Absolute Value and Inequality Practice Problems with Solutions

To solve the inequality $\left|x-5\right| > 11$, we first apply the property of absolute values:

• If $\left|A\right| > B$, then $A > B$ or $A < -B$.

Therefore, for $\left|x-5\right| > 11$, we have two cases to consider:

• Case 1: $x-5 > 11$
• Case 2: $x-5 < -11$

Let's solve each case separately:

Case 1: $x-5 > 11$

Add 5 to both sides to isolate $x$:
$x > 11 + 5$

This simplifies to:

$x > 16$

Case 2: $x-5 < -11$

Add 5 to both sides to isolate $x$:
$x < -11 + 5$

This simplifies to:

$x < -6$

Thus, the solution to the inequality is:

$x > 16$ or $x < -6$

Comparing this result with the given answer choices, the correct one is:

$x>16$ o $x<-6$

Therefore, the solution to the problem is $x > 16$ or $x < -6$.

To solve the inequality $\left|3x + 9\right| < 18$, follow these steps:

• Step 1: Remove the absolute value by expressing it as a double inequality:
  $-18 < 3x + 9 < 18$.

• Step 2: Simplify the inequality:
  First, subtract 9 from all parts:
  $-18 - 9 < 3x + 9 - 9 < 18 - 9$,
  which simplifies to $-27 < 3x < 9$.

• Step 3: Solve for $x$ by dividing the entire inequality by 3:
  $-\frac{27}{3} < \frac{3x}{3} < \frac{9}{3}$,
  resulting in $-9 < x < 3$.

Upon solving, we determine that the solution to the inequality is the interval:

$-9 < x < 3$.

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