Absolute Value Inequalities Practice Problems & Solutions

Let's solve the problem:
Step 1: Recognize that the inequality we are dealing with is $\left|2x-1\right| > -10$.
Step 2: Consider the nature of absolute values: for any real $x$, $\left|2x-1\right|$ is always non-negative (i.e., $\geq 0$).
Step 3: Observe that the right side of the inequality, $-10$, is negative. Therefore, the inequality $\left|2x-1\right| > -10$ is always true because $\left|2x-1\right|$ as a non-negative quantity will always be greater than any negative number.
Step 4: Since the inequality condition always holds true, this means that the statement is valid for all $x$.

Therefore, the correct answer is that the inequality holds for all $x$.

The solution to the problem is For all x.

The absolute value expression \left| x - 5 \right| > -11 inherently suggests that for any real number $x$, the inequality holds.

Since the absolute value of any expression is always non-negative and $-11$ is negative, the condition \left| x - 5 \right| > -11 is always satisfied regardless of the choice of $x$.

Thus, there is no specific limitation or exceptional circumstance that confines $x$ to any particular subset of the real numbers.

This implies that no particular statement about $x$ being greater, less, or constrained to a specific domain can be justified. Therefore, the notion of any statement being "necessarily true" in the conventional sense of constraining $x$ does not apply.

The correct answer, therefore, is: all $x$.

To solve |2x - 5| > 7 , we consider the definition of absolute value inequality |A| > B , which means A > B or A < -B .

Thus, 2x - 5 > 7 or 2x - 5 < -7 .

Let's solve these inequalities separately:

1. 2x - 5 > 7

Add 5 to both sides:

2x > 12

Divide by 2:

x > 6

2. 2x - 5 < -7

Add 5 to both sides:

2x < -2

Divide by 2:

x < -1

Therefore, the solution is x < -1 \text{ or } x > 6 .

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.