Question

$|2|x|-5| < 9$

Answer 1

The solution is the interval $(-7, 7)$.

Answer 2

The given inequality is $|2|x|-5| < 9$.

We use the property that for any real number $a$ and any positive number $b$, the inequality $|a| < b$ is equivalent to $-b < a < b$. In this case, $a = 2|x|-5$ and $b = 9$. Since $9 > 0$, we can apply this property: $-9 < 2|x|-5 < 9$

This is a compound inequality. We need to isolate the term involving $|x|$. Add 5 to all parts of the inequality: $-9 + 5 < 2|x|-5 + 5 < 9 + 5$ $-4 < 2|x| < 14$

Now, divide all parts of the inequality by 2: $\frac{-4}{2} < \frac{2|x|}{2} < \frac{14}{2}$ $-2 < |x| < 7$

This compound inequality is equivalent to two separate inequalities that must both be satisfied:

1. $|x| > -2$
2. $|x| < 7$

Let's solve the first inequality, $|x| > -2$. The absolute value of any real number $x$, denoted by $|x|$, is always non-negative, i.e., $|x| \ge 0$. Since $0 > -2$, the condition $|x| \ge 0$ implies that $|x|$ is always greater than $-2$. Thus, the inequality $|x| > -2$ is true for all real numbers $x$. The solution set for this inequality is $(-\infty, \infty)$.

Now, let's solve the second inequality, $|x| < 7$. For a positive number $c$, the inequality $|x| < c$ is equivalent to $-c < x < c$. Here, $c = 7$. So, $|x| < 7$ is equivalent to $-7 < x < 7$. The solution set for this inequality is the interval $(-7, 7)$.

For the original compound inequality $-2 < |x| < 7$ to hold, both $|x| > -2$ and $|x| < 7$ must hold. The solution set is the intersection of the solution sets of the two inequalities. The intersection of $(-\infty, \infty)$ and $(-7, 7)$ is $(-7, 7)$.

Therefore, the solution to the inequality $|2|x|-5| < 9$ is $-7 < x < 7$.