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Graphing Systems of Linear Inequalities

Solve a System of Linear Inequalities by Graphing

The solution to a single linear inequality is the region on one side of the boundary line that contains all the points that make the inequality true. The solution to a system of two linear inequalities is a region that contains the solutions to both inequalities. To find this region, we will graph each inequality separately and then locate the region where they are both true. The solution is always shown as a graph.

Example 5.52

How to Solve a System of Linear inequalities

Solve the system by graphing.

\(\left\{\begin{array}{l}y \geq 2 x-1 \\ y < x+1\end{array}\right.\)

Solution

Step 1. Graph the first inequality.

Graph the boundary line.

Shade in the side of the boundary line where the inequality is true.

We will graph \(y \geq 2 x-1\).

We graph the line \(y=2 x-1 .\) It is a solid line because the inequality sign is \(\geq\).

We choose \((0,0)\) as a test point It is a solution to \(y \geq 2 x-1\), so we shade in the left side of the boundary line.

\(\left\{\begin{array}{l} y \geq 2 x-1 \\ y < x+1 \end{array}\right.\)

Step 2. On the same grid, graph the second inequality.
Graph the boundary line.

 


Shade in the side of that boundary line where the inequality is true.

We will graph \(y < x+1\) on the same grid.

We graph the line \(y=x+1\). It is a dashed line because the inequality sign is \( < \).

 

Again, we use \((0,0)\) as a test point. It is a solution so we shade in that side of the line \(y=x+1\).

Step 3. The solution is the region where the shading overlaps.

The point where the boundary lines intersect is not a solution because it is not a solution to \(y < x+1\).

The solution is all points in the darker shaded region.

Step 4. Check by choosing a test point.

We'll use \((-1,-1)\) as a test point.

Is \((-1,-1)\) a solution to
\( \begin{array}{rl}y & \geq 2 x-1 ? \\ -1 & \stackrel{?}{ \geq } 2(-1)-1 \\ -1 & \geq-3 \text { true }\end{array} \)

Is \((-1,-1)\) a solution to
\(y < x+1 ?\)
\(-1 \stackrel{?}{ < }-1+1\)
\(-1 < 0 \quad \text{true}\) 

The region containing \((-1,-1)\) is the solution to this system.
Try It 5.103

Solve the system by graphing. \(\left\{\begin{array}{l}y < 3 x+2 \\ y > -x-1\end{array}\right.\)

Try It 5.104

Solve the system by graphing. \(\left\{\begin{array}{l}y < -\frac{1}{2} x+3 \\ y < 3 x-4\end{array}\right.\)

HOW TO

Solve a system of linear inequalities by graphing.
  1. Step 1. Graph the first inequality.
    • Graph the boundary line.
    • Shade in the side of the boundary line where the inequality is true.
  2. Step 2. On the same grid, graph the second inequality.
    • Graph the boundary line.
    • Shade in the side of that boundary line where the inequality is true.
  3. Step 3. The solution is the region where the shading overlaps.
  4. Step 4. Check by choosing a test point.

Example 5.53

Solve the system by graphing. \(\left\{\begin{array}{l}x-y > 3 \\ y < -\frac{1}{5} x+4\end{array}\right.\)

Solution

Graph \(x-y > 3\), by graphing \(x-y=3\) and testing a point.

The intercepts are \(x=3\) and \(y=-3\) and the
boundary line will be dashed.

Test \((0,0)\). It makes the inequality false. So, shade the side that does not contain \((0,0)\) red.

Graph \(y < -\frac{1}{5} x+4\) by graphing \(y=-\frac{1}{5} x+4\) using the slope \(m=-\frac{1}{5}\) and \(y\) -intercept \(b=4\). The boundary line will be dashed.

Test \((0,0)\). It makes the inequality true, so shade the side that contains \((0,0)\) blue.

Choose a test point in the solution and verify that it is a solution to both inequalities.


The point of intersection of the two lines is not included as both boundary lines were dashed. The solution is the area shaded twice which is the darker-shaded region.

