Question

a.
ve the absolute value e
$-2|x + 1| + 3 > 1$
. for which value of $k$
lve each equation.

1. $-3|5x - 2| - 1 = -13$

olve each inequality and

Explanation:

Step1: Isolate absolute value term

Subtract 3 from both sides:
$-2|x+1| > 1 - 3$
$-2|x+1| > -2$

Step2: Divide by negative (flip inequality)

Divide both sides by -2:
$|x+1| < 1$

Step3: Rewrite without absolute value

For $|A| < b$, $-b < A < b$:
$-1 < x+1 < 1$

Step4: Solve for x

Subtract 1 from all parts:
$-1 -1 < x < 1 -1$
$-2 < x < 0$

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Step1: Isolate absolute value term

Add 1 to both sides:
$-3|5x-2| = -13 + 1$
$-3|5x-2| = -12$

Step2: Divide to isolate absolute value

Divide both sides by -3:
$|5x-2| = 4$

Step3: Split into two cases

Case 1: $5x-2 = 4$
Case 2: $5x-2 = -4$

Step4: Solve Case 1

Add 2, divide by 5:
$5x = 4 + 2$
$5x = 6$
$x = \frac{6}{5}$

Step5: Solve Case 2

Add 2, divide by 5:
$5x = -4 + 2$
$5x = -2$
$x = -\frac{2}{5}$

Answer:

For $-2|x+1|+3>1$: $-2 < x < 0$
For $-3|5x-2|-1=-13$: $x = \frac{6}{5}$ and $x = -\frac{2}{5}$