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The elimination method for solving linear systems

• Solve the system IV
• Solve the system V

Another way of solving a linear system is to use the elimination method. In the elimination method you either add or subtract the equations to get an equation in one variable.

When the coefficients of one variable are opposites you add the equations to eliminate a variable and when the coefficients of one variable are equal you subtract the equations to eliminate a variable.

\begin{cases} 3y+2x=6\\ 5y-2x=10 \end{cases}

We can eliminate the \(x\)-variable by addition of the two equations.

\begin{cases} 3y+2x=6 \\ \underline{+\: 5y-2x=10} \end{cases} 

$$=8y\: \: \: \: \; \; \; \; =16$$

$$\begin{matrix} \: \: \: y\: \: \: \: \: \; \; \; \; \; =2 \end{matrix}$$

The value of \(y\) can now be substituted into either of the original equations to find the value of \(x\)

$$3y+2x=6$$

$$3\cdot {\color{green} 2}+2x=6$$

The solution of the linear system is \((0, 2)\).

To avoid errors make sure that all like terms and equal signs are in the same columns before beginning the elimination.

If you don't have equations where you can eliminate a variable by addition or subtraction you directly you can begin by multiplying one or both of the equations with a constant to obtain an equivalent linear system where you can eliminate one of the variables by addition or subtraction.

\begin{cases} 3x+y=9\\ 5x+4y=22 \end{cases}

Begin by multiplying the first equation by \(-4\) so that the coefficients of \(y\) are opposites

\begin{cases} \color{green}{-4} \cdot (3x + y) = \color{green}{-4} \cdot 9\\ 5x + 4y = 22 \end{cases}

$$\Rightarrow$$

\begin{cases}-12x-4y=-36 \\ \underline{+5x+4y=22 }\end{cases}

$$=-7x\: \: \: \: \: \: \: \: \: \: =-14$$

$$\begin{matrix} \: \:\; \:\: x\: \: \: \: \: \: \: \: \: \: \:=2 \end{matrix}$$

Substitute \(x\) in either of the original equations to get the value of \(y\)

$$3\cdot {\color{green} 2}+y=9$$

The solution of the linear system is \((2, 3)\)

Video lesson

Solve the following linear system using the elimination method

\begin{cases} 2y - 4x = 2 \\ y = -x + 4 \end{cases}

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Unit 2: Linear equations, inequalities, and systems

Solving absolute value equations.

• Intro to absolute value equations and graphs (Opens a modal)
• Solving absolute value equations (Opens a modal)
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Solving linear systems with substitution

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Solving linear systems with elimination

• Systems of equations with elimination: x-4y=-18 & -x+3y=11 (Opens a modal)
• Systems of equations with elimination: potato chips (Opens a modal)
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Applying systems of equations

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• System of equations word problem: no solution (Opens a modal)
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• Systems of equations with elimination: TV & DVD (Opens a modal)
• Systems of equations with elimination: apples and oranges (Opens a modal)
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Systems of Linear Equations: Solving by Addition / Elimination

Definitions Graphing Special Cases Substitution Elimination/Addition Gaussian Elimination More Examples

The "addition" method of solving systems of linear equations is also called the "elimination" method. Under either name, this method is similar to the method you probably used when you were first learning how to solve one-variable linear equations .

Suppose, back in the day, they'd given you the equation " x  + 6 = 11 ". To solve this, you would probably have subtracted the six to the other side of the "equals" sign by putting a " −6 " under either side of the equation. Then you'd have drawn a horizontal line underneath (representing an "equals" line) and "added down" to get " x  = 5 " as the solution.

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Solving Systems by Addition

Your work would probably have looked something like this:

x + 6 = 11     −6    −6 x       =   5

You'll do something very similar when you solve systems of linear equations using the addition method. I'll demonstrate with some examples.

• Solve the following system using addition.

