5.3 Solve systems of equations by elimination (Page 2/6)

Elementary algebra Page 2 / 6

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.

How to solve a system of equations by elimination

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

Solution

$This figure has seven rows and three columns. The first row reads, “Step 1. Write both equations in standard form. If any coefficients are fractions, clear them.” It also says, “Both equations are in standard form, A x + B y = C. There are no fractions.” It also gives the two equations as 2x + y = 7 and x – 2y = 6.$

The second row reads, “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.” It also says, “We can eliminate the y’s by multiplying the first equation by 2. Multiply both sides of 2x + y = 7 by 2.” It also shows the steps with equations. Initially the equations are ex + y = 7 and x – 2y = 6. Then they become 2(2x + y) = 2 times 7 and x – 2y = 6. They then become 4x + 2y = 14 and x – 2y = 6.

The third row says, “Step 3: Add the equations resulting from step 2 to eliminate one variable.” It also says, “We add the x’s, y’s, and constants.” It then gives the equation as 5x = 20.

The fourth row says, “Step 4: Solve for the remaining variable.” It also says, “Solve for x.” It gives the equation as x = 4.

The fifth row says, “Step 5: Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.” It also says, “Substitute x = 4 into the second equation, x – 2y = 6. Then solve for y.” It then gives the equations as x – 2y = 6 which becomes 4 – 2y = 6. This is then −2y = 2, and thus, y = −1.

The sixth row says, “Step 6: Write the solution as an order pair.” It also says, “Write it as (x, y).” It gives the ordered pair as (4, −1).

The seventh row says, “Step 7: Check that the ordered pair is a solution to both original equations.” It also says, “Substitute (4, −1) into 2x + y = 7 and x – 2y = 6. Do they make both equations true? Yes!” It then gives the equations. 2x + y = 7 becomes 2 times 4 + −1 = 7 which is 7 = 7. x – 2y = 6 becomes 4 – 2 times −1 = 6 which is 6 = 6. The row then says, “The solution is (4, −1).”

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

$(2, −1)$

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

$(−2, 3)$

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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.
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.
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.

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

Solution


Both equations are in standard form.
The coefficients of y are already opposites.
Add the two equations to eliminate y . The resulting equation has only 1 variable, x .
Solve for x , the remaining variable. Substitute x = 11 into one of the original equations.

Solve for the other variable, y .
Write the solution as an ordered pair.	The ordered pair is (11, −1).
Check that the ordered pair is a solution to both original equations. $\begin{matrix} \begin{matrix} x + y & = & 10 \\ 11 + (- 1) & \overset{?}{=} & 10 \\ 10 & = & 10 ✓ \end{matrix} & \begin{matrix} x - y & = & 12 \\ 11 - (- 1) & \overset{?}{=} & 12 \\ 12 & = & 12 ✓ \end{matrix} \end{matrix}$
	The solution is (11, −1).

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

$(3, −1)$

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

$(−2, 5)$

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

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

Solution


Both equations are in standard form.
None of the coefficients are opposites.
We can make the coefficients of y opposites by multiplying the first equation by −3.
Simplify.
Add the two equations to eliminate y .
Solve for the remaining variable, x . Substitute x = −4 into one of the original equations.

Solve for y .
Write the solution as an ordered pair.	The ordered pair is (−4, −5).
Check that the ordered pair is a solution to both original equations. $\begin{matrix} \begin{matrix} 3 x - 2 y & = & - 2 \\ 3 (- 4) - 2 (- 5) & \overset{?}{=} & - 2 \\ - 12 + 10 & \overset{?}{=} & - 2 \\ - 2 y & = & - 2 ✓ \end{matrix} & \begin{matrix} 5 x - 6 y & = & 10 \\ 3 (- 4) - 6 (- 5) & \overset{?}{=} & 10 \\ - 20 + 30 & \overset{?}{=} & 10 \\ 10 & = & 10 ✓ \end{matrix} \end{matrix}$
	The solution is (−4, −5).

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

$(1, 1)$

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

$(−2, 4)$

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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.

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

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.


Both equations are in standard form. To get opposite coefficients of y , we will multiply the first equation by 2 and the second equation by 3.
Simplify.
Add the two equations to eliminate y .
Solve for x . Substitute x = 0 into one of the original equations.

Solve for y .

Write the solution as an ordered pair.	The ordered pair is (0, −3).
Check that the ordered pair is a solution to both original equations. $\begin{matrix} \begin{matrix} 4 x - 3 y & = & 9 \\ 4 (0) - 3 (- 3) & \overset{?}{=} & 9 \\ 9 & = & 9 ✓ \end{matrix} & \begin{matrix} 7 x + 2 y & = & - 6 \\ 7 (0) + 2 (- 3) & \overset{?}{=} & - 6 \\ - 6 & = & - 6 ✓ \end{matrix} \end{matrix}$
	The solution is (0, −3).

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

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Source: OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2

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