5.5 Graphing Linear Equations and Inequalities

The first number of the coordinate pair is the $x$-coordinate, and the second number is the $y$-coordinate. To plot each point, sketch a vertical line through the $x$-coordinate and a horizontal line through the $y$-coordinate (Figure 5.23). Their intersection is the point.

1. Since $x=-5$, the point is to the left of the $y$-axis. Also, since $y=4$, the point is above the $x$-axis. The point $\left(-5,4\right)$ is in quadrant II.
2. Since $x=-3$, the point is to the left of the $y$-axis. Also, since $y=-4$, the point is below the $x$-axis. The point $\left(-3,-4\right)$ is in quadrant III.
3. Since $x=2$, the point is to the right of the $y$-axis. Since $y=-3$, the point is below the $x$-axis. The point $\left(2,-3\right)$ is in quadrant IV.
4. Since $x=0$, the point whose coordinates are $\left(0,-1\right)$ is on the $y$-axis.
5. Since $x=3$, the point is to the right of the $y$-axis. Since $y=\frac{5}{2}$, which is equal to 2.5, the point is above the $x$-axis. The point $\left(3,\frac{5}{2}\right)$ is in quadrant I.

Figure 5.23

Substitute the $x$- and $y$-values into the equation to check if the ordered pair is a solution to the equation.

1. 
   
   $$\begin{array}{c}\mathrm{A}:(0,-3)\\ \begin{array}{ccc}\hfill y& \hfill =\hfill & 2x-3\hfill \\ \hfill -3& \hfill \stackrel{?}{=}\hfill & 2(0)-3\hfill \\ \hfill -3& \hfill =\hfill & -3✓\hfill \end{array}\\ (0,-3)\text{is a solution.}\end{array}\begin{array}{c}\mathrm{B}:(3,3)\\ \begin{array}{ccc}\hfill y& \hfill =\hfill & 2x-3\hfill \\ \hfill 3& \hfill \stackrel{?}{=}\hfill & 2(3)-3\hfill \\ \hfill 3& \hfill =\hfill & 3✓\hfill \end{array}\\ (3,3)\text{is a solution.}\end{array}\begin{array}{c}\mathrm{C}:(2,-3)\\ \begin{array}{ccc}\hfill y& \hfill =\hfill & 2x-3\hfill \\ \hfill -3& \hfill \stackrel{?}{=}\hfill & 2(2)-3\hfill \\ \hfill -3& \hfill \ne \hfill & 1\hfill \end{array}\\ (2,-3)\text{is not a solution.}\end{array}\begin{array}{c}\mathrm{D}:(-1,-5)\\ \begin{array}{ccc}\hfill y& \hfill =\hfill & 2x-3\hfill \\ \hfill -5& \hfill \stackrel{?}{=}\hfill & 2(-1)-3\hfill \\ \hfill -5& \hfill =\hfill & -5✓\hfill \end{array}\\ (-1,-5)\text{is a solution.}\end{array}$$
2. Plot the points $(0,-3)$, $(3,3)$, $(2,-3)$, and $(-1,-5)$.

Figure 5.25

In Figure 5.25, the points $(0,3)$, $(3,-3)$, and $(-1,-5)$ are on the line $y=2x-3$, and the point $(2,-3)$ is not on the line. The points that are solutions to $y=2x-3$ are on the line, but the point that is not a solution is not on the line.

Find three points that are solutions to the equation. Since this equation has the fraction $\frac{1}{2}$ as a coefficient of $x$, we will choose values of $x$ carefully. We will use zero as one choice and multiples of 2 for the other choices. Why are multiples of two a good choice for values of $x$? By choosing multiples of 2, the multiplication by $\frac{1}{2}$ simplifies to a whole number.

$\begin{array}{lll}x=0\hfill & x=2\hfill & x=4\hfill \\ y=\frac{1}{2}x+3\hfill & y=\frac{1}{2}x+3\hfill & y=\frac{1}{2}x+3\hfill \\ y=\frac{1}{2}(0)+3\hfill & y=\frac{1}{2}(2)+3\hfill & y=\frac{1}{2}(4)+3\hfill \\ y=0+3\hfill & y=1+3\hfill & y=2+3\hfill \\ y=3\hfill & y=4\hfill & y=5\hfill \end{array}$

Plot the points, check that they line up, and draw the line (Figure 5.26).

Figure 5.26

Let $x=\text{the number of gallons of gasoline}$ and let $y=\text{the total cost}$. If gas is $3.53 per gallon, then $y=3.53x$. The two points given are (10, 35.30) and (15, 52.95). Plot the points, check that they line up, and draw the line (Figure 5.27).

Figure 5.27

We can see the point at $x=5$. The $y$-value is found by multiplying 5 by $3.53 to get $17.65. Your neighbor will pay $17.65.

