5.10 Systems of Linear Inequalities in Two Variables

1. Is the ordered pair $\left(-2,4\right)$ a solution?

   We substitute $\mathit{x}\mathbf{=}\mathbf{-}\mathbf{2}$ and $\mathit{y}\mathbf{=}\mathbf{4}$ into both inequalities.

   $$\begin{array}{ccc}\hfill \mathit{x}+4\mathit{y}& \hfill \ge \hfill & 10\hfill \\ \hfill -2+4(4)& \hfill \stackrel{?}{\ge }\hfill & 10\hfill \\ \hfill 14& \hfill \ge \hfill & 10\text{true}\hfill \end{array}\begin{array}{ccc}\hfill 3\mathit{x}-2\mathit{y}& \hfill <\hfill & 12\hfill \\ \hfill 3(-2)-2(4)& \hfill \stackrel{?}{<}\hfill & 12\hfill \\ \hfill -14& \hfill <\hfill & 12\text{true}\end{array}$$

   The ordered pair $\left(-2,4\right)$ made both inequalities true. Therefore $\left(-2,4\right)$ is a solution to this system.

2. Is the ordered pair $\left(3,1\right)$ a solution?

   We substitute $\mathit{x}\mathbf{=}\mathbf{3}$ and $\mathit{y}\mathbf{=}\mathbf{1}$ into both inequalities.

   $$\begin{array}{ccc}\hfill \mathit{x}+4\mathit{y}& \hfill \ge \hfill & 10\hfill \\ \hfill \mathbf{3}+4(\mathbf{1})& \hfill \stackrel{?}{\ge }\hfill & 10\hfill \\ \hfill 7& \hfill \ge \hfill & 10\text{false}\hfill \end{array}\begin{array}{ccc}\hfill 3x-2y& \hfill <\hfill & 12\hfill \\ \hfill 3(\mathbf{3})-2(\mathbf{1})& \hfill \stackrel{?}{<}\hfill & 12\hfill \\ \hfill 7& \hfill <\hfill & 12\text{true}\hfill \end{array}$$

   The ordered pair $\left(3,1\right)$ made one inequality true, but the other one false. Therefore $\left(3,1\right)$ is not a solution to this system.

To solve the system of linear inequalities we look at the graph and find the region that satisfies BOTH inequalities. To do this we pick a test point and check. Let's us pick $\left(-1,-1\right)$.

Is $\left(-1,-1\right)$ a solution to $y\ge 2x-1?$

Is $\left(-1,-1\right)$ a solution to $y<x+1?$

The region containing $\left(-1,-1\right)$ is the solution to the system of linear inequalities. Notice that the solution is all the points in the area shaded twice, which appears as the darkest shaded region.

Graph $x-y>3$ by graphing $x-y=3$ and testing a point (Figure 5.89). The intercepts are $x=3$ and $y=-3$ and the boundary line will be dashed. Test $\left(0,0\right)$ which makes the inequality false so shade the side that does not contain $\left(0,0\right)$.

Figure 5.89

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 $b=4$ (Figure 5.90). The boundary line will be dashed. Test $\left(0,0\right)$ which makes the inequality true, so shade the side that contains $\left(0,0\right)$.

Figure 5.90

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 appears as the darkest shaded region.

Graph $x-2y<5$ by graphing $x-2y=5$ (Figure 5.91) and testing a point. The intercepts are $x=5$ and $y=-2.5$ and the boundary line will be dashed. Test $\left(0,0\right)$, which makes the inequality true, so shade the side that contains $\left(0,0\right)$.

Figure 5.91

Graph $y>-4$ by graphing $y=-4$ and recognizing that it is a horizontal line through $y=-4$ (Figure 5.92). The boundary line will be dashed. Test $\left(0,0\right)$, which makes the inequality true so shade the side that contains $\left(0,0\right)$.

Figure 5.92

The point $\left(0,0\right)$ 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 appears as the darkest shaded region.

Graph $4x+3y\ge 12$, by graphing $4x+3y=12$ (Figure 5.93) and testing a point. The intercepts are $x=3$ and $y=4$ and the boundary line will be solid. Test $\left(0,0\right)$, which makes the inequality false, so shade the side that does not contain $\left(0,0\right)$.

Figure 5.93

Graph $y<-\frac{4}{3}x+1$ by graphing $y=-\frac{4}{3}x+1$ using the slope $m=-\frac{4}{3}$ and $y$-intercept $b=1$ (Figure 5.94). The boundary line will be dashed. Test $\left(0,0\right)$, which makes the inequality true, so shade the side that contains $\left(0,0\right)$.

Figure 5.94

No shared point exists in both shaded regions, so the system has no 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 $y$-intercept $b=-4$ (Figure 5.95). The boundary line will be dashed. Test $\left(0,0\right)$, which makes the inequality true, so shade the side that contains $\left(0,0\right)$.

Figure 5.95

Graph $x-2y<-4$ by graphing $x-2y=-4$ (Figure 5.96) 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. Test $\left(0,0\right)$, which makes the inequality false, so shade the side that does not contain $\left(0,0\right)$.

Figure 5.96

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 $x-2y<-4$.

1. Let $x=$ the number of small photos and $y=\text{the number of large photos}$. To find the system of equations translate the information. They want to have at least 25 photos.

   The number of small plus the number of large should be at least 25.	$x+y\ge 25$
   $4 for each small and $10 for each large must be no more than $200	$4x+10y\le 200$
   The number of small photos must be greater than or equal to 0.	$x\ge 0$
   The number of large photos must be greater than or equal to 0.	$y\ge 0$
   We have our system of equations.	$\{\begin{array}{ccc}\hfill x+y& \hfill \ge \hfill & 25\hfill \\ \hfill 4x+10y& \hfill \le \hfill & 200\hfill \\ \hfill x& \hfill \ge \hfill & 0\hfill \\ \hfill y& \hfill \ge \hfill & 0\hfill \end{array}$

2. Since $x\ge 0$ and $y\ge 0$ (both are greater than or equal to) all solutions will be in the first quadrant. As a result, our graph shows only Quadrant I. To graph $x+y\ge 25$, graph $x+y=25$ as a solid line. Choose $\left(0,0\right)$ as a test point. Since it does not make the inequality true, shade the side that does not include the point $\left(0,0\right)$.

   To graph $4x+10y\le 200$, graph $4x+10y=200$ as a solid line. Choose $\left(0,0\right)$ as a test point. Since it does make the inequality true, shade (bottom left) the side that include the point $\left(0,0\right)$.

   Figure 5.97

   The solution of the system is the region of Figure 5.97 that is shaded the darkest. The boundary line sections that border the darkly shaded section are included in the solution as are the points on the $x$-axis from $\left(25,0\right)$ to $\left(55,0\right)$.

3. To determine if 10 small and 20 large photos would work, we look at the graph to see if the point $\left(10,20\right)$ is in the solution region. We could also test the point to see if it is a solution of both equations. It is not, so the photographer would not display 10 small and 20 large photos.
4. To determine if 20 small and 10 large photos would work, we look at the graph to see if the point $\left(20,10\right)$ is in the solution region. We could also test the point to see if it is a solution of both equations. It is, so the photographer could choose to display 20 small and 10 large photos. Notice that we could also test the possible solutions by substituting the values into each inequality.