Q39. - Algebra 1

Question

Solve each inequality. Then graph the solution set.

$\left|3d-1\right|\mathbf{\le }\mathbf{8}$

Step-by-Step Solution

Answer

The solution for the given inequality $\left|3d-1\right|\le 8$ is $\mathit{d}\in \left[-\frac{\mathbf{7}}{\mathbf{3}},\mathbf{3}\right]$.

The graph of the solution set which is is $d\in \left[-\frac{7}{3},3\right]$:

Step 1. Solve the given inequality | 3 d − 1 | ≤ 8 .

The solution of the given inequality $\left|3d-1\right|\le 8$ is:

Case 1: $3d-1$ is non-negative.

$\begin{array}{c}3d-1\le 8\\ 3d-1+1\le 8+1\\ 3d\le 9\\ \frac{3d}{3}\le \frac{9}{3}\\ d\le 3\\ d\in \left(-\infty ,3\right]\end{array}$

Case 2: $3d-1$ is negative.

$\begin{array}{c}-\left(3d-1\right)\le 8\\ \left(-1\right)\left(-\left(3d-1\right)\right)\ge \left(-1\right)\left(8\right)\\ 3d-1\ge -8\\ 3d-1+1\ge -8+1\\ 3d\ge -7\\ \frac{3d}{3}\ge \frac{-7}{3}\\ d\ge -\frac{7}{3}\\ d\in \left[-\frac{7}{3},\infty \right)\end{array}$

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 $\left|x\right|\le a$ 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 $3d-1\le 8$ and $3d-1\ge -8$ to find the solution of the inequality $\left|3d-1\right|\le 8$.

The intersection of the solutions of the inequalities $3d-1\le 8$ and $3d-1\ge -8$ is:

$d\in \left[-\frac{7}{3},3\right]$

Therefore, the solution of the inequality $\left|3d-1\right|\le 8$ is $d\in \left[-\frac{7}{3},3\right]$.

Step 2. Draw the graph of the solution set which is d ∈ [ − 7 3 , 3 ] .

The graph of the solution set which is $d\in \left[-\frac{7}{3},3\right]$ is: