Chapter 6, Problem 14MCQ

(a)

To determine

Solve the inequality for x.

Answer to Problem 14MCQ

Solution: $x>-\frac{4}{a}\text{and}x\le \frac{4}{a}$

Explanation of Solution

Given:

The inequality equation: $-3<ax+1\le 5$

Concept Used:

The inequality equation: $-3<ax+1\le 5$

Step1: subtract 1 from each − 3 and 5 $\Rightarrow -3-1<ax+1-1\le 5-1$

Step 2: Simplify: $-4<ax\le 4$

Step 3: Divide each side by a : $\frac{-4}{a}<\frac{ax}{a}\le \frac{4}{a}$

Step 4: Simplify.

Calculation:

The inequality equation: $-3<ax+1\le 5$

Solve for x:

  $\begin{array}{l}-3<ax+1\le 5\\ -3-1<ax+1-1\le 5-1\text{[}\text{Subtract}1\text{from}\text{both}\text{sides}\text{]}\\ -4<ax\le 4\\ \frac{-4}{a}<\frac{ax}{a}\le \frac{4}{a}\\ -\frac{4}{a}<x\le \frac{4}{a}\end{array}$ Solution: $x>-\frac{4}{a}\text{and}x\le \frac{4}{a}$

Thus, solution of the equation $-3<ax+1\le 5$ is $x>-\frac{4}{a}\text{and}x\le \frac{4}{a}$

(b)

To determine

Solve the inequality.

Answer to Problem 14MCQ

  $x<-\frac{6}{a}\text{or}x>-5a$

Explanation of Solution

Given:

The inequality equation: $-\frac{1}{a}x+6<1\text{or}2-ax>8$

Concept Used:

The inequality equation: $-\frac{1}{a}x+6<1\text{or}2-ax>8$

Solve the inequalities simultaneously.

Calculation:

The inequality equation: $-\frac{1}{a}x+6<1\text{or}2-ax>8$

Solve for x:

  $\begin{array}{l}-\frac{1}{a}x+6<1\text{or}2-ax>8\\ -\frac{1}{a}x+6-6<1-6\text{or}2-2-ax>8-2\\ -\frac{1}{a}x<-5\text{or}-ax>6\\ -a·\frac{1}{a}x>-a·-5\text{or}\frac{-ax}{-a}<\frac{6}{-a}\\ x>-5a\text{or}x<-\frac{6}{a}\end{array}$ Solution: $x<-\frac{6}{a}\text{or}x>-5a$

Thus, solution: $x<-\frac{6}{a}\text{or}x>-5a$