2 x + y = 9 3 x − y = 16

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When I was solving one-variable linear equations, back in the day, I would "cancel out" an unwanted number by adding its opposite. (In the example above, this would have been the −6 that was added in the second line, in order to cancel out the +6 .) Then I'd draw a horizontal "equals" line under what I'd added to both sides of the original equation, and I'd add down. This would get the variable by itself on one side of the "equals" sign.

I want to do something similar here. I know how to solve linear equations with one variable. Here I've got two. Can I get rid of one of these variables in the system, just as I'd have gotten rid of the −6 in the equation?

Looking at the system of equations they've given me, I see that I've got a + y in the first line, and a − y in the second line. If I added these, they'd cancel out, leaving me with just the variable x . In other words, if I add down, I should end up with a linear equation with just one variable, and I know how to solve those. So let's do that!

I write down the two equations, draw an "equals" bar under them, and add down:

2 x + y = 9 3 x − y = 16 5 x      = 25

Now I have a one-variable linear equation that I already know how to solve. I divide through on both sides by 5 to get x = 5 . This is half of the solution to this system.

(By the way, this adding of the two equations, or two "rows", is called a "row operation".)

To find the other half (that is, to find the y -value), I can plug this x -value back into either one of the original equations, and simplify for the value of y . (This process — of taking a partial solution and plugging it back in to some portion of the original exercise to find the rest of the solution — is called "back-solving".)

I can use either of the original equations to back-solve and find the value of y . The first equation has smaller numbers (and I'm lazy), so I'll back-solve in that one:

2(5) + y = 9   10 + y = 9            y = −1

This gives me the other half of the solution, so my answer is:

( x , y ) = (5, −1)

In case you're wondering how I knew which was the "right" equation to use for the backsolving, I didn't. Because it doesn't matter. Solutions to systems are intersection points; intersection points will, by definition, be on both of the lines; so either equation will work just fine. You'll get the same answer either way.

Check it out: if I'd have used the other equation for the back-solving, here would be my working:

3(5) − y = 16   15 − y = 16         − y = 1            y = −1

...which is the same result as before.

x − 2 y = −9 x + 3 y = 16

Note that the x -terms would cancel out if only they'd had opposite signs. But I can create this opposite-sign cancellation by multiplying either one of the equations by −1 , and then adding down as usual. It doesn't matter which equation I choose, as long as I am careful to multiply the −1 through the entire equation. (That means both sides of the "equals" sign!)

I flipped a coin; I'll multiply the second equation.

(The " −1 R 2 " notation over the arrow in the above image indicates that I multiplied row 2 by −1 . This " R n " notation, indicating that you're doing something with the n -th row, is standard. And this multiplying of a row by a numerical value is another "row operation".)

By setting up the x -terms to cancel out when the equations are added together, I have eliminated that variable. Now I can solve the resulting one-variable equation " −5 y = −25 " to get y = 5 .

To find the corresponding value of x , I plug this y -value back into either of the original equations. Back-solving in the first equation, I get:

x − 2(5) = −9 x − 10 = −9 x = 1

This gives me the other coordinate of the solution point, so my answer is:

( x , y ) = (1, 5)

A very common temptation is to write the solution to a system of equations in the form "(first number I found, second number I found)". Sometimes, though, as in this case, you find the y -value first and then the x -value second, and of course in points the x -value comes first. So just be careful to write the coordinates for your solutions correctly.

2 x − y = 9 3 x + 4 y = −14

Nothing cancels here, but I can multiply to create a cancellation. (As long as I multiply both sides of the equation by the same value, I won't have changed anything in mathematical terms. But I may be able to change things in practical terms, to create a cancellation.) If I multiply the first equation by 4 , this will set up the y -terms to cancel.

Solving this, I get that x = 2 . I'll use the first equation for backsolving, because the coefficients are smaller (and I'm lazy).

2(2) − y = 9 4 − y = 9 − y = 5 y = −5

Now I have the two coordinates of the solution point:

( x , y ) = (2, −5)

4 x − 3 y = 25 −3 x + 8 y = 10

Hmm... As the system stands, nothing cancels. But I know that I can multiply to create a cancellation.

