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Solve the system by elimination. $\{\begin{array}{c}3x4y=\mathrm{9}\hfill \\ 5x+3y=14\hfill \end{array}$
$\left(1,3\right)$
Solve the system by elimination. $\{\begin{array}{c}7x+8y=4\hfill \\ 3x5y=27\hfill \end{array}$
$\left(4,\mathrm{3}\right)$
When the system of equations contains fractions, we will first clear the fractions by multiplying each equation by its LCD.
Solve the system by elimination. $\{\begin{array}{c}x+\frac{1}{2}y=6\hfill \\ \frac{3}{2}x+\frac{2}{3}y=\frac{17}{2}\hfill \end{array}$
In this example, both equations have fractions. Our first step will be to multiply each equation by its LCD to clear the fractions.
To clear the fractions, multiply each equation by its LCD.  
Simplify.  
Now we are ready to eliminate one of the variables. Notice that
both equations are in standard form. 

We can eliminate y multiplying the top equation by −4.  
Simplify and add.
Substitute x = 3 into one of the original equations. 

Solve for y .  
Write the solution as an ordered pair.  The ordered pair is (3, 6). 
Check that the ordered pair is a solution
to both original equations. $\begin{array}{cccc}\begin{array}{ccc}\hfill x+\frac{1}{2}y& =\hfill & 6\hfill \\ \hfill 3+\frac{1}{2}\left(6\right)& \stackrel{?}{=}\hfill & 6\hfill \\ \hfill 3+6& \stackrel{?}{=}\hfill & 6\hfill \\ \hfill 6& =\hfill & 6\phantom{\rule{0.2em}{0ex}}\u2713\hfill \\ \\ \\ \\ \\ \end{array}& & & \begin{array}{ccc}\hfill \frac{3}{2}x+\frac{2}{3}y& =\hfill & \frac{17}{2}\hfill \\ \hfill \frac{3}{2}\left(3\right)+\frac{2}{3}\left(6\right)& \stackrel{?}{=}\hfill & \frac{17}{2}\hfill \\ \hfill \frac{9}{2}+4& \stackrel{?}{=}\hfill & \frac{17}{2}\hfill \\ \hfill \frac{9}{2}+\frac{8}{2}& \stackrel{?}{=}\hfill & \frac{17}{2}\hfill \\ \hfill \frac{17}{2}& =\hfill & \frac{17}{2}\phantom{\rule{0.2em}{0ex}}\u2713\hfill \end{array}\end{array}$ 

The solution is (3, 6). 
Solve the system by elimination. $\{\begin{array}{c}\frac{1}{3}x\frac{1}{2}y=1\hfill \\ \frac{3}{4}xy=\frac{5}{2}\hfill \end{array}$
$\left(6,2\right)$
Solve the system by elimination. $\{\begin{array}{c}x+\frac{3}{5}y=\frac{1}{5}\hfill \\ \frac{1}{2}x\frac{2}{3}y=\frac{5}{6}\hfill \end{array}$
$\left(1,\mathrm{2}\right)$
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.
Solve the system by elimination. $\{\begin{array}{c}3x+4y=12\hfill \\ y=3\frac{3}{4}x\hfill \end{array}$
$\begin{array}{ccc}& & \phantom{\rule{0.8em}{0ex}}\{\begin{array}{ccc}\hfill 3x+4y& =\hfill & 12\hfill \\ \hfill y& =\hfill & 3\frac{3}{4}x\hfill \end{array}\hfill \\ \\ \\ \text{Write the second equation in standard form.}\hfill & & \phantom{\rule{0.8em}{0ex}}\{\begin{array}{ccc}\hfill 3x+4y& =\hfill & 12\hfill \\ \hfill \frac{3}{4}x+y& =\hfill & 3\hfill \end{array}\hfill \\ \\ \\ \begin{array}{c}\text{Clear the fractions by multiplying the}\hfill \\ \text{second equation by 4.}\hfill \end{array}\hfill & & \{\begin{array}{ccc}\hfill 3x+4y& =\hfill & 12\hfill \\ \hfill 4\left(\frac{3}{4}x+y\right)& =\hfill & 4\left(3\right)\hfill \end{array}\hfill \\ \\ \\ \text{Simplify.}\hfill & & \phantom{\rule{1em}{0ex}}\{\begin{array}{ccc}\hfill 3x+4y& =\hfill & 12\hfill \\ \hfill 3x+4y& =\hfill & 12\hfill \end{array}\hfill \\ \\ \\ \begin{array}{c}\text{To eliminate a variable, we multiply the}\hfill \\ \text{second equation by \u22121.}\hfill \end{array}\hfill & & \begin{array}{c}\phantom{\rule{0.2em}{0ex}}\underset{\text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}}{\{\begin{array}{ccc}\hfill 3x+4y& =\hfill & 12\hfill \\ \hfill 3x4y& =\hfill & \mathrm{12}\hfill \end{array}}\hfill \\ \hfill 0=0\hfill \end{array}\hfill \\ \text{Simplify and add.}\hfill & \end{array}$
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.
Solve the system by elimination. $\{\begin{array}{c}5x3y=15\hfill \\ y=\mathrm{5}+\frac{5}{3}x\hfill \end{array}$
infinitely many solutions
Solve the system by elimination. $\{\begin{array}{c}x+2y=6\hfill \\ y=\frac{1}{2}x+3\hfill \end{array}$
infinitely many solutions
Solve the system by elimination. $\{\begin{array}{c}\mathrm{6}x+15y=10\hfill \\ 2x5y=\mathrm{5}\hfill \end{array}$
$\begin{array}{cc}\text{The equations are in standard form.}\hfill & \phantom{\rule{0.1em}{0ex}}\{\begin{array}{ccc}\hfill 6x+15y& =\hfill & 10\hfill \\ \hfill 2x5y& =\hfill & \mathrm{5}\hfill \end{array}\hfill \\ \\ \\ \begin{array}{c}\text{Multiply the second equation by 3 to}\hfill \\ \text{eliminate a variable.}\hfill \end{array}\hfill & \phantom{\rule{0.1em}{0ex}}\{\begin{array}{ccc}\hfill 6x+15y& =\hfill & 10\hfill \\ \hfill 3\left(2x5y\right)& =\hfill & 3\left(\mathrm{5}\right)\hfill \end{array}\hfill \\ \\ \\ \text{Simplify and add.}\hfill & \begin{array}{c}\underset{\text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}}{\{\begin{array}{ccc}\hfill 6x+15y& =\hfill & \phantom{\rule{0.5em}{0ex}}10\hfill \\ \hfill 6x15y& =\hfill & \mathrm{15}\hfill \end{array}}\\ \hfill 0\ne \mathrm{5}\hfill \end{array}\hfill \end{array}$
This statement is false. The equations are inconsistent and so their graphs would be parallel lines.
The system does not have a solution.
Solve the system by elimination. $\{\begin{array}{c}\mathrm{3}x+2y=8\hfill \\ 9x6y=13\hfill \end{array}$
no solution
Solve the system by elimination. $\{\begin{array}{c}7x3y=2\hfill \\ \mathrm{14}x+6y=8\hfill \end{array}$
no solution
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.
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