

\(2q-1=2(\color{OliveGreen}{12}\color{black}{)-1=24-1=23}\)

Joe has \(12\) quarters and \(23\) dimes. Check by multiplying $\(0.25 ⋅ 12 = \)$\(3.00\) and $\(0.10 ⋅ 23 = \)$\(2.30\). Then add to obtain the correct amount: $\(3.00 + \)$\(2.30 = \)$\(5.30\).

Exercise \(\PageIndex{3}\)

A total amount of $\(5,900\) is invested in two accounts. One account earns \(3.5\)% interest and another earns \(4.5\)%. If the interest for \(1\) year is $\(229.50\), then how much is invested in each account?

$\(3,600\) is invested at \(3.5\)% and $\(2,300\) at \(4.5\)%.

Uniform Motion Problems (Distance Problems)

Uniform motion refers to movement at a speed, or rate that does not change. We can determine the distance traveled by multiplying the average rate by the time traveled at that rate with the formula \(D=r⋅t\). Applications involving uniform motion usually have a lot of data, so it helps to first organize the data in a chart and then set up an algebraic equation that models the problem.

Example \(\PageIndex{14}\)

Two trains leave the station at the same time traveling in opposite directions. One travels at \(70\) miles per hour and the other at \(60\) miles per hour. How long does it take for the distance between them to reach \(390\) miles?

First, identify the unknown quantity and organize the data.

Let \(t\) represent the time it takes to separate \(390\) miles.

Figure \(\PageIndex{6}\)

The given information is filled in on the following chart. The time for each train is equal.

Figure \(\PageIndex{7}\)

To avoid introducing two more variables, use the formula \(D=r⋅t\) to fill in the unknown distances traveled by each train.

Distance traveled by train 1: \(D=r\cdot t=70\cdot t\)

Distance traveled by tain 2: \(D=r\cdot t=60\cdot t\)

We can now completely fill in the chart.

Figure \(\PageIndex{8}\)

The algebraic setup is defined by the distance column. The problem asks for the time it takes for the total distance to reach \(390\) miles.

Figure \(\PageIndex{9}\)

Solve for \(t\).

\(\begin{aligned} 70t+60t&=390 \\ 130t&=390 \\ \frac{130t}{\color{Cerulean}{130}}&=\frac{390}{\color{Cerulean}{130}} \\ t&=3 \end{aligned}\)

It takes \(3\) hours for the distance between the trains to reach \(390\) miles.

Example \(\PageIndex{15}\)

A train traveling nonstop to its destination is able to make the trip at an average speed of \(72\) miles per hour. On the return trip, the train makes several stops and is only able to average \(48\) miles per hour. If the return trip takes \(2\) hours longer than the initial trip to the destination, then what is the travel time each way?

Let \(t\) represent the time it takes to arrive at the destination.

Let \(t+2\) represent the time it takes for the return trip.

Figure \(\PageIndex{10}\)

The given information is filled in the following chart:

Figure \(\PageIndex{11}\)

Use the formula \(D=r⋅t\) to fill in the unknown distances.

Distance traveled on the destination: \(D=r\cdot t=72\cdot t\)

Distance traveled on the return trip: \(D=r\cdot t=48\cdot (t+2)\)

Use these expressions to complete the chart.

Figure \(\PageIndex{12}\)

The algebraic setup is again defined by the distance column. In this case, the distance to the destination and back is the same, and the equation is

\(72t=48(t+2)\)

\(\begin{aligned} 72t&=48(t+2) \\ 722&=48t+96 \\ 72t-48t&=48t+96-48t \\ 24t&=96 \\ \frac{24t}{24}&=\frac{96}{24} \\ t&=4 \end{aligned}\)

The return trip takes \(t+2=4+2=6\) hours.

It takes \(4\) hours to arrive at the destination and \(6\) hours to return.

Exercise \(\PageIndex{4}\)

Mary departs for school on a bicycle at an average rate of \(6\) miles per hour. Her sister Kate, running late, leaves \(15\) minutes later and cycles at twice that speed. How long will it take Kate to catch up to Mary? Be careful! Pay attention to the units given in the problem.

It takes \(15\) minutes for Kate to catch up.