In this case, neither variable is an obvious choice for cancellation, so I'll consider the least common multiples of the coefficients. I can multiply the equations (by 3 and 4 , respectively) to convert the x -terms to 12 x 's, or I can multiply them (by 8 and 3 , respectively) to convert the y -terms to 24 y 's. Since I'm lazy and 12 is smaller than 24 , I'll multiply to cancel the x -terms.

(I would get the same answer in the end if I set up the y -terms to cancel. It's not that how I'm doing it is "the right way"; it was just my choice. You could make a different choice, and your choice would be just as correct as mine.)

I will multiply the first row by 3 and the second row by 4 ; then I'll add down and solve.

Solving, I get that y = 5 . Neither equation looks particularly better than the other for back-solving, so I'll flip a coin and use the first equation.

4 x − 3(5) = 25 4 x − 15 = 25 4 x = 40 x = 10

Remembering to put the x -coordinate first in the solution, I get:

( x , y ) = (10, 5)

Usually when you are solving "by addition", you will need to create the cancellation. Warning: The most common mistake is to forget to multiply all the way through the equation, multiplying on both sides of the "equals" sign. Be careful of this; always multiply through the entire equation.

• Solve the following using addition.

12 x −  13 y =   2 −6 x + 6.5 y = −2

I think I'll multiply the second equation by 2 ; this will at least get rid of the decimal place.

Oops! This result isn't true! Zero is never equal to −2 !

All of my steps were correct, but I ended up with garbage. This tells me that my original assumption (being that the system had a solution) must have been wrong. So this is an inconsistent system (that i s, one that graphs as two parallel lines) with no solution (that is, having no intersection point).

no solution: inconsistent system

12 x − 3 y = 6   4 x −    y = 2

I think it'll be simplest to cancel off the y -terms, so I'll multiply the second row by −3 .

Well, yes, zero does equal zero, but...?

I already knew that zero equals zero. This information doesn't add anything to my store of knowledge. In particular, it doesn't help me narrow down my answer to one solution point. All my math was correct, so the issue lies elsewhere.

Then I remember: If the two equations are really the same one equation, then this "zero equals zero" result is the sort of thing I should expect. In fact, this result tells me that this system is a dependent system (that is, one that graphs as just one line) and, solving either of the original equations for " y = ", I find that the solution is the equation of the whole line, namely:

y = 4 x − 2

(Your text may format the answer as " ( s , 4 s − 2) ", or something like that.)

Remember the difference: a nonsense answer (like " 0 = −2 " in the exercise before the last one above) means that you have an inconsistent system with no solution; a useless-but-true answer (like " 0 = 0 " in the last exercise above) means that you have a dependent system where the set of all the points on the whole line is the solution.

Note: Some books use only " x " and " y " for their variables in systems of two equations, but many will also use additional variables. When you write the solution for an x , y -point, you know that the x -coordinate goes first and the y -coordinate goes second. When you are dealing with other variables, assume (unless explicitly told otherwise) that those variables, when used as coordinates of a point, are written in alphabetical order. For instance, if the variables in a given system are a and b , the solution point would be ( a ,  b ) ; it would not be ( b ,  a ) . Remember: Unless otherwise specified, the variables are always written in alphabetical order.

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How to solve systems of equations by Elimination

Step by step tutorial for systems of linear equations (in 2 variables)

Video on Solving by Elimination

What is the Elimination Method?

It is one way to solve a system of equations.

The basic idea is if you have 2 equations, you can sometimes do a single operation and then add the 2 equations in a way that eleiminates 1 of the 2 variables as the example that follows shows.

Elimination Example 1

$$ y = x + 1 \\ y = -x $$

Practice Problems

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4.3: Solve Systems of Equations by Elimination

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Learning Objectives

By the end of this section, you will be able to:

• Solve a system of equations by elimination
• Solve applications of systems of equations by elimination
• Choose the most convenient method to solve a system of linear equations

Before you get started, take this readiness quiz.

