Question

solve and graph. provide answers accurate to the nearest tenth. |-3x + 1| ≥ 8 solution

Explanation:

Step1: Apply absolute - value inequality rule

If \(|u|\geq a\) (\(a\gt0\)), then \(u\leq - a\) or \(u\geq a\). Here \(u=-3x + 1\) and \(a = 8\). So we have two inequalities: \(-3x+1\leq - 8\) or \(-3x + 1\geq8\).

Step2: Solve \(-3x+1\leq - 8\)

Subtract \(1\) from both sides: \(-3x\leq-8 - 1=-9\).
Divide both sides by \(-3\). When dividing by a negative number, the inequality sign flips. So \(x\geq\frac{-9}{-3}=3\).

Step3: Solve \(-3x + 1\geq8\)

Subtract \(1\) from both sides: \(-3x\geq8 - 1 = 7\).
Divide both sides by \(-3\). The inequality sign flips. So \(x\leq\frac{7}{-3}\approx - 2.3\).

Answer:

The solution of the inequality \(| - 3x+1|\geq8\) is \(x\leq - 2.3\) or \(x\geq3\).

To graph:

• Draw a number line.
• Put a closed circle (because the inequality includes equality) at \(x=-2.3\) and shade to the left of \(x =-2.3\).
• Put a closed circle at \(x = 3\) and shade to the right of \(x=3\).