Q40. - Algebra 1

Question

Solve each inequality. Then graph the solution set.

$\left|\frac{4b-2}{3}\right|\mathbf{<}\mathbf{12}$

Step-by-Step Solution

Answer

The solution for the given inequality $\left|\frac{4b-2}{3}\right|<12$ is $\mathit{b}\in \left(\mathbf{-}\mathbf{8}\mathbf{.5}\mathbf{,}\mathbf{9}\mathbf{.5}\right)$.

The graph of the solution set which is $b\in \left(-8.5,9.5\right)$ is:

Step 1. Solve the given inequality | 4 b &#8722; 2 3 | &#60; 12 .

The solution of the given inequality $\left|\frac{4b-2}{3}\right|<12$ is:

Case 1: $\frac{4b-2}{3}$ is non-negative.

$\begin{array}{c}\frac{4b-2}{3}<12\\ \frac{4b-2}{3}\times 3<12\times 3\\ 4b-2<36\\ 4b-2+2<36+2\\ 4b<38\\ \frac{4b}{4}<\frac{38}{4}\\ b<9.5\\ b\in \left(-\infty ,9.5\right)\end{array}$

Case 2: $\frac{4b-2}{3}$ is negative.

$\begin{array}{c}-\left(\frac{4b-2}{3}\right)<12\\ \left(-1\right)\left(-\left(\frac{4b-2}{3}\right)\right)>\left(-1\right)\left(12\right)\\ \frac{4b-2}{3}>-12\\ \frac{4b-2}{3}\times 3>-12\times 3\\ 4b-2>-36\\ 4b-2+2>-36+2\\ 4b>-34\\ \frac{4b}{4}>\frac{-34}{4}\\ b>-8.5\\ b\in \left(-8.5,\infty \right)\end{array}$

The solution of the inequality $\left|x\right|\le a$ is $x\ge -a$ and $x\le a$.

That implies the solution of the inequality is the intersection of the solutions of the inequalities $x\ge -a$ and $x\le a$.

Find the intersection of the solutions of the inequalities $\frac{4b-2}{3}<12$ and $\frac{4b-2}{3}>-12$ to find the solution of the inequality $\left|\frac{4b-2}{3}\right|<12$.

The intersection of the solutions of the inequalities $\frac{4b-2}{3}<12$ and $\frac{4b-2}{3}>-12$ is:

$b\in \left(-8.5,9.5\right)$

Therefore, the solution of the inequality $\left|\frac{4b-2}{3}\right|<12$ is $b\in \left(-8.5,9.5\right)$.

Step 2. Draw the graph of the solution set which is b &#8712; ( &#8722; 8 .5 , 9 .5 ) .

The graph of the solution set which is $b\in \left(-8.5,9.5\right)$ is: