Question

example: $|x|>3$
if $k$ is a real number , $k>0$, then $|u|>k$ is equivalent to $u<-k$ or $u>k$ (also holds for $\geq$)

1. solve $5|2x - 3|-1>9$
2. solve $\frac{1}{5}|3x + 3|\geq3$

Explanation:

Step1: Isolate absolute value term

$5|2x-3| - 1 > 9$
$5|2x-3| > 9 + 1$
$5|2x-3| > 10$

Step2: Simplify the inequality

$|2x-3| > \frac{10}{5}$
$|2x-3| > 2$

Step3: Split into two inequalities

$2x - 3 < -2$ or $2x - 3 > 2$

Step4: Solve first inequality

$2x < -2 + 3$
$2x < 1$
$x < \frac{1}{2}$

Step5: Solve second inequality

$2x > 2 + 3$
$2x > 5$
$x > \frac{5}{2}$

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Step1: Isolate absolute value term

$\frac{1}{5}|3x+3| \geq 3$
$|3x+3| \geq 3 \times 5$
$|3x+3| \geq 15$

Step2: Simplify the inequality

$|3(x+1)| \geq 15$
$3|x+1| \geq 15$
$|x+1| \geq 5$

Step3: Split into two inequalities

$x + 1 \leq -5$ or $x + 1 \geq 5$

Step4: Solve first inequality

$x \leq -5 - 1$
$x \leq -6$

Step5: Solve second inequality

$x \geq 5 - 1$
$x \geq 4$

Answer:

1. $x < \frac{1}{2}$ or $x > \frac{5}{2}$
2. $x \leq -6$ or $x \geq 4$