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7.10: Systems of Linear Inequalities

Difficulty Level: At Grade Created by: CK-12
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Systems of Linear Inequalities 

In the last chapter you learned how to graph a linear inequality in two variables. To do that, you graphed the equation of the straight line on the coordinate plane. The line was solid for \begin{align*}\le\end{align*} or \begin{align*}\ge\end{align*} signs (where the equals sign is included), and the line was dashed for < or > signs (where the equals sign is not included). Then you shaded above the line (if the inequality began with \begin{align*}y > \end{align*}y> or \begin{align*}y \ge\end{align*}y) or below the line (if it began with \begin{align*}y < \end{align*}y< or \begin{align*}y \le\end{align*}y).

In this section, we’ll see how to graph two or more linear inequalities on the same coordinate plane. The inequalities are graphed separately on the same graph, and the solution for the system is the common shaded region between all the inequalities in the system. One linear inequality in two variables divides the plane into two half-planes. A system of two or more linear inequalities can divide the plane into more complex shapes.


Let’s start by solving a system of two inequalities.

Graph a System of Two Linear Inequalities

Solve the following system:

\begin{align*}2x + 3y & \le 18\\ x - 4y & \le 12\end{align*}2x+3yx4y1812

Solving systems of linear inequalities means graphing and finding the intersections. So we graph each inequality, and then find the intersection regions of the solution.

First, let’s rewrite each equation in slope-intercept form. (Remember that this form makes it easier to tell which region of the coordinate plane to shade.) Our system becomes

\begin{align*}3y \le -2x + 18 \qquad \qquad \qquad \qquad y \le - \frac{2}{3}x + 6\\ {\;} \qquad \qquad \qquad \qquad \quad \Rightarrow\!\\ -4y \le -x + 12 \qquad \qquad \qquad \qquad y \ge \frac{x}{4} - 3\end{align*}3y2x+18y23x+64yx+12yx43

Notice that the inequality sign in the second equation changed because we divided by a negative number!

For this first example, we’ll graph each inequality separately and then combine the results.

Here’s the graph of the first inequality:

The line is solid because the equals sign is included in the inequality. Since the inequality is less than or equal to, we shade below the line.

And here’s the graph of the second inequality:

The line is solid again because the equals sign is included in the inequality. We now shade above the line because \begin{align*}y\end{align*}y is greater than or equal to.

When we combine the graphs, we see that the blue and red shaded regions overlap. The area where they overlap is the area where both inequalities are true. Thus that area (shown below in purple) is the solution of the system.

The kind of solution displayed in this example is called unbounded, because it continues forever in at least one direction (in this case, forever upward and to the left).

Systems with No Solution 

There are also situations where a system of inequalities has no solution. For example, let’s solve this system.

\begin{align*}y & \le 2x - 4\\ y & > 2x + 6\end{align*}yy2x4>2x+6

We start by graphing the first line. The line will be solid because the equals sign is included in the inequality. We must shade downwards because \begin{align*}y\end{align*}y is less than.

Next we graph the second line on the same coordinate axis. This line will be dashed because the equals sign is not included in the inequality. We must shade upward because \begin{align*}y\end{align*}y is greater than.

It doesn’t look like the two shaded regions overlap at all. The two lines have the same slope, so we know they are parallel; that means that the regions indeed won’t ever overlap since the lines won’t ever cross. So this system of inequalities has no solution.

But a system of inequalities can sometimes have a solution even if the lines are parallel. For example, what happens if we swap the directions of the inequality signs in the system we just graphed?

To graph the system

\begin{align*}y \ge 2x - 4\!\\ y < 2x + 6\end{align*}y2x4y<2x+6,

we draw the same lines we drew for the previous system, but we shade upward for the first inequality and downward for the second inequality. Here is the result:

You can see that this time the shaded regions overlap. The area between the two lines is the solution to the system.

Graph a System of More Than Two Linear Inequalities

When we solve a system of just two linear inequalities, the solution is always an unbounded region—one that continues infinitely in at least one direction. But if we put together a system of more than two inequalities, sometimes we can get a solution that is bounded—a finite region with three or more sides.

Let’s look at a simple example.