• Simplify −5(6−3a). If you missed this problem, review Exercise 1.10.43 .
• Solve the equation \(\frac{1}{3}x+\frac{5}{8}=\frac{31}{24}\). If you missed this problem, review Exercise 2.5.1 .

We have solved systems of linear equations by graphing and by substitution. Graphing works well when the variable coefficients are small and the solution has integer values. Substitution works well when we can easily solve one equation for one of the variables and not have too many fractions in the resulting expression.

The third method of solving systems of linear equations is called the Elimination Method. When we solved a system by substitution, we started with two equations and two variables and reduced it to one equation with one variable. This is what we’ll do with the elimination method, too, but we’ll have a different way to get there.

Solve a System of Equations by Elimination

The Elimination Method is based on the Addition Property of Equality. The Addition Property of Equality says that when you add the same quantity to both sides of an equation, you still have equality. We will extend the Addition Property of Equality to say that when you add equal quantities to both sides of an equation, the results are equal.

For any expressions a , b , c , and d ,

\[\begin{array}{lc} \text{ if } & a=b \\ \text { and } & c=d \\ \text { then } &a+c =b+d \end{array}\]

To solve a system of equations by elimination, we start with both equations in standard form. Then we decide which variable will be easiest to eliminate. How do we decide? We want to have the coefficients of one variable be opposites, so that we can add the equations together and eliminate that variable.

Notice how that works when we add these two equations together:

\[\begin{array}{l} 3x+y=5 \\ \underline{2x-y=0} \\ 5x\quad\quad=5\end{array}\]

The y ’s add to zero and we have one equation with one variable.

Let’s try another one:

\[\left\{\begin{array}{l}{x+4 y=2} \\ {2 x+5 y=-2}\end{array}\right.\]

This time we don’t see a variable that can be immediately eliminated if we add the equations.

But if we multiply the first equation by −2, we will make the coefficients of x opposites. We must multiply every term on both sides of the equation by −2.

Now we see that the coefficients of the x terms are opposites, so x will be eliminated when we add these two equations.

Add the equations yourself—the result should be −3 y = −6. And that looks easy to solve, doesn’t it? Here is what it would look like.

We’ll do one more:

\[\left\{\begin{array}{l}{4 x-3 y=10} \\ {3 x+5 y=-7}\end{array}\right.\]

It doesn’t appear that we can get the coefficients of one variable to be opposites by multiplying one of the equations by a constant, unless we use fractions. So instead, we’ll have to multiply both equations by a constant.

We can make the coefficients of x be opposites if we multiply the first equation by 3 and the second by −4, so we get 12 x and −12 x .

This gives us these two new equations:

\[\left\{\begin{aligned} 12 x-9 y &=30 \\-12 x-20 y &=28 \end{aligned}\right.\]

When we add these equations,

\[\[\left\{\begin{array}{r}{12 x-9 y=30} \\ {\underline{-12 x-20 y=28}} \\\end{array}\right.\\\quad\qquad {-29 y=58}\]\]

the x ’s are eliminated and we just have −29 y = 58.

Once we get an equation with just one variable, we solve it. Then we substitute that value into one of the original equations to solve for the remaining variable. And, as always, we check our answer to make sure it is a solution to both of the original equations.

Now we’ll see how to use elimination to solve the same system of equations we solved by graphing and by substitution.

Exercise \(\PageIndex{1}\): How to Solve a System of Equations by Elimination

Solve the system by elimination. \(\left\{\begin{array}{l}{2 x+y=7} \\ {x-2 y=6}\end{array}\right.\)

Exercise \(\PageIndex{2}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{3 x+y=5} \\ {2 x-3 y=7}\end{array}\right.\)

(2,−1)

Exercise \(\PageIndex{3}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{4 x+y=-5} \\ {-2 x-2 y=-2}\end{array}\right.\)

(−2,3)

The steps are listed below for easy reference.