 

Try It 5.105

Solve the system by graphing. \(\left\{\begin{array}{l}x+y \leq 2 \\ y \geq \frac{2}{3} x-1\end{array}\right.\)

Try It 5.106

Solve the system by graphing. \(\left\{\begin{array}{l}3 x-2 y \leq 6 \\ y > -\frac{1}{4} x+5\end{array}\right.\)

Example 5.54

Solve the system by graphing. \(\left\{\begin{array}{l}x-2 y < 5 \\ y > -4\end{array}\right.\)

Solution

Graph \(x-2 y < 5\), by graphing \(x-2 y=5\) and testing a point. The intercepts are \(x=5\) and \(y=-2.5\) and the boundary line will be dashed.

Test \((0,0)\). It makes the inequality true. So, shade the side that contains \((0,0)\) red.

Graph \(y > -4\), by graphing \(y=-4\) and recognizing that it is a
horizontal line through \(y=-4\). The boundary line will be dashed.

Test \((0,0)\). It makes the inequality true. So, shade (blue) the side that contains \((0,0)\) blue.

 

The point \((0, 0)\) is in the solution and we have already found it to be a solution of each inequality. The point of intersection of the two lines is not included as both boundary lines were dashed.

The solution is the area shaded twice which is the darker-shaded region.

Try It 5.107

Solve the system by graphing. \(\left\{\begin{array}{l}y \geq 3 x-2 \\ y < -1\end{array}\right.\)

Try It 5.108

Solve the system by graphing. \(\left\{\begin{array}{l}x > -4 \\ x-2 y \leq-4\end{array}\right.\)

Systems of linear inequalities where the boundary lines are parallel might have no solution. We’ll see this in Example 5.55.

Example 5.55

Solve the system by graphing. \(\left\{\begin{array}{l}4 x+3 y \geq 12 \\ y < -\frac{4}{3} x+1\end{array}\right.\)

Solution
Graph \(4 x+3 y \geq 12\), by graphing \(4 x+3 y=12\) and testing a point. The intercepts are \(x=3\) and \(y=4\) and the boundary line will be solid.
Test \((0,0)\). It makes the inequality false. So, shade the side that does not contain \((0,0)\) red.

Graph \(y < -\frac{4}{3} x+1\) by graphing \(y=-\frac{4}{3} x+1\) using
the slope \(m=\frac{4}{3}\) and the \(y\) -intercept \(b=1\). The boundary line will be dashed.

Test \((0,0)\). It makes the inequality true. So, shade the side that contains \((0,0)\) blue.

 

There is no point in both shaded regions, so the system has no solution. This system has no solution.

Try It 5.109

Solve the system by graphing. \(\left\{\begin{array}{l}3 x-2 y \leq 12 \\ y \geq \frac{3}{2} x+1\end{array}\right.\)

Try It 5.110

Solve the system by graphing. \(\left\{\begin{array}{l}x+3 y > 8 \\ y < -\frac{1}{3} x-2\end{array}\right.\)

Example 5.56

Solve the system by graphing. \(\left\{\begin{array}{l}y > \frac{1}{2} x-4 \\ x-2 y < -4\end{array}\right.\)

Solution
Graph \(y > \frac{1}{2} x-4\) by graphing \(y=\frac{1}{2} x-4\) using the slope \(m=\frac{1}{2}\) and the intercept \(b=-4\). The boundary line will be dashed. Test \((0,0)\). It makes the inequality true. So, shade the side that contains \((0,0)\) red.

Graph \(x-2 y < -4\) by graphing \(x-2 y=-4\) and testing a point. The intercepts are \(x=-4\) and \(y=2\) and the boundary line will be dashed.

Choose a test point in the solution and verify that it is a solution to both inequalities.

 

No point on the boundary lines is included in the solution as both lines are dashed.

The solution is the region that is shaded twice, which is also the solution to \(x−2y < −4\).

Try It 5.111

Solve the system by graphing. \(\left\{\begin{array}{l}y \geq 3 x+1 \\ -3 x+y \geq-4\end{array}\right.\)

Try It 5.112

Solve the system by graphing. \(\left\{\begin{array}{l}y \leq-\frac{1}{4} x+2 \\ x+4 y \leq 4\end{array}\right.\)