Chapter 2.3, Problem 57E

To determine

To graph: the given inequality.

Answer to Problem 57E

  

Explanation of Solution

Given information:

  $-6\le 3x-y\le 12$

Calculation:

There are two parts of inequality, $-6\le 3x-y$ and $3x-y\le 12$ .

Make a table for the line $-6=3x-y$ by substituting the value of x and evaluating y as shown below.

$x$	$-10$	$-5$	0	5	10	15	20
$y$	$-24$	$-9$	$6$	21	36	51	66

Make a table for the line $3x-y=12$ by substituting the value of x and evaluating y as shown below.

$x$	$-10$	$-5$	0	5	10	15	20
$y$	$-42$	$-27$	$-12$	3	18	33	48

Now, substitute the test points in the given inequality and find the true test point as follows:

Test Points	$-6\le 3x-y\le 12$	True or false
$\left(0,0\right)$	$\begin{array}{l}-6\le 3\left(0\right)-\left(0\right)\le 12\\ -6\le 0-0\le 12\\ -6\le 0\le 12\end{array}$	True
$\left(0,10\right)$	$\begin{array}{l}-6\le 3\left(0\right)-\left(10\right)\le 12\\ -6\le 0-10\le 12\\ -6\le -10\le 12\end{array}$	False
$\left(0,-10\right)$	$\begin{array}{l}-6\le 3\left(0\right)-\left(-10\right)\le 12\\ -6\le 0+10\le 12\\ -6\le 10\le 12\end{array}$	True
$\left(-10,0\right)$	$\begin{array}{l}-6\le 3\left(-10\right)-\left(0\right)\le 12\\ -6\le -30\le 12\end{array}$	False
$\left(10,0\right)$	$\begin{array}{l}-6\le 3\left(10\right)-\left(0\right)\le 12\\ -6\le 30\le 12\end{array}$	False

Since both the inequalities has less than or equal to sign $(\le )$ , a solid line will be used to graph these points, and only the region of graph having true test point will be shaded as follows: