Question

which two inequalities can be used to find the solution to
$3 + 2|x - 1| < 9$

$\square$ $x - 1 < 3$
$\square$ $-2(x - 1) > 3$
$\square$ $x - 1 < 9$
$\square$ $2(x - 1) < -3$
$\square$ $x - 1 > -3$
$\square$ $x - 1 > -9$

Explanation:

Step1: Isolate absolute value term

Subtract 3 from both sides:
$3 + 2|x-1| - 3 < 9 - 3$
$2|x-1| < 6$

Step2: Simplify the inequality

Divide both sides by 2:
$\frac{2|x-1|}{2} < \frac{6}{2}$
$|x-1| < 3$

Step3: Rewrite absolute value inequality

For $|A| < b$ (where $b>0$), this is equivalent to $-b < A < b$:
$-3 < x-1 < 3$
This splits into two inequalities: $x-1 < 3$ and $x-1 > -3$

Answer:

$x-1 < 3$, $x-1 > -3$