HOW TO SOLVE A SYSTEM OF EQUATIONS BY ELIMINATION.

• Write both equations in standard form. If any coefficients are fractions, clear them.
• Decide which variable you will eliminate.
• Multiply one or both equations so that the coefficients of that variable are opposites.
• Add the equations resulting from Step 2 to eliminate one variable.
• Solve for the remaining variable.
• Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
• Write the solution as an ordered pair.
• Check that the ordered pair is a solution to both original equations.

First we’ll do an example where we can eliminate one variable right away.

Exercise \(\PageIndex{4}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{x+y=10} \\ {x-y=12}\end{array}\right.\)

Exercise \(\PageIndex{5}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{2 x+y=5} \\ {x-y=4}\end{array}\right.\)

(3,−1)

Exercise \(\PageIndex{6}\)

Solve the system by elimination.\(\left\{\begin{array}{l}{x+y=3} \\ {-2 x-y=-1}\end{array}\right.\)

(−2,5)

In Exercise \(\PageIndex{7}\), we will be able to make the coefficients of one variable opposites by multiplying one equation by a constant.

Exercise \(\PageIndex{7}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{3 x-2 y=-2} \\ {5 x-6 y=10}\end{array}\right.\)

Exercise \(\PageIndex{8}\)

Solve the system by elimination.\(\left\{\begin{array}{l}{4 x-3 y=1} \\ {5 x-9 y=-4}\end{array}\right.\)

Exercise \(\PageIndex{9}\)

Solve the system by elimination.\(\left\{\begin{array}{l}{3 x+2 y=2} \\ {6 x+5 y=8}\end{array}\right.\)

(−2,4)

Now we’ll do an example where we need to multiply both equations by constants in order to make the coefficients of one variable opposites.

Exercise \(\PageIndex{10}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{4 x-3 y=9} \\ {7 x+2 y=-6}\end{array}\right.\)

In this example, we cannot multiply just one equation by any constant to get opposite coefficients. So we will strategically multiply both equations by a constant to get the opposites.

Exercise \(\PageIndex{11}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{3 x-4 y=-9} \\ {5 x+3 y=14}\end{array}\right.\)

Exercise \(\PageIndex{12}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{7 x+8 y=4} \\ {3 x-5 y=27}\end{array}\right.\)

(4,−3)

When the system of equations contains fractions, we will first clear the fractions by multiplying each equation by its LCD.

Exercise \(\PageIndex{13}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{x+\frac{1}{2} y=6} \\ {\frac{3}{2} x+\frac{2}{3} y=\frac{17}{2}}\end{array}\right.\)

In this example, both equations have fractions. Our first step will be to multiply each equation by its LCD to clear the fractions.

Exercise \(\PageIndex{14}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{\frac{1}{3} x-\frac{1}{2} y=1} \\ {\frac{3}{4} x-y=\frac{5}{2}}\end{array}\right.\)

Exercise \(\PageIndex{15}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{x+\frac{3}{5} y=-\frac{1}{5}} \\ {-\frac{1}{2} x-\frac{2}{3} y=\frac{5}{6}}\end{array}\right.\)

(1,−2)

In the Solving Systems of Equations by Graphing we saw that not all systems of linear equations have a single ordered pair as a solution. When the two equations were really the same line, there were infinitely many solutions. We called that a consistent system. When the two equations described parallel lines, there was no solution. We called that an inconsistent system.