Key Takeaways

• Simplify the process of solving real-world problems by creating mathematical models that describe the relationship between unknowns. Use algebra to solve the resulting equations.
• Guessing and checking for solutions is a poor practice. This technique might sometimes produce correct answers, but is unreliable, especially when problems become more complex.
• Read the problem several times and search for the key words and phrases. Identify the unknowns and assign variables or expressions to the unknown quantities. Look for relationships that allow you to use only one variable. Set up a mathematical model for the situation and use algebra to solve the equation. Check to see if the solution makes sense and present the solution in sentence form.
• Do not avoid word problems: solving them can be fun and rewarding. With lots of practice you will find that they really are not so bad after all. Modeling and solving applications is one of the major reasons to study algebra.
• Do not feel discouraged when the first attempt to solve a word problem does not work. This is part of the process. Try something different and learn from incorrect attempts.

Exercise \(\PageIndex{5}\) Translate

Translate the following into algebraic equations.

• The sum of a number and \(6\) is \(37\).
• When \(12\) is subtracted from twice some number the result is \(6\).
• Fourteen less than \(5\) times a number is \(1\).
• Twice some number is subtracted from \(30\) and the result is \(50\).
• Five times the sum of \(6\) and some number is \(20\).
• The sum of \(5\) times some number and \(6\) is \(20\).
• When the sum of a number and \(3\) is subtracted from \(10\) the result is \(5\).
• The sum of three times a number and five times that same number is \(24\).
• Ten is subtracted from twice some number and the result is the sum of the number and \(2\).
• Six less than some number is ten times the sum of that number and \(5\).

1. \(x+6=37\)

3. \(5x−14=1\)

5. \(5(x+6)=20\)

7. \(10−(x+3)=5\)

9. \(2x−10=x+2\)

Exercise \(\PageIndex{6}\) Number Problems

Set up an algebraic equation and then solve.

• A larger integer is \(1\) more than twice another integer. If the sum of the integers is \(25\), find the integers.
• If a larger integer is \(2\) more than \(4\) times another integer and their difference is \(32\), find the integers.
• One integer is \(30\) more than another integer. If the difference between the larger and twice the smaller is \(8\), find the integers.
• The quotient of some number and \(4\) is \(22\). Find the number.
• Eight times a number is decreased by three times the same number, giving a difference of \(20\). What is the number?
• One integer is two units less than another. If their sum is \(−22\), find the two integers.
• The sum of two consecutive integers is \(139\). Find the integers.
• The sum of three consecutive integers is \(63\). Find the integers.
• The sum of three consecutive integers is \(279\). Find the integers.
• The difference of twice the smaller of two consecutive integers and the larger is \(39\). Find the integers.
• If the smaller of two consecutive integers is subtracted from two times the larger, then the result is \(17\). Find the integers.
• The sum of two consecutive even integers is \(238\). Find the integers.
• The sum of three consecutive even integers is \(96\). Find the integers.
• If the smaller of two consecutive even integers is subtracted from \(3\) times the larger the result is \(42\). Find the integers.
• The sum of three consecutive even integers is \(90\). Find the integers.
• The sum of two consecutive odd integers is \(68\). Find the integers.
• The sum of two consecutive odd integers is \(180\). Find the integers.
• The sum of three consecutive odd integers is \(57\). Find the integers.
• If the smaller of two consecutive odd integers is subtracted from twice the larger the result is \(23\). Find the integers.
• Twice the sum of two consecutive odd integers is \(32\). Find the integers.
• The difference between twice the larger of two consecutive odd integers and the smaller is \(59\). Find the integers.

1. \(8, 17\)

3. \(22, 52\)

7. \(69, 70\)

9. \(92, 93, 94\)

11. \(15, 16\)

13. \(118, 120\)

15. \(18, 20\)

17. \(33, 35\)

19. \(17, 19, 21\)

21. \(7, 9\)

