How to Solve System of Equations by Graphing

How to Solve System of Equations by Graphing – A system of linear equations is a system made up of two linear equations. To solve the system of equations, you need to find the exact values of x and y that will solve both equations. One good way to do this is to graph each line and see where they intersect.

Before you can graph a linear equation, you need to make sure that it is written in slope-intercept form:

The slope-intercept form of a linear equation is: y = mx + b.

In slope-intercept form, m is the slope of the line and b is the y-intercept, so in the equation we looked at earlier, y = 3x + 5, the slope would be 3 and the y-intercept would be 5.

In Solving Linear Equations and Inequalities we learned how to solve linear equations with one variable. Remember that the solution of an equation is a value of the variable that makes a true statement when substituted into the equation.

Now we will work with systems of linear equations, two or more linear equations grouped together. System of Linear Equations

When two or more linear equations are grouped together, they form a system of linear equations.

We will focus our work here on systems of two linear equations in two unknowns. Later, you may solve larger systems of equations.

An example of a system of two linear equations is shown below. We use a brace to show the two equations are grouped together to form a system of equations.

$\left\{\begin{array}{c}2x+y=7\hfill \\ x-2y=6\hfill \end{array}$

A linear equation in two variables, like 2x + y = 7, has an infinite number of solutions. Its graph is a line. Remember, every point on the line is a solution to the equation and every solution to the equation is a point on the line.

To solve a system of two linear equations, we want to find the values of the variables that are solutions to both equations. In other words, we are looking for the ordered pairs (x, y) that make both equations true. These are called the solutions to a system of equations. Solutions of a System of Equations

Solutions of a system of equations are the values of the variables that make all the equations true. A solution of a system of two linear equations is represented by an ordered pair (x, y).

To determine if an ordered pair is a solution to a system of two equations, we substitute the values of the variables into each equation. If the ordered pair makes both equations true, it is a solution to the system.

Let’s consider the system below:

$\left\{\begin{array}{c}3x-y=7\hfill \\ x-2y=4\hfill \end{array}$

Is the ordered pair $\left(2,-1\right)$ a solution?

This figure begins with a sentence, “We substitute x =2 and y = -1 into both equations.” The first equation shows that 3x minus y equals 7. Then 3 times 2 minus negative, in parentheses, equals 7. Then 7 equals 7 is true. The second equation reads x minus 2y equals 4. Then 2 minus 2 times negative one in parentheses equals 4. Then 4 = 4 is true.

The ordered pair (2, −1) made both equations true. Therefore (2, −1) is a solution to this system.

Let’s try another ordered pair. Is the ordered pair (3, 2) a solution?

This figure begins with the sentence, “We substitute x equals 3 and y equals 2 into both equations.” The first equation reads 3 times x minus 7equals 7. Then, 3 times 3 minus 2 equals 7. Then 7 = 7 is true. The second equation reads x minus 2y equals 4. The n times 2 minus 2 times 2 = 4. Then negative 2 = 4 is false.

The ordered pair (3, 2) made one equation true, but it made the other equation false. Since it is not a solution to both equations, it is not a solution to this system.

Determine whether the ordered pair is a solution to the system: $\left\{\begin{array}{c}x-y=-1\hfill \\ 2x-y=-5\hfill \end{array}$

ⓐ $\left(-2,-1\right)$ ⓑ $\left(-4,-3\right)$ Solution

ⓐ

(−2, −1) does not make both equations true. (−2, −1) is not a solution.

ⓑ

(−4, −3) does not make both equations true. (−4, −3) is a solution.

Determine whether the ordered pair is a solution to the system: $\left\{\begin{array}{c}3x+y=0\hfill \\ x+2y=-5\hfill \end{array}.$

ⓐ $\left(1,-3\right)$ ⓑ $\left(0,0\right)$

ⓐ yes ⓑ no

Determine whether the ordered pair is a solution to the system: $\left\{\begin{array}{c}x-3y=-8\hfill \\ -3x-y=4\hfill \end{array}.$

ⓐ $\left(2,-2\right)$ ⓑ $\left(-2,2\right)$

ⓐ no ⓑ yes

Solve a System of Linear Equations by Graphing

In this chapter we will use three methods to solve a system of linear equations. The first method we’ll use is graphing.

