Question

question
solve the following inequality algebraically.
4|x + 5| + 6 ≤ 46

Explanation:

Step1: Subtract 6 from both sides

To isolate the absolute - value term, we subtract 6 from both sides of the inequality \(4|x + 5|+6\leq46\).
\(4|x + 5|+6 - 6\leq46 - 6\)
\(4|x + 5|\leq40\)

Step2: Divide both sides by 4

We divide both sides of the inequality \(4|x + 5|\leq40\) by 4 to get the absolute - value expression by itself.
\(\frac{4|x + 5|}{4}\leq\frac{40}{4}\)
\(|x + 5|\leq10\)

Step3: Solve the compound inequality

The inequality \(|x + 5|\leq10\) is equivalent to the compound inequality \(- 10\leq x + 5\leq10\) (by the definition of absolute value: if \(|a|\leq b\) (\(b\geq0\)), then \(-b\leq a\leq b\)).

Now, we solve the left - hand side of the compound inequality \(-10\leq x + 5\):
Subtract 5 from both sides: \(-10-5\leq x+5 - 5\), so \(-15\leq x\).

Next, we solve the right - hand side of the compound inequality \(x + 5\leq10\):
Subtract 5 from both sides: \(x+5 - 5\leq10 - 5\), so \(x\leq5\).

Answer:

\(-15\leq x\leq5\)