Exercise \(\PageIndex{16}\)

Solve the system by elimination.\(\left\{\begin{array}{l}{3 x+4 y=12} \\ {y=3-\frac{3}{4} x}\end{array}\right.\)

\(\begin{array} {ll} & \left\{\begin{aligned} 3 x+4 y &=12 \\ y &=3-\frac{3}{4} x \end{aligned}\right. \\\\\text{Write the second equation in standard form.} & \left\{\begin{array}{l}{3 x+4 y=12} \\ {\frac{3}{4} x+y=3}\end{array}\right.\\ \\ \text{Clear the fractions by multiplying thesecond equation by 4.} & \left\{\begin{aligned} 3 x+4 y &=12 \\ 4\left(\frac{3}{4} x+y\right) &=4(3) \end{aligned}\right. \\\\ \text{Simplify.} & \left\{\begin{array}{l}{3 x+4 y=12} \\ {3 x+4 y=12}\end{array}\right.\\\\ \text{To eliminate a variable, we multiply thesecond equation by −1.} & \left\{\begin{array}{c}{3 x+4 y=12} \\ \underline{-3 x-4 y=-12} \end{array}\right.\\ &\qquad\qquad\quad 0=0 \\ \text{Simplify and add.} \end{array}\)

This is a true statement. The equations are consistent but dependent. Their graphs would be the same line. The system has infinitely many solutions.

After we cleared the fractions in the second equation, did you notice that the two equations were the same? That means we have coincident lines.

Exercise \(\PageIndex{17}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{5 x-3 y=15} \\ {y=-5+\frac{5}{3} x}\end{array}\right.\)

infinitely many solutions

Exercise \(\PageIndex{18}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{x+2 y=6} \\ {y=-\frac{1}{2} x+3}\end{array}\right.\)

Exercise \(\PageIndex{19}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{-6 x+15 y=10} \\ {2 x-5 y=-5}\end{array}\right.\)

\(\begin{array} {ll} \text{The equations are in standard form.}& \left\{\begin{aligned}-6 x+15 y &=10 \\ 2 x-5 y &=-5 \end{aligned}\right. \\\\ \text{Multiply the second equation by 3 to eliminate a variable.} & \left\{\begin{array}{l}{-6 x+15 y=10} \\ {3(2 x-5 y)=3(-5)}\end{array}\right. \\\\ \text{Simplify and add.} & \left\{\begin{aligned}{-6 x+15 y =10} \\ \underline{6 x-15 y =-15} \end{aligned}\right. \\ & \qquad \qquad \quad0\neq 5 \end{array}\)

This statement is false. The equations are inconsistent and so their graphs would be parallel lines.

The system does not have a solution.

Exercise \(\PageIndex{20}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{-3 x+2 y=8} \\ {9 x-6 y=13}\end{array}\right.\)

no solution

Exercise \(\PageIndex{21}\)

Solve the system by elimination. \(\left\{\begin{array}{l}{7 x-3 y=-2} \\ {-14 x+6 y=8}\end{array}\right.\)

Solve Applications of Systems of Equations by Elimination

Some applications problems translate directly into equations in standard form, so we will use the elimination method to solve them. As before, we use our Problem Solving Strategy to help us stay focused and organized.

Exercise \(\PageIndex{22}\)

The sum of two numbers is 39. Their difference is 9. Find the numbers.

\(\begin{array} {ll} \textbf{Step 1. Read}\text{ the problem}& \\ \textbf{Step 2. Identify} \text{ what we are looking for.} & \text{We are looking for two numbers.} \\\textbf{Step 3. Name} \text{ what we are looking for.} & \text{Let n = the first number.} \\ & \text{ m = the second number} \\\textbf{Step 4. Translate} \text{ into a system of equations.}& \\ & \text{The sum of two numbers is 39.} \\ & n+m=39\\ & \text{Their difference is 9.} \\ & n−m=9 \\ \\ \text{The system is:} & \left\{\begin{array}{l}{n+m=39} \\ {n-m=9}\end{array}\right. \\\\ \textbf{Step 5. Solve} \text{ the system of equations. } & \\ \text{To solve the system of equations, use} \\ \text{elimination. The equations are in standard} \\ \text{form and the coefficients of m are} & \\ \text{opposites. Add.} & \left\{\begin{array}{l}{n+m=39} \\ \underline{n-m=9}\end{array}\right. \\ &\quad 2n\qquad=48 \\ \\\text{Solve for n.} & n=24 \\ \\ \text{Substitute n=24 into one of the original} &n+m=39 \\ \text{equations and solve form.} & 24+m=39 \\ & m=15 \\ \textbf{Step 6. Check}\text{ the answer.} & \text{Since 24+15=39 and 24−15=9, the answers check.}\\ \textbf{Step 7. Answer} \text{ the question.} & \text{The numbers are 24 and 15.} \end{array}\)

