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Solve the system by substitution. $\{\begin{array}{c}3x+y=5\hfill \\ 2x+4y=\mathrm{10}\hfill \end{array}$
We need to solve one equation for one variable. Then we will substitute that expression into the other equation.
Solve for
y .
Substitute into the other equation. 

Replace the y with −3 x + 5.  
Solve the resulting equation for x .  
 
Substitute x = 3 into 3 x + y = 5 to find y .  
 
The ordered pair is (3, −4).  
Check the ordered pair in both equations:
$\begin{array}{cccc}\begin{array}{ccc}\hfill 3x+y& =\hfill & 5\hfill \\ \hfill 3\xb73+(4)& \stackrel{?}{=}\hfill & 5\hfill \\ \hfill 94& \stackrel{?}{=}\hfill & 5\hfill \\ \hfill 5& =\hfill & 5\phantom{\rule{0.2em}{0ex}}\u2713\hfill \end{array}& & & \begin{array}{ccc}\hfill 2x+4y& =\hfill & 10\hfill \\ \hfill 2\xb73+4(4)& =\hfill & 10\hfill \\ \hfill 616& \stackrel{?}{=}\hfill & 10\hfill \\ \hfill 10& =\hfill & 10\phantom{\rule{0.2em}{0ex}}\u2713\hfill \end{array}\end{array}$ 

The solution is (3, −4). 
Solve the system by substitution. $\{\begin{array}{c}4x+y=2\hfill \\ 3x+2y=\mathrm{1}\hfill \end{array}$
$\left(1,\mathrm{2}\right)$
Solve the system by substitution. $\{\begin{array}{c}x+y=4\hfill \\ 4xy=2\hfill \end{array}$
$\left(2,6\right)$
In [link] it was easiest to solve for y in the first equation because it had a coefficient of 1. In [link] it will be easier to solve for x .
Solve the system by substitution. $\{\begin{array}{c}x2y=\mathrm{2}\hfill \\ 3x+2y=34\hfill \end{array}$
We will solve the first equation for $x$ and then substitute the expression into the second equation.
Solve for
x .
Substitute into the other equation. 

Replace the x with 2 y − 2.  
Solve the resulting equation for y .  
Substitute y = 5 into x − 2 y = −2 to find x . 

The ordered pair is (8, 5).  
Check the ordered pair in both equations:
$\begin{array}{cccc}\begin{array}{ccc}\hfill x2y& =\hfill & 2\hfill \\ \hfill 82\xb75& \stackrel{?}{=}\hfill & 2\hfill \\ \hfill 810& \stackrel{?}{=}\hfill & 2\hfill \\ \hfill 2& =\hfill & 2\phantom{\rule{0.2em}{0ex}}\u2713\hfill \end{array}& & & \begin{array}{ccc}\hfill 3x+2y& =\hfill & 34\hfill \\ \hfill 3\xb78+2\xb75& \stackrel{?}{=}\hfill & 34\hfill \\ \hfill 24+10& \stackrel{?}{=}\hfill & 34\hfill \\ \hfill 34& =\hfill & 34\phantom{\rule{0.2em}{0ex}}\u2713\hfill \end{array}\end{array}$ 

The solution is (8, 5). 
Solve the system by substitution. $\{\begin{array}{c}x5y=13\hfill \\ 4x3y=1\hfill \end{array}$
$\left(\mathrm{2},\mathrm{3}\right)$
Solve the system by substitution. $\{\begin{array}{c}x6y=\mathrm{6}\hfill \\ 2x4y=4\hfill \end{array}$
$\left(6,2\right)$
When both equations are already solved for the same variable, it is easy to substitute!
Solve the system by substitution. $\{\begin{array}{c}y=\mathrm{2}x+5\hfill \\ y=\frac{1}{2}x\hfill \end{array}$
Since both equations are solved for y , we can substitute one into the other.
Substitute $\frac{1}{2}x$ for y in the first equation.  
Replace the y with $\frac{1}{2}x.$  
Solve the resulting equation. Start
by clearing the fraction. 

Solve for x .  
Substitute x = 2 into y = $\frac{1}{2}x$ to find y . 

