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 \((0,0)\text{.}\)