The graph of a linear equation is a line. Each point on the line is a solution to the equation. For a system of two equations, we will graph two lines. Then we can see all the points that are solutions to each equation. And, by finding what the lines have in common, we’ll find the solution to the system.

Most linear equations in one variable have one solution, but we saw that some equations, called contradictions, have no solutions and for other equations, called identities, all numbers are solutions.

Similarly, when we solve a system of two linear equations represented by a graph of two lines in the same plane, there are three possible cases, as shown in (Figure):

This figure shows three x y-coordinate planes. The first plane shows two lines which intersect at one point. Under the graph it says, “The lines intersect. Intersecting lines have one point in common. There is one solution to this system.” The second x y-coordinate plane shows two parallel lines. Under the graph it says, “The lines are parallel. Parallel lines have no points in common. There is no solution to this system.” The third x y-coordinate plane shows one line. Under the graph it says, “Both equations give the same line. Because we have just one line, there are infinitely many solutions.”

For the first example of solving a system of linear equations in this section and in the next two sections, we will solve the same system of two linear equations. But we’ll use a different method in each section. After seeing the third method, you’ll decide which method was the most convenient way to solve this system. How to Solve a System of Linear Equations by Graphing

Solve the system by graphing: $\left\{\begin{array}{c}2x+y=7\hfill \\ x-2y=6\hfill \end{array}.$ Solution

This table has four rows and three columns. The first column acts as the header column. The first row reads, “Step 1. Graph the first equation.” Then it reads, “To graph the first line, write the equation in slope-intercept form.” The equation reads 2x + y = 7 and becomes y = -2x + 7 where m = -2 and b = 7. Then it shows a graph of the equations 2x + y = 7. The equation x – 2y = 6 is also listed.

The second row reads, “Step 2. Graph the second equation on the same rectangular coordinate system.” Then it says, “To graph the second line, use intercepts.” This is followed by the equation x – 2y = 6 and the ordered pairs (0, -3) and (6, 0). The last column of this row shows a graph of the two equations.

The third row reads, “Step 3. Determine whether the lines intersect, are parallel, or are the same line.” Then “Look at the graph of the lines.” Finally it reads, “The lines intersect.”

The fourth row reads, “Step 4. Identify the solution to the system. If the lines intersect, identify the point of intersection. Check to make sure it is a solution to both equations. This is the solution to the system. If the lines are parallel, the system has no solution. If the lines are the same, the system has an infinite number of solutions.” Then it reads, “Since the lines intersect, find the point of intersection. Check the point in both equations.” Finally it reads, “The lines intersect at (4, -1). It then uses substitution to show that, “The solution is (4, -1).”

Solve each system by graphing: $\left\{\begin{array}{c}x-3y=-3\hfill \\ x+y=5\hfill \end{array}.$

$\left(3,2\right)$

Solve each system by graphing: $\left\{\begin{array}{c}-x+y=1\hfill \\ 3x+2y=12\hfill \end{array}.$

$\left(2,3\right)$

The steps to use to solve a system of linear equations by graphing are shown below. To solve a system of linear equations by graphing.

Graph the first equation.
Graph the second equation on the same rectangular coordinate system.
Determine whether the lines intersect, are parallel, or are the same line.
Identify the solution to the system.
- If the lines intersect, identify the point of intersection. Check to make sure it is a solution to both equations. This is the solution to the system.
- If the lines are parallel, the system has no solution.
- If the lines are the same, the system has an infinite number of solutions.

Solve the system by graphing: $\left\{\begin{array}{c}y=2x+1\hfill \\ y=4x-1\hfill \end{array}.$ Solution

Both of the equations in this system are in slope-intercept form, so we will use their slopes and y-intercepts to graph them. $\left\{\begin{array}{c}y=2x+1\hfill \\ y=4x-1\hfill \end{array}$

Solve each system by graphing: $\left\{\begin{array}{c}y=2x+2\hfill \\ y=\text{−}x-4\hfill \end{array}.$

$\left(-2,-2\right)$

Solve each system by graphing: $\left\{\begin{array}{c}y=3x+3\hfill \\ y=\text{−}x+7\hfill \end{array}.$

$\left(1,6\right)$

Both equations in (Figure) were given in slope–intercept form. This made it easy for us to quickly graph the lines. In the next example, we’ll first re-write the equations into slope–intercept form.