Find the solution to the following system of inequalities.

\begin{align*}3x - y & < 4\\ 4y + 9x & < 8\\ x & \ge 0\\ y & \ge 0\end{align*}3xy4y+9xxy<4<800

Let’s start by writing our inequalities in slope-intercept form.

\begin{align*}y & > 3x - 4\\ y & < - \frac{9}{4}x + 2\\ x & \ge 0\\ y & \ge 0\end{align*}yyxy>3x4<94x+200

Now we can graph each line and shade appropriately. First we graph \begin{align*}y > 3x - 4\end{align*}y>3x4 :

Next we graph \begin{align*}y < - \frac{9}{4}x + 2\end{align*}y<94x+2 :

Finally we graph \begin{align*}x \ge 0\end{align*}x0 and \begin{align*}y \ge 0\end{align*}y0, and we’re left with the region below; this is where all four inequalities overlap.

The solution is bounded because there are lines on all sides of the solution region. In other words, the solution region is a bounded geometric figure, in this case a triangle.

Notice, too, that only three of the lines we graphed actually form the boundaries of the region. Sometimes when we graph multiple inequalities, it turns out that some of them don’t affect the overall solution; in this case, the solution would be the same even if we’d left out the inequality \begin{align*}y > 3x - 4\end{align*}y>3x4. That’s because the solution region of the system formed by the other three inequalities is completely contained within the solution region of that fourth inequality; in other words, any solution to the other three inequalities is automatically a solution to that one too, so adding that inequality doesn’t narrow down the solution set at all.

But that wasn’t obvious until we actually drew the graph!



Example 1

Write the system of inequalities shown below.

There are two boundary lines, so there are two inequalities. Write each one in slope-intercept form.

\begin{align*}y & \le \frac{1}{4} x + 7\\ y & \ge -\frac{5}{2} x - 5\end{align*}yy14x+752x5


  1. Consider the system \begin{align*}y < 3x - 5\!\\ y > 3x - 5\end{align*}y<3x5y>3x5 Is it consistent or inconsistent? Why?
  2. Consider the system \begin{align*}y \le 2x + 3\!\\ y \ge 2x + 3\end{align*}y2x+3y2x+3 Is it consistent or inconsistent? Why?
  3. Consider the system \begin{align*}y \le -x + 1\!\\ y > -x + 1\end{align*}yx+1y>x+1 Is it consistent or inconsistent? Why?
  4. In example 3 in this lesson, we solved a system of four inequalities and saw that one of the inequalities, \begin{align*}y > 3x - 4\end{align*}y>3x4, didn’t affect the solution set of the system.
    1. What would happen if we changed that inequality to \begin{align*}y < 3x - 4\end{align*}y<3x4?
    2. What’s another inequality that we could add to the original system without changing it? Show how by sketching a graph of that inequality along with the rest of the system.
    3. What’s another inequality that we could add to the original system to make it inconsistent? Show how by sketching a graph of that inequality along with the rest of the system.
  5. Recall the compound inequalities in one variable that we worked with back in chapter 6. Compound inequalities with “and” are simply systems like the ones we are working with here, except with one variable instead of two.
    1. Graph the inequality \begin{align*}x > 3\end{align*}x>3 in two dimensions. What’s another inequality that could be combined with it to make an inconsistent system?
    2. Graph the inequality \begin{align*}x \le 4\end{align*}x4 on a number line. What two-dimensional system would have a graph that looks just like this one?

Find the solution region of the following systems of inequalities.

  1. \begin{align*}x - y < -6\!\\ 2y \ge 3x + 17\end{align*}xy<62y3x+17
  2. \begin{align*}4y - 5x < 8\!\\ -5x \ge 16 - 8y\end{align*}4y5x<85x168y
  3. \begin{align*}5x - y \ge 5\!\\ 2y - x \ge -10\end{align*}5xy52yx10
  4. \begin{align*}5x + 2y \ge -25\!\\ 3x - 2y \le 17\!\\ x - 6y \ge 27\end{align*}5x+2y253x2y17x6y27
  5. \begin{align*}2x - 3y \le 21\!\\ x + 4y \le 6\!\\ 3x + y \ge -4\end{align*}2x3y21x+4y63x+y4
  6. \begin{align*}12x - 7y < 120\!\\ 7x - 8y \ge 36\!\\ 5x + y \ge 12\end{align*}12x7y<1207x8y365x+y12

Review (Answers)

To view the Review answers, open this PDF file and look for section 7.10. 

Texas Instruments Resources

In the CK-12 Texas Instruments Algebra I FlexBook® resource, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9617.

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Feasible region

The feasible region is the inner region within a set of four inequalities on a graph.

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Difficulty Level:
At Grade
Date Created:
Oct 01, 2012
Last Modified:
Aug 16, 2016
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