Question

1. solve the compound inequality and graph the solution set on the number line provided. -7 < -3r - 4 ≤ 8
2. consider the following mathematical statement: “four times the difference of z and 2 is at most 16.

a) write an inequality to model the statement.
b) solve the inequality for z. show all of your work.

1. consider the following mathematical statement: “twenty less than three times a number x is less than or equal to twice the same number increased by 5.”

a) write an inequality to model the statement.
b) solve the inequality for x. show all of your work.

Explanation:

8.

Step1: Add 4 to all parts

Add 4 to each part of the compound - inequality $-7 < - 3r-4\leqslant8$.
$-7 + 4< - 3r-4 + 4\leqslant8 + 4$
$-3 < - 3r\leqslant12$

Step2: Divide all parts by - 3

When dividing an inequality by a negative number, the direction of the inequality signs changes.
$\frac{-3}{-3}>\frac{-3r}{-3}\geqslant\frac{12}{-3}$
$1 > r\geqslant - 4$

To graph on the number - line: Draw a number line. Mark a open circle at 1 (because $r$ is less than 1, not equal to 1) and a closed circle at - 4 (because $r$ is greater than or equal to - 4). Then shade the region between - 4 and 1.

Step1: Distribute 4

Use the distributive property $a(b - c)=ab - ac$. Here, $a = 4$, $b = z$, and $c = 2$.
$4(z - 2)=4z-8$, so the inequality becomes $4z-8\leqslant16$.

Step2: Add 8 to both sides

$4z-8 + 8\leqslant16 + 8$
$4z\leqslant24$

Step3: Divide both sides by 4

$\frac{4z}{4}\leqslant\frac{24}{4}$
$z\leqslant6$

Step1: Subtract 2x from both sides

$3x-20-2x\leqslant2x + 5-2x$
$x-20\leqslant5$

Step2: Add 20 to both sides

$x-20 + 20\leqslant5 + 20$
$x\leqslant25$

Answer:

$-4\leqslant r<1$

9.

a)

The difference of $z$ and 2 is $z - 2$. Four times this difference is $4(z - 2)$. The phrase "is at most 16" means $\leqslant16$. So the inequality is $4(z - 2)\leqslant16$.

b)