Graphing Systems of Linear Inequalities - Process & Examples

Graphing Systems of Linear Inequalities

A system of linear inequalities is two or more inequalities that use the same variables with the same values.

The graph of a system of inequalities shows all of the solutions that make both inequalities true.

There are a few things to remember when you are graphing inequalities:

1. Get your inequality into slope intercept form ($y=mx+b$). You can use an equal sign or the given inequality symbol.

1. Graph the line. If the inequality uses $<$ or $>$, the line should be dotted. If the inequality uses $\leq$ or $\geq$, the line should be solid. Keep in mind, you want to graph the functions on a coordinate plane, not a number line.

1. Shade the portion of the graph that shows the solutions. Use test points to determine which side to shade. The shaded portion will only be on one side of the line.

Let's look at an example: $$\begin{cases} y < 4x + 1 \\ 2x + 3y \geq 6 \end{cases}$$

STEP 1: Get inequalities into slope-intercept form ($y=mx+b$)

The first inequality in our system is already in slope intercept form.

The second inequality needs to be rearranged slightly to be in slope-intercept form, as it's currently in standard form:

Once we rearrange the second inequality, we get: $$y \ge -\frac{2}{3}x + 2$$ STEP 2: Graph the lines

We're going to graph each straight line as if they had an equal sign rather than an inequality.

The dotted blue line represents the first inequality in the system and the red line represents the second.

Notice how we're using a dotted line for the first inequality, but a solid line for the second. This is because the first inequality ($y<4x+1$) uses a $<$ sign. It is a strict inequality, so the values on the actual line are not solutions to this inequality. The second inequality uses $\geq$, so the line is solid, and the values on the line are solutions to this inequality.

STEP 3: Shade the portion of the graph to indicate range of solutions

The blue part of the graph shows the possible solutions for the first inequality ($y<4x+1$). The red part of the graph shows the possible solutions for the second inequality ($2x+3y \geq 6$). The purple area shows where these two colors (ranges of solutions) overlap. The overlapping shaded area is the solution region.

The solutions to the system of inequalities lie in the region where the graphs overlap.

Use test points to determine which side of each line should be shaded. Pick a point on either side of each of your lines, and plug those values into your inequality.

If the inequality is true, then you should shade the side of the line where the test point is. If the inequality is false when you plug in the test point, you should shade the other side of the line.

Let's look at a few examples of these test points.

Take a look at the blue line in the graph. The point $(-4, 0)$ is on the left side of the line. Let's plug in $(-4, 0)$ to the inequality and see if it produces a true statement: $$\begin{align}0&<4(-4) + 1 \\ 0&<-15\end{align}$$

This statement is not true since $0$ is not less than $-15$. This means that we need to shade the other side of the blue line. (Note that this is, in fact, the area we have shaded blue).

Now take a look at the red line in the graph. The point $(-2, 4)$ is located on one side of the line. Let's plug in $(-2, 4)$ to the inequality and see if it produces a true statement: $$\begin{align} 2x + 3y &\geq 6 \\ 2(-2) + 3(4) &\geq 6 \\ -4 + 12 &\geq 6 \\ 8 &\geq 6 \end{align}$$

This statement is true since $8$ is greater than or equal to $6$. This means that we need to shade this side of the red line. (Note that this is the area we have shaded red).

The purple area shows where these two colors (ranges of solutions) overlap. The solutions to the system of inequalities lie in the region where the graphs overlap.

Once you've mastered this section, take it a step further by going from a graph to inequality functions or practicing with word problems!

(Video) Solving Systems of Inequalities

by mathman1024

This video by Friendly Math 101 shows how to graph systems of linear inequalities.

$\large{\text{Summary}}$

Just like with a system of linear equations, we want to graph all the inequalities on one graph. The areas where the shaded regions overlap is the solution. Any point inside that region satisfies all your inequalities.

Somethings to remember before we begin:

$<$ or $\leq$ means we shade below the line.

$>$ or $\geq$ means we shade above the line.

$<$ or $>$ means we use a dashed line.

The first system she looks at is the following: $$\begin{cases}y>3x-3\\y\leq\frac12x+2\end{cases}$$ We graph each inequality, including the shaded regions.

For the purposes of graphing the line, simply pretend the inequality sign is an equal sign and graph as normal using ordered pairs.

As you can see we get an overlapping region, which is the purple.

(Video) Solving a System of Inequalities by Graphing and Shading

by Brian McLogan

This video by Brian McLogan shows you how to interpret inequality graphs.

$\large{\text{Summary}}$

He starts with these two problems: $$\begin{align} y&\ge x-3\\ 4&>x+y \end{align}$$

Both inequalities need to be in slope-intercept form. The first inequality, $y\ge x-3,$ is already in slope-intercept, but $4>x+y$ needs to be changed: $$4>x+y\\ y<-x+4$$

Now, our rearranged system of linear inequalities is:

$$\begin{align} y&\ge x-3\\ y&<-x+4 \end{align}$$

He then graphs both the lines. Note that $y<-x+4$ is a dotted line, while $y\ge x-3$ is a solid line. This is due to the $<$ and $\ge$ signs, respectively.

All of the area that is shaded blue satisfies the inequality $y<-x+4$ and all of the area shaded red satisfies $y\ge x-3$. The overlap between those areas (shaded a faint purple) are all of the points where both inequalities would be satisfied.