Question

Solve the following inequality:

$||2x+3|-4| \le 10$

Answer 1

$\left[-\frac{17}{2}, \frac{11}{2}\right]$

Answer 2

The given inequality is $||2x+3|-4| \le 10$.

We use the property that for any expression $A$ and a non-negative number $B$, the inequality $|A| \le B$ is equivalent to $-B \le A \le B$.

Applying this property to the given inequality, where $A = |2x+3|-4$ and $B = 10$, we get: $-10 \le |2x+3|-4 \le 10$

Now, we solve this compound inequality for $|2x+3|$. Add 4 to all parts of the inequality: $-10 + 4 \le |2x+3|-4 + 4 \le 10 + 4$ $-6 \le |2x+3| \le 14$

This compound inequality can be split into two separate inequalities:

1. $|2x+3| \ge -6$
2. $|2x+3| \le 14$

Let's solve the first inequality, $|2x+3| \ge -6$. The absolute value of any real number is always non-negative, i.e., $|y| \ge 0$ for any real number $y$. Since $0 \ge -6$, the inequality $|2x+3| \ge -6$ is true for all real values of $x$. The solution set for this inequality is $\mathbb{R}$.

Now, let's solve the second inequality, $|2x+3| \le 14$. Using the property $|A| \le B \iff -B \le A \le B$ again, where $A = 2x+3$ and $B = 14$, we get: $-14 \le 2x+3 \le 14$

Now, we solve this compound inequality for $x$. Subtract 3 from all parts: $-14 - 3 \le 2x+3 - 3 \le 14 - 3$ $-17 \le 2x \le 11$

Now, divide all parts by 2: $\frac{-17}{2} \le \frac{2x}{2} \le \frac{11}{2}$ $-\frac{17}{2} \le x \le \frac{11}{2}$

The solution set for this inequality is the interval $\left[-\frac{17}{2}, \frac{11}{2}\right]$.

The solution to the original inequality is the intersection of the solution sets of the two separate inequalities. The intersection of $\mathbb{R}$ and $\left[-\frac{17}{2}, \frac{11}{2}\right]$ is $\left[-\frac{17}{2}, \frac{11}{2}\right]$.

Therefore, the solution set for the inequality $||2x+3|-4| \le 10$ is $x \in \left[-\frac{17}{2}, \frac{11}{2}\right]$.