Solve Systems of Linear Equations with Two Variables

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We’ll solve both of these equations for $y$ so that we can easily graph them using their slopes and y-intercepts.

Solve the first equation for y.
Find the slope and y-intercept.
Solve the second equation for y.
Find the slope and y-intercept.
Graph the lines.
Determine the point of intersection.	The lines intersect at $(\mathrm{-1},2).$
Check the solution in both equations.
	The solution is $(\mathrm{-1},2).$

Find the slope and y-intercept of the first equation.
Find the intercepts of the second equation.
Graph the lines.
	The lines are the same! Since every point on the line makes both equations true, there are infinitely many ordered pairs that make both equations true. There are infinitely many solutions to this system.

If you write the second equation in slope-intercept form, you may recognize that the equations have the same slope and same y-intercept.

ⓐ We will compare the slopes and intercepts of the two lines.

$\begin{array}{cccc}\begin{array}{}\\ \\ \\ \\ \hfill \text{The first equation is already in slope-intercept form.}\end{array}\hfill & & & \begin{array}{c}\{\begin{array}{ccc}\hfill y& =\hfill & 3x-1\hfill \\ \hfill 6x-2y& =\hfill & 12\hfill \end{array}\hfill \\ y=3x-1\hfill \end{array}\hfill \end{array}$

$\begin{array}{cccc}\begin{array}{c}\text{Write the second equation in slope-intercept form.}\hfill \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \text{Find the slope and intercept of each line.}\hfill \\ \\ \\ \\ \end{array}\hfill & & & \begin{array}{cccccccc}& & & & & \hfill 6x-2y& =\hfill & 12\hfill \\ & & & & & \hfill -2y& =\hfill & \mathrm{-6}x+12\hfill \\ & & & & & \hfill \frac{\mathrm{-2}y}{\mathrm{-2}}& =\hfill & \frac{\mathrm{-6}x+12}{\mathrm{-2}}\hfill \\ & & & & & \hfill y& =\hfill & 3x-6\hfill \\ \hfill y& =\hfill & 3x-1\hfill & & & \hfill y& =\hfill & 3x-6\hfill \\ \hfill m& =\hfill & 3\hfill & & & \hfill m& =\hfill & 3\hfill \\ \hfill b& =\hfill & \mathrm{-1}\hfill & & & \hfill b& =\hfill & \mathrm{-6}\hfill \end{array}\hfill \\ & & & \begin{array}{c}\text{Since the slopes are the same and}y\text{-intercepts are}\hfill \\ \text{different, the lines are parallel.}\hfill \end{array}\hfill \end{array}$

A system of equations whose graphs are parallel lines has no solution and is inconsistent and independent.

ⓑ We will compare the slope and intercepts of the two lines.

$\begin{array}{cccc}& & & \{\begin{array}{ccc}\hfill 2x+y& =\hfill & \mathrm{-3}\hfill \\ \hfill x-5y& =\hfill & 5\hfill \end{array}\hfill \\ \begin{array}{c}\text{Write both equations in slope–intercept form.}\hfill \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \end{array}\hfill & & & \begin{array}{cccccccc}\hfill 2x+y& =\hfill & -3\hfill & & & \hfill x-5y& =\hfill & 5\hfill \\ \hfill y& =\hfill & \mathrm{-2}x-3\hfill & & & \hfill \mathrm{-5}y& =\hfill & \text{−}x+5\hfill \\ & & & & & \hfill \frac{-5y}{-5}& =\hfill & \frac{-x+5}{-5}\hfill \\ & & & & & \hfill y& =\hfill & \frac{1}{5}x-1\hfill \end{array}\hfill \\ \begin{array}{c}\text{Find the slope and intercept of each line.}\hfill \\ \\ \\ \\ \\ \\ \end{array}\hfill & & & \begin{array}{cccccccccccc}\hfill y& =\hfill & -2x-3\hfill & & & & & & & \hfill y& =\hfill & \frac{1}{5}x-1\hfill \\ \hfill m& =\hfill & -2\hfill & & & & & & & \hfill m& =\hfill & \frac{1}{5}\hfill \\ \hfill b& =\hfill & \mathrm{-3}\hfill & & & & & & & \hfill b& =\hfill & \mathrm{-1}\hfill \end{array}\hfill \\ & & & \text{Since the slopes are different, the lines intersect.}\hfill \end{array}$

A system of equations whose graphs are intersect has 1 solution and is consistent and independent.

We need to solve one equation for one variable. We will solve the first equation for y.

Solve the first equation for y. Substitute $-2x+2$ for y in the second equation.
Replace the y with $-2x+2.$
Solve the equation for x.
Substitute $x=\frac{5}{4}$ into $4x+2y=4$ to find y.
	The ordered pair is $\left(\frac{5}{4},-\frac{1}{2}\right).$
Check the ordered pair in both equations.
	The solution is $\left(\frac{5}{4},-\frac{1}{2}\right).$

In this example, we cannot multiply just one equation by any constant to get opposite coefficients. So we will strategically multiply both equations by different constants 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 $\left(0,\mathrm{-3}\right).$
Check that the ordered pair is a solution to both original equations.
	The solution is $\left(0,\mathrm{-3}\right).$

In this example, both equations have fractions. Our first step will be to multiply each equation by the LCD of all the fractions in the equation 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 $y$ by 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 $\left(3,6\right).$
Check that the ordered pair is a solution to both original equations.
	The solution is $\left(3,6\right).$

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

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.

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Since both equations are in standard form, using elimination will be most convenient.

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Since one equation is already solved for y, using substitution will be most convenient.