Inequalities with Absolute Values: Solving the problem

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.

To solve this inequality, we use the principle that if $|A| > B$, then $A > B$ or $A < -B$. Here's a detailed step-by-step solution:

• Set up the two inequalities from the absolute value expression:
  1. $4x + 12 > 16$
  2. $4x + 12 < -16$
• For the first inequality, $4x + 12 > 16$:
  - Subtract 12 from both sides to isolate the term with $x$:
  $4x > 16 - 12$
  $4x > 4$
  - Divide both sides by 4:
  $x > 1$
• For the second inequality, $4x + 12 < -16$:
  - Subtract 12 from both sides:
  $4x < -16 - 12$
  $4x < -28$
  - Divide both sides by 4:
  $x < -7$
• The solution is the union of both results:
  $x > 1$ or $x < -7$

Therefore, the correct answer among the given choices is $x > 1$ or $x < -7$.

To solve the inequality $\left|5x-10\right| > 15$, we rewrite it as two separate inequalities:

• First inequality: $5x - 10 > 15$
• Second inequality: $5x - 10 < -15$

Let's solve each one:

For the first inequality $5x - 10 > 15$:
Add 10 to both sides: $5x > 25$
Divide both sides by 5: $x > 5$

For the second inequality $5x - 10 < -15$:
Add 10 to both sides: $5x < -5$
Divide both sides by 5: $x < -1$

Combining these solutions, we have:
$x > 5$ or $x < -1$

Therefore, the correct statement regarding the solution set is: $x > 5$ or $x < -1$.

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 the absolute value inequality $|2x - 4| < 8$, we begin by removing the absolute value expression. This gives us a compound inequality:

$-8 < 2x - 4 < 8$.

We will solve this compound inequality by handling each part separately:

• Start with the left inequality: $-8 < 2x - 4$.
• Add 4 to both sides to isolate the term with $x$: $-8 + 4 < 2x$.
• Simplify: $-4 < 2x$.
• Finally, divide both sides by 2: $-2 < x$.

• Now, solve the right inequality: $2x - 4 < 8$.
• Add 4 to both sides to isolate the term with $x$: $2x - 4 + 4 < 8 + 4$.
• Simplify: $2x < 12$.
• Finally, divide both sides by 2: $x < 6$.

Combining the two solutions from the parts, we find:

$-2 < x < 6$.

The solution indicates that $x$ must be greater than -2 and less than 6. This form matches answer choice 4. Therefore, the correct solution is:

$-2 < x < 6$.

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 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 $\left|x + 4\right| > 13$, we use the property of absolute values, which says that for $\left|a\right| > b$, it implies $a > b$ or $a < -b$.

Applying this to our problem, we have:

• $x + 4 > 13$ or $x + 4 < -13$.

Now, let's solve each inequality separately:

First inequality: $x + 4 > 13$

Subtract 4 from both sides to isolate $x$:

$x > 13 - 4$

$x > 9$

Second inequality: $x + 4 < -13$

Subtract 4 from both sides to isolate $x$:

$x < -13 - 4$

$x < -17$

Therefore, the solution to the inequality $\left|x + 4\right| > 13$ is $x > 9$ or $x < -17$.

The correct answer choice is:

• $x > 9$ or $x < -17$.

To solve the inequality $|x - 4| < 8$, we will break it down into two separate inequalities.

• First, recognize that $|x - 4| < 8$ means the expression $x - 4$ can vary between -8 and 8 without violating the inequality constraint.
• This gives us two inequalities to solve: $-8 < x - 4$ and $x - 4 < 8$.

Let's solve each inequality:
1. For $-8 < x - 4$:
- Add 4 to both sides to isolate $x$:
$-8 + 4 < x \rightarrow -4 < x$ 2. For $x - 4 < 8$:
- Add 4 to both sides to isolate $x$:
$x < 8 + 4 \rightarrow x < 12$

By combining these results, we obtain the solution:
$-4 < x < 12$

Therefore, the range of $x$ that satisfies the inequality $|x - 4| < 8$ is $-4 < x < 12$.

Hence, the correct statement from the given choices is $\boxed{-4 < x < 12}$.

To solve the inequality $|x + 2| < 3$, we will apply the property of absolute values by rewriting it without the absolute value sign as follows:

Step 1: Transform the absolute value inequality
Using the rule $|A| < B$ implies $-B < A < B$, we write

$-3 < x + 2 < 3$.

Step 2: Solve this compound inequality. We do this by isolating $x$ as follows:

• Subtract 2 from all parts: $-3 - 2 < x + 2 - 2 < 3 - 2$.
• This simplifies to: $-5 < x < 1$.

Thus, the inequality $|x + 2| < 3$ is solved as $-5 < x < 1$.

The correct solution is contained in choice 3: $-5 < x < 1$.

To solve the inequality $|x-3| \leq 5$, we need to consider the definition of absolute value inequalities. The inequality $|a| \leq b$ translates to $-b \leq a \leq b$.

Applying this to our expression $|x-3| \leq 5$, we have:

$-5 \leq x-3 \leq 5$.

We add 3 to all parts of the inequality to isolate $x$:

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

This simplifies to $-2 \leq x \leq 8$.

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 $|x+5| < 2$, apply the property of absolute values which states that $|a| < b$ translates to $-b < a < b$.

Therefore, $-2 < x+5 < 2$.

Subtract 5 from all parts of the inequality to isolate $x$:

$-2 - 5 < x+5 - 5 < 2 - 5$

This simplifies to $-7 < x < -3$.

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 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 $|2x - 5| > 7$, 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 $\left| 3x - 4 \right| \leq 5$, 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.

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 $\left| 5x + 3 \right| \leq 7$, consider both cases:$5x + 3 \leq 7$ and $5x + 3 \geq -7$.

1. Solving $5x + 3 \leq 7$:

$5x + 3 \leq 7$

Subtract 3 from both sides:

$5x \leq 4$

Divide both sides by 5:

$x \leq 0.8$

2. Solving $5x + 3 \geq -7$:

$5x + 3 \geq -7$

Subtract 3 from both sides:

$5x \geq -10$

Divide both sides by 5:

$x \geq -2$

Combining both results, we find $-2 \leq x \leq 0.8$.

Let's solve the inequality $\left|3x - 1\right| < 5x$:

1. Remove the absolute value by considering two cases:

   1. $3x - 1 < 5x$
      Solve for $x$:

   2. $3x - 1 < 5x$

      Subtract $3x$ from both sides:
      $-1 < 2x$

      Divide both sides by $2$:
      $-\frac{1}{2} < x$

      Rewriting this inequality yields: $x > -\frac{1}{2}$

   3. $-(3x - 1) < 5x$
      Simplifying yields:

   4. $-3x + 1 < 5x$

      Add $3x$ to both sides:
      $1 < 8x$

      Divide both sides by $8$:
      $\frac{1}{8} < x$

      This implies $x > \frac{1}{8}$

2. Combine results:

3. Both conditions imply $x > \frac{1}{8}$. Thus, the solution is $x > \frac{1}{8}$.