Absolute Value and Inequality - Examples, Exercises and Solutions

To solve the inequality $|x + 3| \leq 5$, we first consider the definition of absolute value inequality $|A| \leq B$, which is equivalent to $-B \leq A \leq B$.

Applying this definition, we have:

$-5 \leq x + 3 \leq 5$

Next, we isolate x by subtracting 3 from all parts of the inequality:

$-5 - 3 \leq x + 3 - 3 \leq 5 - 3$

This simplifies to:

$-8 \leq x \leq 2$

To solve the inequality $|2x + 1| > 7$, we split it into two separate inequalities:

$2x + 1 > 7$ or $2x + 1 < -7$.

For the first inequality $2x + 1 > 7$, subtract 1 from both sides:

$2x > 6$

Divide by 2:

$x > 3$

For the second inequality $2x + 1 < -7$, subtract 1 from both sides:

$2x < -8$

Divide by 2:

$x < -4$

Therefore, the solution is $x > 3$ or $x < -4$.

To solve the inequality $|3x - 2| \geq 4$, we separate it into:

$3x - 2 \geq 4$ or $3x - 2 \leq -4$.

For $3x - 2 \geq 4$, add 2 to both sides:

$3x \geq 6$

Divide by 3:

$x \geq 2$

For $3x - 2 \leq -4$, add 2 to both sides:

$3x \leq -2$

Divide by 3:

$x \leq -\frac{2}{3}$

Therefore, the solution is $x \geq 2$ or $x \leq -\frac{2}{3}$.

To solve \left| 2x + 1 \right| > 3 , consider the two cases for the absolute value: 2x + 1 > 3 and 2x + 1 < -3 .

1. Solving 2x + 1 > 3 :

2x + 1 > 3

Subtract 1 from both sides:

2x > 2

Divide both sides by 2:

x > 1

2. Solving 2x + 1 < -3 :

2x + 1 < -3

Subtract 1 from both sides:

2x < -4

Divide both sides by 2:

x < -2

Thus, the solution is x < -2 \text{ or } x > 1 .

To solve $\left| 3x - 4 \right| \leq 5$, we should consider two scenarios for the absolute value: $3x - 4 \leq 5$ and $3x - 4 \geq -5$.

1. Solving $3x - 4 \leq 5$:

$3x - 4 \leq 5$

Add 4 to both sides:

$3x \leq 9$

Divide both sides by 3:

$x \leq 3$

2. Solving $3x - 4 \geq -5$:

$3x - 4 \geq -5$

Add 4 to both sides:

$3x \geq -1$

Divide both sides by 3:

$x \geq -\frac{1}{3}$

Combining both results, we have $-\frac{1}{3} \leq x \leq 3$, which is the correct answer.