Solve the system by graphing: $\left\{\begin{array}{c}3x+y=-1\hfill \\ 2x+y=0\hfill \end{array}.$ Solution

We’ll solve both of these equations for so that we can easily graph them using their slopes and y-intercepts. $\left\{\begin{array}{c}3x+y=-1\hfill \\ 2x+y=0\hfill \end{array}$

\begin{array}{c}\begin{array}{ccc}\hfill 3x+y& =\hfill & -1\hfill \\ \hfill y& =\hfill & -3x-1\hfill \\ \\ \hfill m& =\hfill & -3\hfill \\ \hfill b& =\hfill & -1\hfill \\ \\ \\ \hfill 2x+y& =\hfill & 0\hfill \\ \hfill y& =\hfill & -2x\hfill \\ \\ \hfill m& =\hfill & -2\hfill \\ \hfill b& =\hfill & 0\hfill \\ \hfill \end{array}\end{array}

Solve each system by graphing: $\left\{\begin{array}{c}-x+y=1\hfill \\ 2x+y=10\hfill \end{array}.$

$\left(3,4\right)$

Solve each system by graphing: $\left\{\begin{array}{c}2x+y=6\hfill \\ x+y=1\hfill \end{array}.$

$\left(5,-4\right)$

Usually when equations are given in standard form, the most convenient way to graph them is by using the intercepts. We’ll do this in (Figure).

Solve the system by graphing: $\left\{\begin{array}{c}x+y=2\hfill \\ x-y=4\hfill \end{array}.$ Solution

We will find the x– and y-intercepts of both equations and use them to graph the lines.

\begin{array}{cccc}\begin{array}{ccc}\hfill x+y& =\hfill & 2\hfill \\ 0+y& =\hfill & 2\hfill \\ y& =\hfill & 2\hfill \end{array}& & & \begin{array}{ccc}\hfill x+y& =\hfill & 2\hfill \\ x+0& =\hfill & 2\hfill \\ x& =\hfill & 2\hfill \end{array}\end{array}

Solve each system by graphing: $\left\{\begin{array}{c}x+y=6\hfill \\ x-y=2\hfill \end{array}.$

$\left(4,2\right)$

Solve each system by graphing: $\left\{\begin{array}{c}x+y=2\hfill \\ x-y=-8\hfill \end{array}.$

$\left(5,-3\right)$

Do you remember how to graph a linear equation with just one variable? It will be either a vertical or a horizontal line.

Solve the system by graphing: $\left\{\begin{array}{c}y=6\hfill \\ 2x+3y=12\hfill \end{array}.$ Solution

Solve each system by graphing: $\left\{\begin{array}{c}y=-1\hfill \\ x+3y=6\hfill \end{array}.$

$\left(9,-1\right)$

Solve each system by graphing: $\left\{\begin{array}{c}x=4\hfill \\ 3x-2y=24\hfill \end{array}.$

$\left(4,-6\right)$

In all the systems of linear equations so far, the lines intersected and the solution was one point. In the next two examples, we’ll look at a system of equations that has no solution and at a system of equations that has an infinite number of solutions.

Solve the system by graphing: $\left\{\begin{array}{c}y=\frac{1}{2}x-3\hfill \\ x-2y=4\hfill \end{array}.$ Solution

Solve each system by graphing: $\left\{\begin{array}{c}y=-\frac{1}{4}x+2\hfill \\ x+4y=-8\hfill \end{array}.$

no solution

Solve each system by graphing: $\left\{\begin{array}{c}y=3x-1\hfill \\ 6x-2y=6\hfill \end{array}.$

no solution

Solve the system by graphing: $\left\{\begin{array}{c}y=2x-3\hfill \\ -6x+3y=-9\hfill \end{array}.$ Solution

Solve each system by graphing: $\left\{\begin{array}{c}y=-3x-6\hfill \\ 6x+2y=-12\hfill \end{array}.$

infinitely many solutions

Solve each system by graphing: $\left\{\begin{array}{c}y=\frac{1}{2}x-4\hfill \\ 2x-4y=16\hfill \end{array}.$

infinitely many solutions

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

When we graphed the second line in the last example, we drew it right over the first line. We say the two lines are coincident. Coincident lines have the same slope and same y-intercept. Coincident Lines

Coincident lines have the same slope and same y-intercept.

Determine the Number of Solutions of a Linear System

There will be times when we will want to know how many solutions there will be to a system of linear equations, but we might not actually have to find the solution. It will be helpful to determine this without graphing.

We have seen that two lines in the same plane must either intersect or are parallel. The systems of equations in (Figure) through (Figure) all had two intersecting lines. Each system had one solution.

A system with parallel lines, like (Figure), has no solution. What happened in (Figure)? The equations have coincident lines, and so the system had infinitely many solutions.

We’ll organize these results in (Figure) below:

This table has two columns and four rows. The first row labels each column “Graph” and “Number of solutions.” Under “Graph” are “2 intersecting lines,” “Parallel lines,” and “Same line.” Under “Number of solutions” are “1,” “None,” and “Infinitely many.”

Parallel lines have the same slope but different y-intercepts. So, if we write both equations in a system of linear equations in slope–intercept form, we can see how many solutions there will be without graphing! Look at the system we solved in (Figure).

$\begin{array}{cccc}& & & \hfill \phantom{\rule{0.1em}{0ex}}\left\{\phantom{\rule{0.1em}{0ex}}\begin{array}{ccc}\hfill y& =\hfill & \frac{1}{2}x-3\hfill \\ \hfill x-2y& =\hfill & 4\hfill \end{array}\hfill \\ \text{The first line is in slope-intercept form.}\hfill & & & \text{If we solve the second equation for}\phantom{\rule{0.2em}{0ex}}y,\phantom{\rule{0.2em}{0ex}}\text{we get}\hfill \\ \hfill y=\frac{1}{2}x-3\hfill & & & \hfill \phantom{\rule{1em}{0ex}}\begin{array}{ccc}\hfill x-2y& =\hfill & 4\hfill \\ \hfill -2y& =\hfill & \text{−}x+4\hfill \\ \hfill y& =\hfill & \frac{1}{2}x-2\hfill \end{array}\hfill \\ \hfill m=\frac{1}{2},b=-3\hfill & & & \hfill m=\frac{1}{2},b=-2\hfill \end{array}$

The two lines have the same slope but different y-intercepts. They are parallel lines.

(Figure) shows how to determine the number of solutions of a linear system by looking at the slopes and intercepts.

This table is entitled “Number of Solutions of a Linear System of Equations.” There are four columns. The columns are labeled, “Slopes,” “Intercepts,” “Type of Lines,” “Number of Solutions.” Under “Slopes” are “Different,” “Same,” and “Same.” Under “Intercepts,” the first cell is blank, then the words “Different” and “Same” appear. Under “Types of Lines” are the words, “Intersecting,” “Parallel,” and “Coincident.” Under “Number of Solutions” are “1 point,” “No Solution,” and “Infinitely many solutions.”

Let’s take one more look at our equations in (Figure) that gave us parallel lines.

$\left\{\begin{array}{c}y=\frac{1}{2}x-3\hfill \\ x-2y=4\hfill \end{array}$

When both lines were in slope-intercept form we had:

$y=\frac{1}{2}x-3\phantom{\rule{2em}{0ex}}y=\frac{1}{2}x-2$

Do you recognize that it is impossible to have a single ordered pair $\left(x,y\right)$ that is a solution to both of those equations?

We call a system of equations like this an inconsistent system. It has no solution.

A system of equations that has at least one solution is called a consistent system. Consistent and Inconsistent Systems

A consistent system of equations is a system of equations with at least one solution.

An inconsistent system of equations is a system of equations with no solution.

We also categorize the equations in a system of equations by calling the equations independent or dependent. If two equations are independent equations, they each have their own set of solutions. Intersecting lines and parallel lines are independent.

If two equations are dependent, all the solutions of one equation are also solutions of the other equation. When we graph two dependent equations, we get coincident lines. Independent and Dependent Equations

Two equations are independent if they have different solutions.

Two equations are dependent if all the solutions of one equation are also solutions of the other equation.

Let’s sum this up by looking at the graphs of the three types of systems. See (Figure) and (Figure).

This figure shows three x y coordinate planes in a horizontal row. The first shows two lines intersecting. The second shows two parallel lines. The third shows two coincident lines.

