Solve the inequality. Graph the solution set and write it using interval notation. $| 4 - 3 x | \leq 13$
A) $x \in \left( - \infty , \frac { 17 } { 3 } \right]$;
B) $x \in ( - \infty , - 3 ) \cup \left( \frac { 17 } { 3 } , \infty \right)$;
C) $x \in [ - 3 , \infty )$;
D) $x \in \left[ - 3 , \frac { 17 } { 3 } \right]$;
E) no solution;

Question

Solve the inequality. Graph the solution set and write it using interval notation. $| 4 - 3 x | \leq 13$
A) $x \in \left( - \infty , \frac { 17 } { 3 } \right]$; 
B) $x \in ( - \infty , - 3 ) \cup \left( \frac { 17 } { 3 } , \infty \right)$; 
C) $x \in [ - 3 , \infty )$; 
D) $x \in \left[ - 3 , \frac { 17 } { 3 } \right]$; 
E) no solution;

Quizplus · Accepted Answer

First, we solve the equation without the absolute value:
$$4-3x\leq 13$$
$$-3x\leq 9$$
$$x\geq -3 $$

And

$$-(4-3x)\leq 13$$
$$-4+3x\leq 13$$
$$3x\leq 17$$
$$x\leq \frac{17}{3} $$

So, the solution is $-3\leq x \leq \frac{17}{3}$, which can be written in interval notation as $\left[-3,\frac{17}{3}\right]$. Since the inequality is "less than or equal to" the absolute value, we take the "mirror image" of the solution set across the $x$-axis to account for the absolute value. This means that we also have the solution $3x-4\leq 13$, which simplifies to $3x\leq 17$, or $x\leq \frac{17}{3}$. Putting both solutions together, we get $-3\leq x\leq \frac{17}{3}$, or $\left[-3,\frac{17}{3}\right]$. Therefore, the correct choice is (D).

Solve the Inequality ∣4−3x∣≤13| 4 - 3 x | \leq 13∣4−3x∣≤13

Solve the Inequality $| 4 - 3 x | \leq 13$