1. Learning objectives

Learning objectives

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

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.

2. Solve a System of Equations by Elimination

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}{lrll} \mathrm{if} & a & = & b \\ \mathrm{and} & c & = & d \\ \mathrm{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 \\ \underset{\mathrm{\_\_\_\_\_\_\_\_\_}}{2x-y=0} \\ 5x\,=5 \\ \end{array}

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

Let’s try another one:

\mathrm{\{}\begin{array}{l} x+4y=2 \\ 2x+5y=-2 \\ \end{array}

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:

\mathrm{\{}\begin{array}{l} 4x-3y=10 \\ 3x+5y=-7 \\ \end{array}

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:

\mathrm{\{}\begin{array}{l} \,12x-9y=30 \\ -12x-20y=28 \\ \end{array}

When we add these equations,

\begin{array}{c} \mathrm{\{}\begin{array}{l} \,12x-9y=30 \\ \underset{\mathrm{\_\_\_\_\_\_\_\_\_\_\_\_\_}}{-12x-20y=28} \\ \end{array} \\ -29y=58 \\ \end{array}

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.

3. Example 5.25

Example 5.25

How to Solve a System of Equations by Elimination

Solve the system by elimination. $\mathrm{\{}\begin{array}{l} 2x+y=7 \\ x-2y=6 \\ \end{array}$

The steps are listed below for easy reference.

How to solve a system of equations by elimination.

Step 1. Write both equations in standard form. If any coefficients are fractions, clear them.
Step 2. Make the coefficients of one variable opposites. Decide which variable you will eliminate. Multiply one or both equations so that the coefficients of that variable are opposites.
Step 3. Add the equations resulting from Step 2 to eliminate one variable.
Step 4. Solve for the remaining variable.
Step 5. Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
Step 6. Write the solution as an ordered pair.
Step 7. 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.

4. Example 5.26

Example 5.26

Solve the system by elimination. $\mathrm{\{}\begin{array}{l} x+y=10 \\ x-y=12 \\ \end{array}$

Solution

In Example 5.27, we will be able to make the coefficients of one variable opposites by multiplying one equation by a constant.

5. Example 5.27

Example 5.27

Solve the system by elimination. $\mathrm{\{}\begin{array}{l} 3x-2y=-2 \\ 5x-6y=10 \\ \end{array}$

Solution

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.

6. Example 5.28

Example 5.28

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

Solution

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.

What other constants could we have chosen to eliminate one of the variables? Would the solution be the same?

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

7. Example 5.29

Example 5.29

Solve the system by elimination. $\mathrm{\{}\begin{array}{l} x+\frac{1}{2}y=6 \\ \frac{3}{2}x+\frac{2}{3}y=\frac{17}{2} \\ \end{array}$

Solution

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

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.

8. Example 5.30

Example 5.30

Solve the system by elimination. $\mathrm{\{}\begin{array}{l} 3x+4y=12 \\ y=3-\frac{3}{4}x \\ \end{array}$

Solution


	$\mathrm{\{}\begin{array}{l} 3x+4y=12 \\ y=3-\frac{3}{4}x \\ \end{array}$
Write the second equation in standard form.	$\mathrm{\{}\begin{array}{rll} 3x+4y & = & 12 \\ \frac{3}{4}x+y & = & 3 \\ \end{array}$
Clear the fractions by multiplying the second equation by 4.	$\mathrm{\{}\begin{array}{rll} 3x+4y & = & 12 \\ 4(\frac{3}{4}x+y) & = & 4(3) \\ \end{array}$
Simplify.	$\mathrm{\{}\begin{array}{rll} 3x+4y & = & 12 \\ 3x+4y & = & 12 \\ \end{array}$
To eliminate a variable, we multiply the second equation by $-1$ . Simplify and add.	$\begin{array}{lcccc} \,\underset{\mathrm{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}}{\mathrm{\{}\begin{array}{rll} 3x+4y & = & 12 \\ -3x-4y & = & -12 \\ \end{array}} \\ 0=0 \\ \end{array}$

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.

9. Example 5.31

Example 5.31

Solve the system by elimination. $\mathrm{\{}\begin{array}{l} -6x+15y=10 \\ 2x-5y=-5 \\ \end{array}$

Solution


The equations are in standard form.	$\mathrm{\{}\begin{array}{rll} -6x+15y & = & 10 \\ 2x-5y & = & -5 \\ \end{array}$
Multiply the second equation by 3 to eliminate a variable.	$\mathrm{\{}\begin{array}{rll} -6x+15y & = & 10 \\ 3(2x-5y) & = & 3(-5) \\ \end{array}$
Simplify and add.	$\begin{array}{ccccc} \underset{\mathrm{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}}{\mathrm{\{}\begin{array}{rll} -6x+15y & = & \,10 \\ 6x-15y & = & -15 \\ \end{array}} \\ 0\ne-5 \\ \end{array}$

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

The system does not have a solution.

10. Solve Applications of Systems of Equations by Elimination

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.

11. Example 5.32

Example 5.32

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

Solution


Step 1. Read the problem.
Step 2. Identify what we are looking for.	We are looking for two numbers.
Step 3. Name what we are looking for. Choose a variable to represent that quantity.	Let $n=$ the first number. $m=$ the second number.
Step 4. Translate into a system of equations.The system is:	The sum of two numbers is 39. $n+m=39$ Their difference is 9. $\begin{array}{c} n-m=9 \\ \mathrm{\{}\begin{array}{l} n+m=39 \\ n-m=9 \\ \end{array} \\ \end{array}$
Step 5. Solve the system of equations. To solve the system of equations, use elimination. The equations are in standard form and the coefficients of $m$ are opposites. Add. Solve for $n$ . Substitute $n=24$ into one of the original equations and solve for $m$ .	$\begin{array}{c} \underset{\mathrm{\_\_\_\_\_\_\_\_\_\_\_\_}}{\mathrm{\{}\begin{array}{l} n+m=39 \\ n-m=9 \\ \end{array}} \\ 2n\,=48 \\ \,n=24 \\ n+m=39 \\ 24+m=39 \\ m=15 \\ \end{array}$
Step 6. Check the answer.	Since $24+15=39$ and $24-15=9$ , the answers check.
Step 7. Answer the question.	The numbers are 24 and 15.

12. Example 5.33

Example 5.33

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?

Solution

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

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.

14. Example 5.34

Example 5.34

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

ⓐ $\mathrm{\{}\begin{array}{l} 3x+8y=40 \\ 7x-4y=-32 \\ \end{array}$ ⓑ $\mathrm{\{}\begin{array}{l} 5x+6y=12 \\ y=\frac{2}{3}x-1 \\ \end{array}$

Solution

ⓐ $\begin{array}{ccl} & & \mathrm{\{}\begin{array}{l} 3x+8y=40 \\ 7x-4y=-32 \\ \end{array} \\ \end{array}$ Since both equations are in standard form, using elimination will be most convenient.

ⓑ $\begin{array}{ccl} & & \mathrm{\{}\begin{array}{l} 5x+6y=12 \\ y=\frac{2}{3}x-1 \\ \end{array} \\ \end{array}$

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

15. Section 5.3 Exercises

Section 5.3 Exercises

Solve a System of Equations by Elimination

In the following exercises, solve the systems of equations by elimination.

Solve Applications of Systems of Equations by Elimination

In the following exercises, translate to a system of equations and solve.

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

In the following exercises, decide whether it would be more convenient to solve the system of equations by substitution or elimination.

Everyday Math

Writing Exercises

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

16. Practice 1

Solve the system by elimination. $\mathrm{\{}\begin{array}{l} 3x+y=5 \\ 2x-3y=7 \\ \end{array}$

Solve the system by elimination. $\mathrm{\{}\begin{array}{l} 4x+y=-5 \\ -2x-2y=-2 \\ \end{array}$

Solve the system by elimination. $\mathrm{\{}\begin{array}{l} 2x+y=5 \\ x-y=4 \\ \end{array}$

17. Practice 2

Solve the system by elimination. $\mathrm{\{}\begin{array}{l} x+y=3 \\ -2x-y=-1 \\ \end{array}$

Solve the system by elimination. $\mathrm{\{}\begin{array}{l} 4x-3y=1 \\ 5x-9y=-4 \\ \end{array}$

Solve the system by elimination. $\mathrm{\{}\begin{array}{l} 3x+2y=2 \\ 6x+5y=8 \\ \end{array}$

18. Practice 3

Solve the system by elimination. $\mathrm{\{}\begin{array}{l} 3x-4y=-9 \\ 5x+3y=14 \\ \end{array}$

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

Solve the system by elimination. $\mathrm{\{}\begin{array}{l} \frac{1}{3}x-\frac{1}{2}y=1 \\ \frac{3}{4}x-y=\frac{5}{2} \\ \end{array}$

19. Practice 4

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

Solve the system by elimination. $\mathrm{\{}\begin{array}{l} 5x-3y=15 \\ y=-5+\frac{5}{3}x \\ \end{array}$

Solve the system by elimination. $\mathrm{\{}\begin{array}{l} x+2y=6 \\ y=-\frac{1}{2}x+3 \\ \end{array}$

20. Practice 5

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

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

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

21. Practice 6

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

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?

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?

22. Practice 8

$\mathrm{\{}\begin{array}{l} 3x-y=-7 \\ 4x+2y=-6 \\ \end{array}$

$\mathrm{\{}\begin{array}{l} x+y=-8 \\ x-y=-6 \\ \end{array}$

$\mathrm{\{}\begin{array}{l} -7x+6y=-10 \\ x-6y=22 \\ \end{array}$

23. Practice 9

$\mathrm{\{}\begin{array}{l} 5x+2y=1 \\ -5x-4y=-7 \\ \end{array}$

$\mathrm{\{}\begin{array}{l} 3x-4y=-11 \\ x-2y=-5 \\ \end{array}$

$\mathrm{\{}\begin{array}{l} 6x-5y=-75 \\ -x-2y=-13 \\ \end{array}$

24. Practice 10

$\mathrm{\{}\begin{array}{l} 2x-5y=7 \\ 3x-y=17 \\ \end{array}$

$\mathrm{\{}\begin{array}{l} 7x+y=-4 \\ 13x+3y=4 \\ \end{array}$

$\mathrm{\{}\begin{array}{l} 3x-5y=-9 \\ 5x+2y=16 \\ \end{array}$

25. Practice 11

$\mathrm{\{}\begin{array}{l} 4x+7y=14 \\ -2x+3y=32 \\ \end{array}$

$\mathrm{\{}\begin{array}{l} 3x+8y=-3 \\ 2x+5y=-3 \\ \end{array}$

$\mathrm{\{}\begin{array}{l} 3x+8y=67 \\ 5x+3y=60 \\ \end{array}$

26. Practice 12

$\mathrm{\{}\begin{array}{l} \frac{1}{3}x-y=-3 \\ x+\frac{5}{2}y=2 \\ \end{array}$

$\mathrm{\{}\begin{array}{l} x+\frac{1}{3}y=-1 \\ \frac{1}{2}x-\frac{1}{3}y=-2 \\ \end{array}$

$\mathrm{\{}\begin{array}{l} 2x+y=3 \\ 6x+3y=9 \\ \end{array}$

27. Practice 13

$\mathrm{\{}\begin{array}{l} -3x-y=8 \\ 6x+2y=-16 \\ \end{array}$

$\mathrm{\{}\begin{array}{l} 3x+2y=6 \\ -6x-4y=-12 \\ \end{array}$

$\mathrm{\{}\begin{array}{l} -11x+12y=60 \\ -22x+24y=90 \\ \end{array}$

28. Practice 14

$\mathrm{\{}\begin{array}{l} 5x-3y=15 \\ y=\frac{5}{3}x-2 \\ \end{array}$

The sum of two numbers is 65. Their difference is 25. Find the numbers.

The sum of two numbers is −27. Their difference is −59. Find the numbers.

29. Practice 15

Andrea is buying some new shirts and sweaters. She is able to buy 3 shirts and 2 sweaters for $114 or she is able to buy 2 shirts and 4 sweaters for$ 164. How much does a shirt cost? How much does a sweater cost?

The total amount of sodium in 2 hot dogs and 3 cups of cottage cheese is 4720 mg. The total amount of sodium in 5 hot dogs and 2 cups of cottage cheese is 6300 mg. How much sodium is in a hot dog? How much sodium is in a cup of cottage cheese?

ⓐ $\mathrm{\{}\begin{array}{l} 8x-15y=-32 \\ 6x+3y=-5 \\ \end{array}$ ⓑ $\mathrm{\{}\begin{array}{l} x=4y-3 \\ 4x-2y=-6 \\ \end{array}$

30. Practice 7

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

ⓐ $\mathrm{\{}\begin{array}{l} 4x-5y=-32 \\ 3x+2y=-1 \\ \end{array}$ ⓑ $\mathrm{\{}\begin{array}{l} x=2y-1 \\ 3x-5y=-7 \\ \end{array}$

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

ⓐ $\mathrm{\{}\begin{array}{l} y=2x-1 \\ 3x-4y=-6 \\ \end{array}$ ⓑ $\mathrm{\{}\begin{array}{l} 6x-2y=12 \\ 3x+7y=-13 \\ \end{array}$

$\mathrm{\{}\begin{array}{l} -3x+y=-9 \\ x-2y=-12 \\ \end{array}$

31. Practice 16

ⓐ $\mathrm{\{}\begin{array}{l} y=4x+9 \\ 5x-2y=-21 \\ \end{array}$ ⓑ $\mathrm{\{}\begin{array}{l} 9x-4y=24 \\ 3x+5y=-14 \\ \end{array}$

Norris can row 3 miles upstream against the current in 1 hour, the same amount of time it takes him to row 5 miles downstream, with the current. Solve the system. $\mathrm{\{}\begin{array}{l} r-c=3 \\ r+c=5 \\ \end{array}$

ⓐ for $r$ , his rowing speed in still water.

ⓑ Then solve for $c$ , the speed of the river current.

Solve the system $\mathrm{\{}\begin{array}{l} x+y=10 \\ 5x+8y=56 \\ \end{array}$