This table has four columns and four rows. The columns are labeled, “Lines,” “Intersecting,” “Parallel,” and “Coincident.” In the first row under the labeled column “lines” it reads “Number of solutions.” Reading across, it tell us that an intersecting line contains 1 point, a parallel line provides no solution, and a coincident line has infinitely many solutions. A consistent/inconsistent line has consistent lines if they are intersecting, inconsistent lines if they are parallel and consistent if the lines are coincident. Finally, dependent and independent lines are considered independent if the lines intersect, they are also independent if the lines are parallel, and they are dependent if the lines are coincident.

Without graphing, determine the number of solutions and then classify the system of equations: $\left\{\begin{array}{c}y=3x-1\hfill \\ 6x-2y=12\hfill \end{array}.$ Solution

\left\{\begin{array}{c}y=3x-1\hfill \\ 6x-2y=12\hfill \end{array}.

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

Without graphing, determine the number of solutions and then classify the system of equations.

$\left\{\begin{array}{c}y=-2x-4\hfill \\ 4x+2y=9\hfill \end{array}$

no solution, inconsistent, independent

Without graphing, determine the number of solutions and then classify the system of equations.

$\left\{\begin{array}{c}y=\frac{1}{3}x-5\hfill \\ x-3y=6\hfill \end{array}$

no solution, inconsistent, independent

Without graphing, determine the number of solutions and then classify the system of equations: $\left\{\begin{array}{c}2x+y=-3\hfill \\ x-5y=5\hfill \end{array}.$ Solution

\begin{array}{ccc}\hfill 2x+y& =& -3\hfill \\ \hfill y& =& -2x-3\hfill \end{array}

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

Without graphing, determine the number of solutions and then classify the system of equations.

$\left\{\begin{array}{c}3x+2y=2\hfill \\ 2x+y=1\hfill \end{array}$

one solution, consistent, independent

Without graphing, determine the number of solutions and then classify the system of equations.

$\left\{\begin{array}{c}x+4y=12\hfill \\ -x+y=3\hfill \end{array}$

one solution, consistent, independent

Without graphing, determine the number of solutions and then classify the system of equations. $\left\{\begin{array}{c}3x-2y=4\hfill \\ y=\frac{3}{2}x-2\hfill \end{array}$ Solution

\left\{\begin{array}{c}3x-2y=4\hfill \\ y=\frac{3}{2}x-2\hfill \end{array}

\begin{array}{ccc}\hfill 3x-2y& =\hfill & 4\hfill \\ \hfill -2y& =\hfill & -3x+4\hfill \\ \hfill \frac{-2y}{-2}& =\hfill & \frac{-3x+4}{-2}\hfill \\ \hfill y& =\hfill & \frac{3}{2}x-2\hfill \end{array}

A system of equations whose graphs are coincident lines has infinitely many solutions and is consistent and dependent.

Without graphing, determine the number of solutions and then classify the system of equations.

$\left\{\begin{array}{c}4x-5y=20\hfill \\ y=\frac{4}{5}x-4\hfill \end{array}$

infinitely many solutions, consistent, dependent

Without graphing, determine the number of solutions and then classify the system of equations.

$\left\{\begin{array}{c}-2x-4y=8\hfill \\ y=-\frac{1}{2}x-2\hfill \end{array}$

infinitely many solutions, consistent, dependent

Conclusion

In order to investigate situations such as that of the skateboard manufacturer, we need to recognize that we are dealing with more than one variable and likely more than one equation. A system of linear equations consists of two or more linear equations made up of two or more variables such that all equations in the system are considered simultaneously.

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Find the slope and y-intercept of the first equation.
Find the slope and y-intercept of the first equation.
Graph the two lines.
Determine the point of intersection.	The lines intersect at (1, 3).

Check the solution in both equations.	$\begin{array}{cccc}\begin{array}{ccc}\hfill y& =\hfill & 2x+1\hfill \\ \hfill 3& \stackrel{?}{=}\hfill & 2·1+1\hfill \\ \hfill 3& =\hfill & 3\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}& & & \begin{array}{ccc}\hfill y& =\hfill & 4x-1\hfill \\ \hfill 3& \stackrel{?}{=}\hfill & 4·1-1\hfill \\ \hfill 3& =\hfill & 3\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}\end{array}$
	The solution is (1, 3).

