The statements $x+y>4,2x+3y<7,y\ge x,$ and $x+y\le 9$ are examples of linear inequalities in the variables x and y. As with equations, a solution of an inequality in two variables x and y is an ordered pair (a, b) that results in a true statement when x is replaced with a and y is replaced with b in the inequality. For example, the ordered pair (3, 2) is a solution of the inequality $4x+5y>11$ because $4(3)+5(2)=22>11$ is a true statement. The ordered pair (1, 1) is not a solution of the inequality $4x+5y>11$ because $4(1)+5(1)=9>11$ is a false statement. The set of all solutions of an inequality is called the solution set of the inequality. The graph of an inequality in two variables is the graph of the solution set of the inequality.

Example 1 Graphing a Linear Inequality

Graph the inequality $2x+y>6.$

Solution

First, graph the equation $2x+y=6.$ See Figure 5.12(a). Notice that if we solve this equation for y, we have $y=-2x+6.$ Any point P(a, b) whose y-coordinate, b, is greater than $6-2a$ must be above this line. Therefore, the graph of $y>-2x+6,$ which is equivalent to $2x+y>6,$ consists of all points (x, y) in the plane that lie above the line $2x+y=6.$ The graph of the inequality $2x+y>6$ is shaded in Figure 5.12(b).

Practice Problem 1

1. Graph the inequality $3x+y<6.$

In Figure 5.12(b), the graph of the equation $2x+y=6$ is drawn as a dashed line to indicate that points on the line are not included in the solution set of the inequality $2x+y>6.$ In Figure 5.13(a), however, the graph of the inequality $2x+y\ge 6$ shows $2x+y=6$ as a solid line. The solid line indicates that the points on the line are included in the solution set. Note that the region below the line $2x+y=6$ in Figure 5.13(b) is the graph of the inequality $2x+y<6.$

In general, the graph of a linear inequality’s corresponding equation (found by changing the inequality symbol to the equality sign) is a line that separates the plane into two regions, each called a half plane. The line itself is called the boundary of each region. All of the points in one region satisfy the inequality, and none of the points in the other region are solutions. You can therefore test one point not on the boundary, called a test point, to determine which region represents the solution set.

Practice Problem 2

1. Graph the linear inequality $y\le -2x+4.$

Example 3 Graphing Inequalities

Sketch the graph of each inequality.

1. $x\ge 2$

2. $y<3$

3. $x+y<4$

Solution

Steps	Inequality
1. Change the inequality to an equality.	a. $x\ge 2$ $x=2$	b. $y<3$ $y=3$	c. $x+y<4$ $x+y=4$
2. Graph the equation in Step 1 with a dashed line $(<\text{ or }>)$ or a solid line $(\le \text{ or }\ge ).$
3. Select a test point.	Test (0, 0) in $x\ge 2;0\ge 2$ is a false statement. The region not containing (0, 0), together with the vertical line, is the solution set.	Test (0, 0) in $y<3;0<3$ is a true statement. The region containing (0, 0) is the solution set.	Test (0, 0) in $x+y<4;$ $0+0<4$ is a true statement. The region containing (0, 0) is the solution set.
4. Shade the solution set.

An ordered pair (a, b) is a solution of a system of inequalities involving two variables if it is a solution of each of its inequalities. For example, (0, 1) is a solution of the system

because both $2(0)+1\le 1$ and $0-3(1)<0$ are true statements. The solution set of a system of inequalities is the intersection of the solution sets of all inequalities in the system. To find the graph of this solution set, graph the solution set of each inequality in the system (with different-color shadings) on the same coordinate plane. The region common to all of the solution sets (the one where all the shaded parts overlap) is the solution set. If there are no points common to all solutions, then the system is inconsistent, and the answer is $\varnothing$.

Example 4 Graphing a System of Two Inequalities

Graph the solution set of the system of inequalities.

$$\{\begin{array}{rcrcll}2x& +& 3y& >& 6& (1)\\ y& -& x& \le & 0& (2)\end{array}$$

Solution

Graph each inequality separately in the same coordinate plane.

