# How do you solve abs(3x) ≤ abs(2x - 5)?

Jun 2, 2018

$1 \le x \le 5$

#### Explanation:

$| 3 x | \le | 2 x - 5 |$

There are four solutions;

$1 s t$

$- 3 x \le - \left(2 x - 5\right)$

$2 n d$

$3 x \le 2 x - 5$

$3 r d$

$3 x \le - \left(2 x - 5\right)$

$4 t h$

$- 3 x \le 2 x - 5$

From the $1 s t$

$- 3 x \le - \left(2 x - 5\right)$

$- 3 x \le - 2 x + 5$

Add $2 x$ to both sides;

$- 3 x + 2 x \le - 2 x + 5 + 2 x$

$- x \le 5$

Multiply through by Minus $\left(-\right)$

$- \left(- x\right) \le - \left(5\right)$

Note: When you divide or multiplty an inequality sign by a negative value, the sign changes..

$x \ge - 5$

From the $2 n d$

$3 x \le 2 x - 5$

Subtracting both sides by $2 x$;

$3 x - 2 x \le 2 x + 5 - 2 x$

$x \le 5$

From the $3 r d$

$3 x \le - \left(2 x - 5\right)$

Removing the bracket;

$3 x \le - 2 x + 5$

Add $2 x$ to both sides;

$3 x + 2 x \le - 2 x + 5 + 2 x$

$5 x \le 5$

Dividing both sides by the coefficient of $x$;

$\frac{5 x}{5} \le \frac{5}{5}$

$\frac{\cancel{5} x}{\cancel{5}} \le \frac{\cancel{5}}{\cancel{5}}$

$x \le 5$

From the $4 t h$

$- 3 x \le 2 x - 5$

Subtracting $2 x$ from both sides;

$- 3 x - 2 x \le 2 x - 5 - 2 x$

$- 5 x \le - 5$

Dividing both sides by the coefficient of $x$;

$\frac{- 5 x}{- 5} \le \frac{- 5}{- 5}$

$\frac{\cancel{- 5} x}{\cancel{- 5}} \le \frac{\cancel{- 5}}{\cancel{- 5}}$

$x \ge 1$

Hence the possible ranges are;

$x \ge - 5$

$x \le 5$

$x \le 5$

$x \ge 1$

Therefore;

$1 \le x \le 5$