Absolute Value and Inequality Practice Problems with Solutions

Let's solve the equation $-\left|x\right|=15$. Since the expression inside the absolute value can be either positive or negative, we consider two scenarios:

1. $-x = 15$: This implies $x = -15$.

2. $-(-x) = 15$: This simplifies to $x = 15$.

Thus, the solutions are $x = -15$ and $x = 15$.

Let's solve the equation $-\left|x\right|=8$. The expression inside the absolute value can take two forms:

1. $-x = 8$: This gives $x = -8$.

2. $-(-x) = 8$: This simplifies to $x = 8$.

Therefore, the solutions are $x = -8$ and $x = 8$.

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 $x$.

The equation is $\left|x\right|=3$, which implies that $x$ can be 3 or -3. Hence, the solutions are $x=3$ and $x=-3$.