Chapter 1.8, Problem 19E

To determine

To graph: the given inequality.

Answer to Problem 19E

  

Explanation of Solution

Given information:

  $-8\le 2x+y<6$

Calculation:

The inequality has two parts as shown below.

  $\begin{array}{l}-8\le 2x+y\\ 2x+y<6\end{array}$

Rearrange both the inequalities in terms of y as follows:

  $\begin{array}{l}-8-2x\le y\\ y<6-2x\end{array}$

Create a table by plugging the different values x in the line equation $-8-2x=y$ to evaluate the value for y.

$x$	$y$
-5	2
-3	-2
-1	-6
0	-8
1	-10
3	-14
5	-15

Since the inequality has greater than or equal to sign $(\le )$ , solid line will be used to graph a line $-8-2x=y$ .

Put the test points $\left(-5,0\right)$ and $\left(0,0\right)$ in the inequality and check the true test point as follows:

Test point $\left(x,y\right)$	Inequality, $-8-2x\le y$	True/False test point
$\left(-5,0\right)$	$\begin{array}{c}-8-2\left(-5\right)\le 0\\ -8+10\le 0\\ 2\le 0\end{array}$	False
$\left(0,0\right)$	$\begin{array}{c}-8-2\left(0\right)\le 0\\ -8\le 0\end{array}$	True

Create a table by plugging the different values x in the line equation $y=6-2x$ to evaluate the value for y.

$x$	$y$
-5	16
-3	12
-1	8
0	6
1	4
3	0
5	-4

Since the inequality has less than sign $(<)$ , dotted line will be used to graph a line $y=6-2x$ .

Put the test points $\left(3,3\right)$ and $\left(0,0\right)$ in the inequality and check the true test point as follows:

Test point $\left(x,y\right)$	Inequality, $y<6-2x$	True/False test point
$\left(3,3\right)$	$\begin{array}{l}3<6-2\left(3\right)\\ 3<6-6\\ 3<0\end{array}$	False
$\left(0,0\right)$	$\begin{array}{l}0<6-2\left(0\right)\\ 0<6\end{array}$	True

The graph for the given inequality is shown below.