Question

solve the absolute value inequality $2|x + 11| + 17 > 3$. provide the answer using interval notation. (1 point)
the solution to $2|x + 11| + 17 > 3$ is (\\(\square\\)).

Explanation:

Step1: Isolate the absolute value term

Subtract 17 from both sides of the inequality \(2|x + 11| + 17>3\).
\(2|x + 11|+17 - 17>3 - 17\)
Simplify to get \(2|x + 11|>- 14\).

Step2: Divide by the coefficient of the absolute value

Divide both sides by 2: \(\frac{2|x + 11|}{2}>\frac{-14}{2}\)
Simplify to \(|x + 11|>-7\).

Step3: Analyze the absolute value inequality

Recall that the absolute value of any real number is always non - negative (i.e., \(|a|\geq0\) for all real numbers \(a\)). So, for any real number \(x\), \(|x + 11|\) is either 0 or a positive number. Since 0 and all positive numbers are greater than - 7, the inequality \(|x + 11|>-7\) is true for all real numbers \(x\).

In interval notation, the set of all real numbers is \((-\infty,\infty)\).

Answer:

\((-\infty,\infty)\)