ORCCAOpen Resources for Community College Algebra

Example4.10.2Office Supplies

Michael has a budget of $\$133.00$ to purchase some staplers and markers for the office supply closet. Each stapler costs $\$19.00\text{,}$ and each marker costs $\$1.75\text{.}$ If we use variable names so that he will purchase $x$ staplers and $y$ markers. Write and plot a linear inequality to model the relationship between the number of staplers and markers Michael can purchase. Keep in mind that Michael might not spend all of the $\$133.00\text{.}$

The cost of buying $x$ staplers would be $19x$ dollars. Similarly, the cost of buying $y$ markers would be $1.75y$ dollars. Since whatever Michael spends needs to be no more than $133$ dollars, we have the inequality

This is a standard-form inequality, similar to Equation (4.7.1). Next, let's graph it.

The first method to graph the inequality is to graph the corresponding equation, $19x+1.75y=133\text{.}$ Its $x$- and $y$-intercepts can be found this way:

So the intercepts are $(7,0)$ and $(0,76)\text{,}$ and we can plot the line in Figure 4.10.3.

The points on this line represent ways in which Michael can spend exactly all of the $\$133\text{.}$ But what does a point like $(2,40)$ in Figure 4.10.4, which is not on the line, mean in this context? That would mean Michael bought $2$ staplers and $40$ markers, spending $19\cdot2+1.75\cdot40=108$ dollars. That is within Michael's budget.

In fact, any point on the lower left side of this line represents a total purchase within Michael's budget. The shading in Figure 4.10.5 captures all solutions to $19x+1.75y\leq133\text{.}$ Some of those solutions have negative $x$- and $y$-values, which make no sense in context. So in Figure 4.10.6, we restrict the shading to solutions which make physical sense.

Example4.10.7

Is the point $(1,2)$ a solution of $y\gt2x+1\text{?}$

In the inequality $y\gt2x+1\text{,}$ substitute $x$ with $1$ and $y$ with $2\text{,}$ and we will see whether the inequality is true:

Since $2\gt5$ is not true, $(1,2)$ is not a solution of $y\gt2x+1\text{.}$

Example4.10.8

Graph $y\gt2x+1\text{.}$

There are two steps to graphing this linear inequality in two variables.

1. Graph the line $y=2x+1\text{.}$ Because the inequality symbol is $\gt$ (instead of $\ge$), the line should be dashed (instead of solid).
2. Next, we need to decide whether to shade the region above $y=2x+1$ or below it. We will choose a point to test whether $y\gt2x+1$ is true. As long as the line doesn't cross $(0,0)\text{,}$ we will use $(0,0)$ to test because the number $0$ is the easiest number for calculation.
   $$\begin{align*} y\amp\gt2x+1\\ 0\amp\stackrel{?}{\gt}2(0)+1\\ 0\amp\stackrel{no}{\gt}1 \end{align*}$$
   Because $0\gt1$ is not true, the point $(0,0)$ is not a solution and should not be shaded. As a result, we shade the region without $(0,0)\text{.}$

Example4.10.11

Graph $y\leq -\frac{5}{3}x+2\text{.}$

There are two steps to graphing this linear inequality in two variables.

1. Graph the line $y= -\frac{5}{3}x+2\text{.}$ Because the inequality symbol is $\leq$ (instead of $\lt$), the line should be solid.
2. Next, we need to decide whether to shade the region above $y= -\frac{5}{3}x+2$ or below it. We will choose a point to test whether $y\leq -\frac{5}{3}x+2$ is true there. Using $(0,0)$ as a test point:
   $$\begin{align*} y\amp\leq -\frac{5}{3}x+2\\ 0\amp\stackrel{?}{\leq}-\frac{5}{3}(0)+2\\ 0\amp\stackrel{\checkmark}{\leq}2 \end{align*}$$
   Because $0\leq2$ is true, the point $(0,0)$ is a solution. As a result, we shade the region with $(0,0)\text{.}$