Solve the Absolute Value Inequality: |2x + 4| > 3x + 1

Given:

$\left|2x + 4\right| > 3x + 1$

Which of the following statements is necessarily true?

Firstly, let's analyze the inequality: $\left|2x + 4\right| > 3x + 1$ .

We split it into two separate inequalities:

(1) $2x + 4 > 3x + 1$ and (2) $2x + 4 < -(3x + 1)$ .

For inequality (1):

$2x + 4 > 3x + 1$

Subtract $3x$ from both sides:

$2x + 4 - 3x > 1$

$-x + 4 > 1$

Subtract 4 from both sides:

$-x > -3$

Divide both sides by $-1$ and flip the inequality:

$x < 3$

For inequality (2):

$2x + 4 < -3x - 1$

Add $3x$ to both sides:

$5x + 4 < -1$

Subtract 4 from both sides:

$5x < -5$

Divide both sides by 5:

$x < -1$

When we combine the solutions, we see that the answer is $x < 3$

$x < 3$

Question