Solve the first equation for y. Find the slope and y-intercept. Solve the second equation for y. Find the slope and y-intercept.	$\begin{array}{c}\begin{array}{ccc}\hfill 3x+y& =\hfill & -1\hfill \\ \hfill y& =\hfill & -3x-1\hfill \\ \\ \hfill m& =\hfill & -3\hfill \\ \hfill b& =\hfill & -1\hfill \\ \\ \\ \hfill 2x+y& =\hfill & 0\hfill \\ \hfill y& =\hfill & -2x\hfill \\ \\ \hfill m& =\hfill & -2\hfill \\ \hfill b& =\hfill & 0\hfill \\ \hfill \end{array}\end{array}$
Graph the lines.
Determine the point of intersection.	The lines intersect at (−1, 2).
Check the solution in both equations.	$\begin{array}{cccc}\begin{array}{ccc}\hfill 3x+y& =\hfill & -1\hfill \\ \hfill 3\left(-1\right)+2& \stackrel{?}{=}\hfill & -1\hfill \\ \hfill -1& =\hfill & -1\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}& & & \begin{array}{ccc}\hfill 2x+y& =\hfill & 0\hfill \\ \hfill 2\left(-1\right)+2& \stackrel{?}{=}\hfill & 0\hfill \\ \hfill 0& =\hfill & 0\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}\end{array}$
	The solution is (−1, 2).


To find the intercepts, let x = 0 and solve for y, then let y = 0 and solve for x.	$\begin{array}{cccc}\begin{array}{ccc}\hfill x+y& =\hfill & 2\hfill \\ 0+y& =\hfill & 2\hfill \\ y& =\hfill & 2\hfill \end{array}& & & \begin{array}{ccc}\hfill x+y& =\hfill & 2\hfill \\ x+0& =\hfill & 2\hfill \\ x& =\hfill & 2\hfill \end{array}\end{array}$

To find the intercepts, let x = 0 then let y = 0.	$\begin{array}{cccc}\begin{array}{ccc}\hfill x-y& =\hfill & 4\hfill \\ \hfill 0-y& =\hfill & 4\hfill \\ \hfill -y& =\hfill & 4\hfill \\ \hfill y& =\hfill & -4\hfill \end{array}& & & \begin{array}{ccc}\hfill x-y& =\hfill & 4\hfill \\ \hfill x-0& =\hfill & 4\hfill \\ \hfill x& =\hfill & 4\hfill \\ \\ \\ \end{array}\end{array}$
Graph the line.
Determine the point of intersection.	The lines intersect at (3, −1).
Check the solution in both equations.	$\begin{array}{cccccccc}\hfill x+y& =\hfill & 2\hfill & & & \hfill x-y& =\hfill & 4\hfill \\ 3+\left(-1\right)\hfill & \stackrel{?}{=}\hfill & 2\hfill & & & \hfill 3-\left(-1\right)& \stackrel{?}{=}\hfill & 4\hfill \\ \hfill 2& =\hfill & 2✓\hfill & & & \hfill 4& =\hfill & 4✓\hfill \end{array}$ The solution is (3, −1).


We know the first equation represents a horizontal line whose y-intercept is 6.
The second equation is most conveniently graphed using intercepts.
To find the intercepts, let x = 0 and then y = 0.
Graph the lines.
Determine the point of intersection.	The lines intersect at (−3, 6).
Check the solution to both equations.	$\begin{array}{cccccccc}\hfill y& =\hfill & 6\hfill & & & \hfill 2x+3y& =\hfill & 12\hfill \\ \hfill 6& \stackrel{?}{=}\hfill & 6✓\hfill & & & \hfill 2\left(-3\right)+3\left(6\right)& \stackrel{?}{=}\hfill & 12\hfill \\ \hfill 2& =\hfill & 2\hfill & & & \hfill -6+18& \stackrel{?}{=}\hfill & 12\hfill \\ & & & & & \hfill 12& =\hfill & 12✓\hfill \end{array}$
	The solution is (−3, 6).


To graph the first equation, we will use its slope and y-intercept.


To graph the second equation, we will use the intercepts.

Graph the lines.
Determine the point of intersection.	The lines are parallel.
	Since no point is on both lines, there is no ordered pair that makes both equations true. There is no solution to this system.