Inequality 1	Inequality 2
$$2x+3y>6$$	$$y-x\le 0$$
Step 1 $2x+3y=6$ Step 2 Sketch $2x+3y=6$ as a dashed line by joining the points (0, 2) and (3, 0). Step 3 Test (0, 0) in $$\begin{array}{rcl}2x+3y& >& 6\\ 2\cdot 0+3\cdot 0& \stackrel{?}{>}& 6\\ 0& \stackrel{?}{>}& 6\text{False}\end{array}$$ Step 4 Shade as in Figure 5.14(a) .	Step 1 $y-x=0$ Step 2 Sketch $y-x=0$ as a solid line by joining the points (0, 0) and (1, 1). Step 3 Test (1, 0) in $$\begin{array}{rcl}y-x& \le & 0\\ 0-1& \stackrel{?}{\le }& 0\\ -1& \stackrel{?}{\le }& 0\text{True}\end{array}$$ Step 4 Shade as in Figure 5.14(b) .

The graph of the solution set of the system of inequalities (1) and (2) (the region where the shadings overlap) is shown shaded in purple in Figure 5.14(c).

Practice Problem 4

1. Graph the solution set of the system of inequalities.

   $$\{\begin{array}{r}3x-y>3\\ x-y\ge 1\end{array}$$

The point $P({\displaystyle \frac{6}{5}},{\displaystyle \frac{6}{5}})$ in Figure 5.14(c) is the point of intersection of the two lines $2x+3y=6$ and $y-x=0.$ Such a point is called a corner point, or vertex, of the solution set. It is found by solving the system of equations $\{\begin{array}{l}2x+3y=6\\ y-x=0\end{array}.$

In this case, however, the point P is not in the solution set of the given system because the ordered pair $({\displaystyle \frac{6}{5}},{\displaystyle \frac{6}{5}})$ satisfies only inequality (2), but not inequality (1), of Example 4. We show P as an open circle in Figure 5.14(c).

Example 5 Solving a System of Three Linear Inequalities

Sketch the graph and label the vertices of the solution set of the system of linear inequalities.

$$\{\begin{array}{rcrcll}3x& +& 2y& \le & 11& (1)\\ x& -& y& \le & 2& (2)\\ 7x& -& 2y& \ge & -1& (3)\end{array}$$

Solution

First, on the same coordinate plane, sketch the graphs of the three linear equations that correspond to the three inequalities. Because either $\le$ or $\ge$ occurs in all three inequalities, all of the equations are graphed as solid lines. See Figure 5.15.

Notice that (0, 0) satisfies each of the inequalities but none of the equations. You can use (0, 0) as the test point for each inequality.

Make the following conclusions:

1. Because (0, 0) lies below the line $3x+2y=11,$ the region on or below the line $3x+2y=11$ is in the solution set of inequality (1).

2. Because (0, 0) is above the line $x-y=2,$ the region on or above the line $x-y=2$ is in the solution set of inequality (2).

3. Finally, because (0, 0) lies below the line $7x-2y=-1,$ the region on or below the line $7x-2y=-1$ belongs to the solution set of inequality (3).

Side Note

In working with linear inequalities, you must use the test point to graph the solution set. Do not assume that just because the symbol $\le$ or $<$ appears in the inequality, the graph of the inequality is below the line.

The region common to the regions in (i), (ii), and (iii) (including the boundaries) is the solution set of the given system of inequalities. The solution set is the shaded region in Figure 5.16, including the sides of the triangle.

Figure 5.16 also shows the vertices of the solution set. These vertices are obtained by solving each pair of equations in the system. Because all vertices are solutions of the given system, they are shown as (solid) points.

1. To find the vertex (3, 1), solve the system

   $$\{\begin{array}{rcrcll}3x& +& 2y& =& 11& (1)\\ x& -& y& =& 2& (2)\end{array}$$

   Solve equation (2) for x to obtain $x=2+y.$ Substitute $x=2+y$ in equation (1).

   $$\begin{array}{rcll}3(2+y)+2y& =& 11& \\ 6+3y+2y& =& 11& \text{Distributive property}\\ y& =& 1& \text{Solve for }y.\end{array}$$

   Back-substitute $y=1$ in equation (2) to find $x=3.$ The vertex is the point (3, 1).

