Finding the Solution to Inequalities by Graphing

Graphing Systems of Linear Inequalities

To graph a linear inequality in two variables (say, $x$ and $y$ ), first get $y$ alone on one side. Then consider the related equation obtained by changing the inequality sign to an equality sign. The graph of this equation is a line.

If the inequality is strict ( $<$ or $>$ ), graph a dashed line. If the inequality is not strict ( $\le$ or $\ge$ ), graph a solid line.

Finally, pick one point that is not on either line ( $\left(0,0\right)$ is usually the easiest) and decide whether these coordinates satisfy the inequality or not. If they do, shade the half-plane containing that point. If they don't, shade the other half-plane.

Graph each of the inequalities in the system in a similar way. The solution of the system of inequalities is the intersection region of all the solutions in the system.

Example 1:

Solve the system of inequalities by graphing:

$\begin{array}{l}y\le x-2\\ y>-3x+5\end{array}$

First, graph the inequality $y\le x-2$ . The related equation is $y=x-2$ .

Since the inequality is $\le$ , not a strict one, the border line is solid.

Graph the straight line.

Consider a point that is not on the line - say, $\left(0,0\right)$ - and substitute in the inequality $y\le x-2$ .

$\begin{array}{l}0\le 0-2\\ 0\le -2\end{array}$

This is false. So, the solution does not contain the point $\left(0,0\right)$ . Shade the lower half of the line.

Similarly, draw a dashed line for the related equation of the second inequality $y>-3x+5$ which has a strict inequality. The point $\left(0,0\right)$ does not satisfy the inequality, so shade the half that does not contain the point $\left(0,0\right)$ .

The solution of the system of inequalities is the intersection region of the solutions of the two inequalities.

Example 2:

Solve the system of inequalities by graphing:

$\begin{array}{l}2x+3y\ge 12\\ 8x-4y>1\\ x<4\end{array}$

Rewrite the first two inequalities with $y$ alone on one side.

$\begin{array}{cccc}\begin{array}{l}3y\ge -2x+12\\ y\ge -\frac{2}{3}x+4\end{array}& & & \begin{array}{l}-4y>-8x+1\\ y<2x-\frac{1}{4}\end{array}\end{array}$

Now, graph the inequality $y\ge -\frac{2}{3}x+4$ . The related equation is $y=-\frac{2}{3}x+4$ .

Since the inequality is $\ge$ , not a strict one, the border line is solid.

Graph the straight line.

Consider a point that is not on the line - say, $\left(0,0\right)$ - and substitute in the inequality.

$\begin{array}{l}0\ge -\frac{2}{3}\left(0\right)+4\\ 0\ge 4\end{array}$

This is false. So, the solution does not contain the point $\left(0,0\right)$ . Shade upper half of the line.

Similarly, draw a dashed line of related equation of the second inequality $y<2x-\frac{1}{4}$ which has a strict inequality. The point $\left(0,0\right)$ does not satisfy the inequality, so shade the half that does not contain the point $\left(0,0\right)$ .

Draw a dashed vertical line $x=4$ which is the related equation of the third inequality.

Here point $\left(0,0\right)$ satisfies the inequality, so shade the half that contains the point.

The solution of the system of inequalities is the intersection region of the solutions of the three inequalities.

Finding the Solution to Inequalities by Graphing

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