5.3 Solve systems of equations by elimination  (Page 3/6)

Solution

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. $\begin{array}{cccc}\begin{array}{ccc}\hfill x+\frac{1}{2}y& =\hfill & 6\hfill \\ \hfill 3+\frac{1}{2}\left(6\right)& \stackrel{?}{=}\hfill & 6\hfill \\ \hfill 3+6& \stackrel{?}{=}\hfill & 6\hfill \\ \hfill 6& =\hfill & 6✓\hfill \\ \\ \\ \\ \\ \end{array}& & & \begin{array}{ccc}\hfill \frac{3}{2}x+\frac{2}{3}y& =\hfill & \frac{17}{2}\hfill \\ \hfill \frac{3}{2}\left(3\right)+\frac{2}{3}\left(6\right)& \stackrel{?}{=}\hfill & \frac{17}{2}\hfill \\ \hfill \frac{9}{2}+4& \stackrel{?}{=}\hfill & \frac{17}{2}\hfill \\ \hfill \frac{9}{2}+\frac{8}{2}& \stackrel{?}{=}\hfill & \frac{17}{2}\hfill \\ \hfill \frac{17}{2}& =\hfill & \frac{17}{2}✓\hfill \end{array}\end{array}$
	The solution is (3, 6).

Solution

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

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.

Solution

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

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

The system does not have a solution.