1. 
   $\begin{array}{ll}(0,0)& y>x+4\\ \text{Substitute}0\text{for}x\text{and}0\text{for}y& 0\stackrel{?}{>}0+4\\ \text{Simplify}\text{.}& 0\cancel{>}4\\ & (0,0)\text{is not a solution to}y>x+4.\end{array}$
2. 
   $\begin{array}{ll}(1,6)& y>x+4\\ \text{Substitute}1\text{for}x\text{and}6\text{for}y& 6\stackrel{?}{>}1+4\\ \text{Simplify}\text{.}& 6>5\\ & (1,6)\text{is a solution to}y>x+4.\end{array}$
3. 
   $\begin{array}{ll}(2,6)& y>x+4\\ \text{Substitute}2\text{for}x\text{and}6\text{for}y& 6\stackrel{?}{>}2+4\\ \text{Simplify}\text{.}& 6\cancel{>}6\\ & (2,6)\text{is not a solution to}y>x+4.\end{array}$
4. 
   $\begin{array}{ll}(-5,-15)& y>x+4\\ \text{Substitute}-5\text{for}x\text{and}-15\text{for}y& -15\stackrel{?}{>}-5+4\\ \text{Simplify}\text{.}& -15\cancel{>}-1\\ & \text{So},(-5,-15)\text{is not a solution to}y>x+4.\end{array}$
5. 
   $\begin{array}{ll}(-8,12)& y>x+4\\ \text{Substitute}-8\text{for}x\text{and}12\text{for}y& 12\stackrel{?}{>}-8+4\\ \text{Simplify}\text{.}& 12>-4\\ & \text{So},(-8,12)\text{is a solution to}y>x+4.\end{array}$

The line $y=2x-1$ is the boundary line. On one side of the line are the points with $y>2x-1$ and on the other side of the line are the points with $y<2x-1$. Let us test the point $(0,0)$ and see which inequality describes its position relative to the boundary line. At $(0,0)$, which inequality is true: $y>2x-1$ or $y<2x-1$?

Since $y>2x-1$ is true, the side of the line with $(0,0)$, is the solution. The shaded region shows the solution of the inequality $y>2x-1$. Since the boundary line is graphed with a dashed line, the inequality does not include the equal sign. The graph shows the inequality $y>2x-1$.

We could use an $y$ point as a test point, provided it is not on the line. Why did we choose $(0,0)$? Because it is the easiest to evaluate. You may want to pick a point on the other side of the boundary line and check that $y<2x-1$.

1. Let $x$ be the number of hours she works at the job in food service and let $y$ be the number of hours she works tutoring. She earns $10 per hour at the job in food service and $15 an hour tutoring. At each job, the number of hours multiplied by the hourly wage will give the amount earned at that job.
   $$\underset{\text{10}x}{\underbrace{\text{Amount earned at the food service job}}}\underset{\text{+}}{\text{plus}}\underset{\text{15}y}{\underbrace{\text{the amount earned tutoring}}}\underset{\ge \text{240}}{\underbrace{\text{is at least}}}$$
2. Graph the inequality:

   Step 1: Graph the boundary line $10x+15y=240$

   Create a table of values

   $x$	$y$
   0	$10\left(0\right)+15y=240->y=16$
   6	$10\left(6\right)+15y=240->y=12$
   12	$10\left(12\right)+15y=240->y=8$

   Step 2: Pick a test point. Let us pick $\left(0,0\right)$ again:

   $10\left(0\right)+15\left(0\right)\ge 240$?

   $0\ge 240$ is false and not a solution so the shading happens on the other side of the boundary line (Figure 5.38).

   Figure 5.38
3. From the graph, we see that the ordered pairs $\left(15,10\right)$, $\left(0,16\right)$, $\left(24,0\right)$ represent three of infinitely many solutions. Check the values in the inequality.
   $$\begin{array}{c}(15,10)\\ \begin{array}{ccc}\hfill 10x+15y& \hfill \ge \hfill & 240\hfill \\ \hfill 10(15)+15(10)& \hfill \stackrel{?}{\ge }\hfill & 240\hfill \\ \hfill 300& \hfill \ge \hfill & 240\text{True}\hfill \end{array}\end{array}\begin{array}{c}(0,16)\\ \begin{array}{ccc}\hfill 10x+15y& \hfill \ge \hfill & 240\hfill \\ \hfill 10(0)+15(16)& \hfill \stackrel{?}{\ge }\hfill & 240\hfill \\ \hfill 240& \hfill \ge \hfill & 240\text{True}\hfill \end{array}\end{array}\begin{array}{c}(24,0)\\ \begin{array}{ccc}\hfill 10x+15y& \hfill \ge \hfill & 240\hfill \\ \hfill 10(24)+15(0)& \hfill \stackrel{?}{\ge }\hfill & 240\hfill \\ \hfill 240& \hfill \ge \hfill & 240\text{True}\hfill \end{array}\end{array}$$

For Hilaria, it means that to earn at least $240, she can work 15 hours tutoring and 10 hours at her food service job, earn all her money tutoring for 16 hours, or earn all her money while working 24 hours at the job in food service.