Exercise \(\PageIndex{7}\) Geometry Problems

• If the perimeter of a square is \(48\) inches, then find the length of each side.
• The length of a rectangle is \(2\) inches longer than its width. If the perimeter is \(36\) inches, find the length and width.
• The length of a rectangle is \(2\) feet less than twice its width. If the perimeter is \(26\) feet, find the length and width.
• The width of a rectangle is \(2\) centimeters less than one-half its length. If the perimeter is \(56\) centimeters, find the length and width.
• The length of a rectangle is \(3\) feet less than twice its width. If the perimeter is \(54\) feet, find the dimensions of the rectangle.
• If the length of a rectangle is twice as long as the width and its perimeter measures \(72\) inches, find the dimensions of the rectangle.
• The perimeter of an equilateral triangle measures \(63\) centimeters. Find the length of each side.
• An isosceles triangle whose base is one-half as long as the other two equal sides has a perimeter of \(25\) centimeters. Find the length of each side.
• Each of the two equal legs of an isosceles triangle are twice the length of the base. If the perimeter is \(105\) centimeters, then how long is each leg?
• A triangle has sides whose measures are consecutive even integers. If the perimeter is \(42\) inches, find the measure of each side.
• A triangle has sides whose measures are consecutive odd integers. If the perimeter is \(21\) inches, find the measure of each side.
• A triangle has sides whose measures are consecutive integers. If the perimeter is \(102\) inches, then find the measure of each side.
• The circumference of a circle measures \(50π\) units. Find the radius.
• The circumference of a circle measures \(10π\) units. Find the radius.
• The circumference of a circle measures \(100\) centimeters. Determine the radius to the nearest tenth.
• The circumference of a circle measures \(20\) centimeters. Find the diameter rounded off to the nearest hundredth.
• The diameter of a circle measures \(5\) inches. Determine the circumference to the nearest tenth.
• The diameter of a circle is \(13\) feet. Calculate the exact value of the circumference.

1. \(12\) inches

3. Width: \(5\) feet; length: \(8\) feet

5. Width: \(10\) feet; length: \(17\) feet

7. \(21\) centimeters

9. \(21\) centimeters, \(42\) centimeters, \(42\) centimeters

11. \(5\) inches, \(7\) inches, \(9\) inches

13. \(25\) units

15. \(15.9\) centimeters

17. \(15.7\) inches

Exercise \(\PageIndex{8}\) Percent and Money Problems

• Calculate the simple interest earned on a \(2\)-year investment of $\(1,550\) at a \(8\frac{3}{4}\)% annual interest rate.
• Calculate the simple interest earned on a \(1\)-year investment of $\(500\) at a \(6\)% annual interest rate.
• For how many years must $\(10,000\) be invested at an \(8\frac{1}{2}\)% annual interest rate to yield $\(4,250\) in simple interest?
• For how many years must $\(1,000\) be invested at a \(7.75\)% annual interest rate to yield $\(503.75\) in simple interest?
• At what annual interest rate must $\(2,500\) be invested for \(3\) years in order to yield $\(412.50\) in simple interest?
• At what annual interest rate must $\(500\) be invested for \(2\) years in order to yield $\(93.50\) in simple interest?
• If the simple interest earned for \(1\) year was $\(47.25\) and the annual rate was \(6.3\)%, what was the principal?
• If the simple interest earned for \(2\) years was $\(369.60\) and the annual rate was \(5\frac{1}{4}\)%, what was the principal?
• Joe invested last year’s $\(2,500\) tax return in two different accounts. He put most of the money in a money market account earning \(5\)% simple interest. He invested the rest in a CD earning \(8\)% simple interest. How much did he put in each account if the total interest for the year was $\(138.50\)?
• James invested $\(1,600\) in two accounts. One account earns \(4.25\)% simple interest and the other earns \(8.5\)%. If the interest after \(1\) year was $\(85\), how much did he invest in each account?
• Jane has her $\(5,400\) savings invested in two accounts. She has part of it in a CD at \(3\)% annual interest and the rest in a savings account that earns \(2\)% annual interest. If the simple interest earned from both accounts is $\(140\) for the year, then how much does she have in each account?
• Marty put last year’s bonus of $\(2,400\) into two accounts. He invested part in a CD with \(2.5\)% annual interest and the rest in a money market fund with \(1.3\)% annual interest. His total interest for the year was $\(42.00\). How much did he invest in each account?
• Alice puts money into two accounts, one with \(2\)% annual interest and another with \(3\)% annual interest. She invests \(3\) times as much in the higher yielding account as she does in the lower yielding account. If her total interest for the year is $\(27.50\), how much did she invest in each account?
• Jim invested an inheritance in two separate banks. One bank offered \(5.5\)% annual interest rate and the other \(6\frac{1}{4}\)%. He invested twice as much in the higher yielding bank account than he did in the other. If his total simple interest for \(1\) year was $\(4,860\), then what was the amount of his inheritance?
• If an item is advertised to cost $\(29.99\) plus \(9.25\)% tax, what is the total cost?
• If an item is advertised to cost $\(32.98\) plus \(8\frac{3}{4}\)% tax, what is the total cost?
• An item, including an \(8.75\)% tax, cost $\(46.49\). What is the original pretax cost of the item?
• An item, including a \(5.48\)% tax, cost $\(17.82\). What is the original pretax cost of the item?
• If a meal costs $\(32.75\), what is the total after adding a \(15\)% tip?
• How much is a \(15\)% tip on a restaurant bill that totals $\(33.33\)?
• Ray has a handful of dimes and nickels valuing $\(3.05\). He has \(5\) more dimes than he does nickels. How many of each coin does he have?
• Jill has \(3\) fewer half-dollars than she has quarters. The value of all \(27\) of her coins adds to $\(9.75\). How many of each coin does Jill have?
• Cathy has to deposit $\(410\) worth of five- and ten-dollar bills. She has \(1\) fewer than three times as many tens as she does five-dollar bills. How many of each bill does she have to deposit?
• Billy has a pile of quarters, dimes, and nickels that values $\(3.75\). He has \(3\) more dimes than quarters and \(5\) more nickels than quarters. How many of each coin does Billy have?
• Mary has a jar with one-dollar bills, half-dollar coins, and quarters valuing $\(14.00\). She has twice as many quarters than she does half-dollar coins and the same amount of half-dollar coins as one-dollar bills. How many of each does she have?
• Chad has a bill-fold of one-, five-, and ten-dollar bills totaling $\(118\). He has \(2\) more than \(3\) times as many ones as he does five-dollar bills and \(1\) fewer ten- than five-dollar bills. How many of each bill does Chad have?

