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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.
Solve the system by elimination. $\{\begin{array}{c}2x+y=7\hfill \\ x2y=6\hfill \end{array}$
Solve the system by elimination. $\{\begin{array}{c}3x+y=5\hfill \\ 2x3y=7\hfill \end{array}$
$\left(2,\mathrm{1}\right)$
Solve the system by elimination. $\{\begin{array}{c}4x+y=\mathrm{5}\hfill \\ \mathrm{2}x2y=\mathrm{2}\hfill \end{array}$
$\left(\mathrm{2},3\right)$
The steps are listed below for easy reference.
First we’ll do an example where we can eliminate one variable right away.
Solve the system by elimination. $\{\begin{array}{c}x+y=10\hfill \\ xy=12\hfill \end{array}$
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. $\begin{array}{cccc}\begin{array}{ccc}\hfill x+y& =\hfill & 10\hfill \\ \hfill 11+\left(1\right)& \stackrel{?}{=}\hfill & 10\hfill \\ \hfill 10& =\hfill & 10\phantom{\rule{0.2em}{0ex}}\u2713\hfill \end{array}& & & \begin{array}{ccc}\hfill xy& =\hfill & 12\hfill \\ \hfill 11\left(1\right)& \stackrel{?}{=}\hfill & 12\hfill \\ \hfill 12& =\hfill & 12\phantom{\rule{0.2em}{0ex}}\u2713\hfill \end{array}\end{array}$ 

The solution is (11, −1). 
Solve the system by elimination. $\{\begin{array}{c}2x+y=5\hfill \\ xy=4\hfill \end{array}$
$\left(3,\mathrm{1}\right)$
Solve the system by elimination. $\{\begin{array}{c}x+y=3\hfill \\ \mathrm{2}xy=\mathrm{1}\hfill \end{array}$
$\left(\mathrm{2},5\right)$
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. $\{\begin{array}{c}3x2y=\mathrm{2}\hfill \\ 5x6y=10\hfill \end{array}$
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. $\begin{array}{cccc}\begin{array}{ccc}\hfill 3x2y& =\hfill & 2\hfill \\ \hfill 3\left(4\right)2\left(5\right)& \stackrel{?}{=}\hfill & 2\hfill \\ \hfill 12+10& \stackrel{?}{=}\hfill & 2\hfill \\ \hfill 2y& =\hfill & 2\phantom{\rule{0.2em}{0ex}}\u2713\hfill \end{array}& & & \begin{array}{ccc}\hfill 5x6y& =\hfill & 10\hfill \\ \hfill 3\left(4\right)6\left(5\right)& \stackrel{?}{=}\hfill & 10\hfill \\ \hfill 20+30& \stackrel{?}{=}\hfill & 10\hfill \\ \hfill 10& =\hfill & 10\phantom{\rule{0.2em}{0ex}}\u2713\hfill \end{array}\end{array}$ 

The solution is (−4, −5). 
Solve the system by elimination. $\{\begin{array}{c}4x3y=1\hfill \\ 5x9y=\mathrm{4}\hfill \end{array}$
$\left(1,1\right)$
Solve the system by elimination. $\{\begin{array}{c}3x+2y=2\hfill \\ 6x+5y=8\hfill \end{array}$
$\left(\mathrm{2},4\right)$
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. $\{\begin{array}{c}4x3y=9\hfill \\ 7x+2y=\mathrm{6}\hfill \end{array}$
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. $\begin{array}{cccc}\begin{array}{ccc}\hfill 4x3y& =\hfill & 9\hfill \\ \hfill 4\left(0\right)3\left(3\right)& \stackrel{?}{=}\hfill & 9\hfill \\ \hfill 9& =\hfill & 9\phantom{\rule{0.2em}{0ex}}\u2713\hfill \end{array}& & & \begin{array}{ccc}\hfill 7x+2y& =\hfill & 6\hfill \\ \hfill 7\left(0\right)+2\left(3\right)& \stackrel{?}{=}\hfill & 6\hfill \\ \hfill 6& =\hfill & 6\phantom{\rule{0.2em}{0ex}}\u2713\hfill \end{array}\end{array}$ 

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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