2. Solve the system of equations $\{\begin{array}{rcrcr}x& -& y& =& 2\\ 7x& -& 2y& =& -1\end{array}$ by the substitution method to find the vertex $(-1,-3).$

3. Solve the system of equations $\{\begin{array}{l}3x+2y=11\\ 7x-2y=-1\end{array}$ by the elimination method to find the vertex (1, 4).

Practice Problem 5

1. Sketch the graph (and label the corner points) of the solution set of the system of inequalities.

   $$\{\begin{array}{rcrcr}2x& +& 3y& <& 16\\ 4x& +& 2y& \ge & 16\\ 2x& -& y& \le & 8\end{array}$$

We follow the same steps that we used in solving a linear inequality.

The procedure for solving a nonlinear system of inequalities is identical to the procedure for solving a linear system of inequalities.

Example 7 Solving a Nonlinear System of Inequalities

Graph the solution set of the system of inequalities.

$$\{\begin{array}{ll}y\le 4-x^2& (1)\\ y\ge {\displaystyle {\displaystyle \frac{3}{2}}}x-3& (2)\\ y\ge -6x-3& (3)\end{array}$$

Solution

Graph each inequality separately in the same coordinate plane. Because (0, 0) is not a solution of any of the corresponding equations, use (0, 0) as a test point for each inequality.

Inequality (1)	Inequality (2)	Inequality (3)
$y\le 4-x^2$	$y\ge {\displaystyle \frac{3}{2}}x-3$	$y\ge -6x-3$
Step 1 $y=4-x^2$	$y={\displaystyle \frac{3}{2}}x-3$	$y=-6x-3$
Step 2 The graph of $y=4-x^2$ is a parabola opening down with vertex (0, 4). Sketch it as a solid curve.	The graph of $y={\displaystyle \frac{3}{2}}x-3$ is a line through the points $(0,-3)$ and (2, 0). Sketch a solid line.	The graph of $y=-6x-3$ is a line through the points $(0,-3)$ and $(-{\displaystyle \frac{1}{2}},0).$ Sketch a solid line.
Step 3 Testing (0, 0) in $y\le 4-x^2$ gives $0\le 4,$ a true statement. The solution set is below the parabola.	Testing (0, 0) in $y\ge {\displaystyle \frac{3}{2}}x-3$ gives $0\ge -3,$ a true statement. The solution set is above the line.	Testing (0, 0) in $y\ge -6x-3$ gives $0\ge -3,$ a true statement. The solution set is above the line.
Step 4 Shade the region as in Figure 5.17(a).	Shade the region as in Figure 5.17(b).	Shade the region as in Figure 5.17(c).

The region common to all three graphs is the graph of the solution set of the given system of inequalities. See Figure 5.17(d).

Use the substitution method to locate the points of intersection A, B, and C of the corresponding equations. See Figure 5.17(d). Solve all pairs of corresponding equations.

The point A(2, 0) is the solution of the system

$$\{\begin{array}{ll}y=4-x^2& \text{Boundary of inequality }(1)\\ y={\displaystyle {\displaystyle \frac{3}{2}}}x-3& \text{Boundary of inequality }(2)\end{array}$$

The point $B(0,-3)$ is the solution of the system

$$\{\begin{array}{ll}y={\displaystyle {\displaystyle \frac{3}{2}}}x-3& \text{Boundary of inequality }(2)\\ y=-6x-3& \text{Boundary of inequality }(3)\end{array}$$

Finally, the point $C(-1,3)$ is the solution of the system

$$\{\begin{array}{ll}y=-6x-3& \text{Boundary of inequality }(1)\\ y=4-x^2& \text{Boundary of inequality }(2)\end{array}$$

Practice Problem 7

1. Graph the solution set of the system of inequalities.

   $$\{\begin{array}{l}y\ge x^2+1\\ y\le -x+13\\ y<4x+13\end{array}$$