1. $\(271.25\)

3. \(5\) years

5. \(5.5\)%

7. $\(750.00\)

9. Joe invested $\(2,050\) in the money market account and $\(450\) in the CD.

11. Jane has $\(3,200\) in the CD and $\(2,200\) in savings.

13. Alice invested $\(250\) at \(2\)% and $\(750\) at a \(3\)%.

15. $\(32.76\)

17. $\(42.75\)

19. $\(37.66\)

21. He has \(17\) nickels and \(22\) dimes.

23. Cathy has \(12\) fives and \(35\) ten-dollar bills.

25. Mary has \(7\) one-dollar bills, \(7\) half-dollar coins, and \(14\) quarters.

Exercise \(\PageIndex{9}\) Uniform Motion (Distance Problems)

Set up an algebraic equation then solve.

• Two cars leave a location traveling in opposite directions. If one car averages \(55\) miles per hour and the other averages \(65\) miles per hour, then how long will it take for them to separate a distance of \(300\) miles?
• Two planes leave the airport at the same time traveling in opposite directions. The average speeds for the planes are \(450\) miles per hour and \(395\) miles per hour. How long will it take the planes to be a distance of \(1,478.75\) miles apart?
• Bill and Ted are racing across the country. Bill leaves \(1\) hour earlier than Ted and travels at an average rate of \(60\) miles per hour. If Ted intends to catch up at a rate of \(70\) miles per hour, then how long will it take?
• Two brothers leave from the same location, one in a car and the other on a bicycle, to meet up at their grandmother’s house for dinner. If one brother averages \(30\) miles per hour in the car and the other averages \(12\) miles per hour on the bicycle, then it takes the brother on the bicycle \(1\) hour less than \(3\) times as long as the other in the car. How long does it take each of them to make the trip?
• A commercial airline pilot flew at an average speed of \(350\) miles per hour before being informed that his destination airfield may be closed due to poor weather conditions. In an attempt to arrive before the storm, he increased his speed \(400\) miles per hour and flew for another \(3\) hours. If the total distance flown was \(2,950\) miles, then how long did the trip take?
• Two brothers drove the \(2,793\) miles from Los Angeles to New York. One of the brothers, driving during the day, was able to average \(70\) miles per hour, and the other, driving at night, was able to average \(53\) miles per hour. If the brother driving at night drove \(3\) hours less than the brother driving in the day, then how many hours did they each drive?
• Joe and Ellen live \(21\) miles apart. Departing at the same time, they cycle toward each other. If Joe averages \(8\) miles per hour and Ellen averages \(6\) miles per hour, how long will it take them to meet?
• If it takes \(6\) minutes to drive to the automobile repair shop at an average speed of \(30\) miles per hour, then how long will it take to walk back at an average rate of \(4\) miles per hour?
• Jaime and Alex leave the same location and travel in opposite directions. Traffic conditions enabled Alex to average \(14\) miles per hour faster than Jaime. After \(1\frac{1}{2}\) hours they are \(159\) miles apart. Find the speed at which each was able to travel.
• Jane and Holly live \(51\) miles apart and leave at the same time traveling toward each other to meet for lunch. Jane traveled on the freeway at twice the average speed as Holly. They were able to meet in a half hour. At what rate did each travel?

