Question

4 ) solve and graph the inequality
4|x - 5| + 1 ≤ 21
isolate the absolute
value expression
set up two inequalities to solve

Explanation:

Step1: Isolate the absolute value expression

Subtract \(1\) from both sides of the inequality \(4|x - 5|+1\leq21\).

$$
LATEXBLOCK0
$$

Divide both sides by \(4\).

$$
LATEXBLOCK1
$$

Step2: Set up two inequalities

For \(|u|\leq a\) (\(a\geq0\)), the solution is \(-a\leq u\leq a\). Here \(u = x - 5\) and \(a = 5\).
So we have \(-5\leq x - 5\leq5\).
Add \(5\) to all parts of the compound - inequality.
For the left - hand side: \(-5+5\leq x-5 + 5\), which gives \(0\leq x\).
For the right - hand side: \(x-5 + 5\leq5 + 5\), which gives \(x\leq10\).

Answer:

The solution of the inequality \(4|x - 5|+1\leq21\) is \(0\leq x\leq10\). On the number - line, we mark closed circles at \(x = 0\) and \(x = 10\) (since the inequality includes equality) and shade the region between them.