Exercise \(\PageIndex{23}\)

The sum of two numbers is 42. Their difference is 8. Find the numbers.

The numbers are 25 and 17.

Exercise \(\PageIndex{24}\)

The sum of two numbers is −15. Their difference is −35. Find the numbers.

The numbers are −25 and 10.

Exercise \(\PageIndex{25}\)

Joe stops at a burger restaurant every day on his way to work. Monday he had one order of medium fries and two small sodas, which had a total of 620 calories. Tuesday he had two orders of medium fries and one small soda, for a total of 820 calories. How many calories are there in one order of medium fries? How many calories in one small soda?

Exercise \(\PageIndex{26}\)

Malik stops at the grocery store to buy a bag of diapers and 2 cans of formula. He spends a total of $37. The next week he stops and buys 2 bags of diapers and 5 cans of formula for a total of $87. How much does a bag of diapers cost? How much is one can of formula?

The bag of diapers costs $11 and the can of formula costs $13.

Exercise \(\PageIndex{27}\)

To get her daily intake of fruit for the day, Sasha eats a banana and 8 strawberries on Wednesday for a calorie count of 145. On the following Wednesday, she eats two bananas and 5 strawberries for a total of 235 calories for the fruit. How many calories are there in a banana? How many calories are in a strawberry?

There are 105 calories in a banana and 5 calories in a strawberry.

Choose the Most Convenient Method to Solve a System of Linear Equations

When you will have to solve a system of linear equations in a later math class, you will usually not be told which method to use. You will need to make that decision yourself. So you’ll want to choose the method that is easiest to do and minimizes your chance of making mistakes.

Exercise \(\PageIndex{28}\)

For each system of linear equations decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

• \(\left\{\begin{array}{l}{3 x+8 y=40} \\ {7 x-4 y=-32}\end{array}\right.\)
• \(\left\{\begin{array}{l}{5 x+6 y=12} \\ {y=\frac{2}{3} x-1}\end{array}\right.\)

1. \(\left\{\begin{array}{l}{3 x+8 y=40} \\ {7 x-4 y=-32}\end{array}\right.\)

• Since both equations are in standard form, using elimination will be most convenient.

2. \(\left\{\begin{array}{l}{5 x+6 y=12} \\ {y=\frac{2}{3} x-1}\end{array}\right.\)

Since one equation is already solved for y , using substitution will be most convenient.

Exercise \(\PageIndex{29}\)

For each system of linear equations, decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

• \(\left\{\begin{array}{l}{4 x-5 y=-32} \\ {3 x+2 y=-1}\end{array}\right.\)
• \(\left\{\begin{array}{l}{x=2 y-1} \\ {3 x-5 y=-7}\end{array}\right.\)
• Since one equation is already solved for xx, using substitution will be most convenient.

Exercise \(\PageIndex{30}\)

• \(\left\{\begin{array}{l}{y=2 x-1} \\ {3 x-4 y=-6}\end{array}\right.\)
• \(\left\{\begin{array}{l}{6 x-2 y=12} \\ {3 x+7 y=-13}\end{array}\right.\)
• Since one equation is already solved for yy, using substitution will be most convenient;

Access these online resources for additional instruction and practice with solving systems of linear equations by elimination.

• Instructional Video-Solving Systems of Equations by Elimination
• Instructional Video-Solving by Elimination
• Instructional Video-Solving Systems by Elimination

Key Concepts

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