1. \(2.5\) hours

3. \(6\) hours

5. \(8\) hours

7. \(1\frac{1}{2}\) hours

9. Jaime: \(46\) miles per hour; Alex: \(60\) miles per hour

Exercise \(\PageIndex{10}\) Discussion Board Topics

• Discuss ideas for calculating taxes and tips mentally.
• Research historical methods for representing unknowns.
• Research and compare simple interest and compound interest. What is the difference?
• Discuss why algebra is a required subject.
• Research ways to show that a repeating decimal is rational. Share your findings on the discussion board.

1. Answers may vary

3. Answers may vary

5. Answers may vary

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Mathematics > Algebraic Geometry

Title: positive moments forever: undecidable and decidable cases.

Abstract: Is there an algorithm to determine attributes such as positivity or non-zeroness of linear recurrence sequences? This long-standing question is known as Skolem's problem. In this paper, we study the complexity of an equivalent problem, namely the (generalized) moment membership problem for matrices. We show that this problem is decidable for orthogonal, unitary and real eigenvalue matrices, and undecidable for matrices over certain commutative and non-commutative polynomial rings. Our results imply that the positivity problem for simple unitary linear recurrence sequences is decidable, and is undecidable for linear recurrence sequences over the ring of commutative polynomials. As a byproduct, we prove a free version of Polya's theorem.

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Unit 10: Advanced: Algebra

About this unit.

Ready for a challenge? This unit covers the hardest algebra questions on the SAT Math test. Work through each skill, taking quizzes and the unit test to level up your mastery progress.

Solving linear equations and inequalities: advanced

• Solving linear equations and linear inequalities | Lesson (Opens a modal)
• Solving linear equations and linear inequalities — Basic example (Opens a modal)
• Solving linear equations and linear inequalities — Harder example (Opens a modal)
• Solving linear equations and inequalities: advanced Get 3 of 4 questions to level up!

Linear equation word problems: advanced

• Understanding linear relationships | Lesson (Opens a modal)
• Linear equation word problems — Basic example (Opens a modal)
• Linear equation word problems — Harder example (Opens a modal)
• Linear equation word problems: advanced Get 3 of 4 questions to level up!

Linear relationship word problems: advanced

• Interpreting linear functions — Basic example (Opens a modal)
• Linear function word problems — Basic example (Opens a modal)
• Linear relationship word problems: advanced Get 3 of 4 questions to level up!

Graphs of linear equations and functions: advanced

• Graphs of linear equations and functions | Lesson (Opens a modal)
• Graphing linear equations — Basic example (Opens a modal)
• Graphing linear equations — Harder example (Opens a modal)
• Graphs of linear equations and functions: advanced Get 3 of 4 questions to level up!

Solving systems of linear equations: advanced

• Solving systems of linear equations | Lesson (Opens a modal)
• Solving systems of linear equations — Basic example (Opens a modal)
• Solving systems of linear equations — Harder example (Opens a modal)
• Solving systems of linear equations: advanced Get 3 of 4 questions to level up!

Systems of linear equations word problems: advanced

• Systems of linear equations word problems | Lesson (Opens a modal)
• Systems of linear equations word problems — Basic example (Opens a modal)
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Linear inequality word problems: advanced

• Linear inequality word problems | Lesson (Opens a modal)
• Linear inequality word problems — Basic example (Opens a modal)
• Systems of linear inequalities word problems — Harder example (Opens a modal)
• Linear inequality word problems: advanced Get 3 of 4 questions to level up!

Graphs of linear systems and inequalities: advanced

• Graphs of linear systems and inequalities | Lesson (Opens a modal)
• Graphs of linear systems and inequalities — Basic example (Opens a modal)
• Graphs of linear systems and inequalities — Harder example (Opens a modal)
• Graphs of linear systems and inequalities: advanced Get 3 of 4 questions to level up!