The ordered pair is (2,1).  
Check the ordered pair in both equations:
$\begin{array}{cccc}\begin{array}{ccc}\hfill y& =\hfill & \frac{1}{2}x\hfill \\ \hfill 1& \stackrel{?}{=}\hfill & \frac{1}{2}\xb72\hfill \\ \hfill 1& =\hfill & 1\phantom{\rule{0.2em}{0ex}}\u2713\hfill \end{array}& & & \begin{array}{ccc}\hfill y& =\hfill & 2x+5\hfill \\ \hfill 1& \stackrel{?}{=}\hfill & 2\xb72+5\hfill \\ \hfill 1& =\hfill & 4+5\hfill \\ \hfill 1& =\hfill & 1\phantom{\rule{0.2em}{0ex}}\u2713\hfill \end{array}\end{array}$ 

The solution is (2,1). 
Solve the system by substitution. $\{\begin{array}{c}y=3x16\hfill \\ y=\frac{1}{3}x\hfill \end{array}$
$\left(6,2\right)$
Solve the system by substitution. $\{\begin{array}{c}y=\text{\u2212}x+10\hfill \\ y=\frac{1}{4}x\hfill \end{array}$
$\left(8,2\right)$
Be very careful with the signs in the next example.
Solve the system by substitution. $\{\begin{array}{c}4x+2y=4\hfill \\ 6xy=8\hfill \end{array}$
We need to solve one equation for one variable. We will solve the first equation for y .
Solve the first equation for y .  
Substitute −2 x + 2 for y in the second equation.  
Replace the y with −2 x + 2.  
Solve the equation for x .  
 
Substitute $x=\frac{5}{4}$ into 4 x + 2 y = 4 to find y . 

The ordered pair is $\left(\frac{5}{4},\frac{1}{2}\right).$  
Check the ordered pair in both equations.
$\begin{array}{cccc}\begin{array}{ccc}\hfill 4x+2y& =\hfill & 4\hfill \\ \hfill 4\left(\frac{5}{4}\right)+2\left(\frac{1}{2}\right)& \stackrel{?}{=}\hfill & 4\hfill \\ \hfill 51& \stackrel{?}{=}\hfill & 4\hfill \\ \hfill 4& =\hfill & 4\phantom{\rule{0.2em}{0ex}}\u2713\hfill \\ \\ \\ \\ \\ \end{array}\hfill & & & \begin{array}{ccc}\hfill 6xy& =\hfill & 8\hfill \\ \hfill 6\left(\frac{5}{4}\right)\left(\frac{1}{2}\right)& \stackrel{?}{=}\hfill & 8\hfill \\ \hfill \frac{15}{4}\left(\frac{1}{2}\right)& \stackrel{?}{=}\hfill & 8\hfill \\ \hfill \frac{16}{2}& \stackrel{?}{=}\hfill & 8\hfill \\ \hfill 8& =\hfill & 8\phantom{\rule{0.2em}{0ex}}\u2713\hfill \end{array}\hfill \end{array}$ 

The solution is $\left(\frac{5}{4},\frac{1}{2}\right).$ 
Solve the system by substitution. $\{\begin{array}{c}x4y=\mathrm{4}\hfill \\ \mathrm{3}x+4y=0\hfill \end{array}$
$\left(2,\frac{3}{2}\right)$
Solve the system by substitution. $\{\begin{array}{c}4xy=0\hfill \\ 2x3y=5\hfill \end{array}$
$\left(\frac{1}{2},\mathrm{2}\right)$
In [link] , it will take a little more work to solve one equation for x or y .
Solve the system by substitution. $\{\begin{array}{c}4x3y=6\hfill \\ 15y20x=\mathrm{30}\hfill \end{array}$
We need to solve one equation for one variable. We will solve the first equation for x .
Solve the first equation for x .  
Substitute $\frac{3}{4}y+\frac{3}{2}$ for x in the second equation.  
Replace the x with $\frac{3}{4}y+\frac{3}{2}.$  
Solve for y .  
Since 0 = 0 is a true statement, the system is consistent. The equations are dependent. The graphs of these two equations would give the same line. The system has infinitely many solutions.
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