Question

warm - up: solve each equation.

1. |x - 3|+2 = 18
2. 3|9 + x|-16 = 20
3. - 20|x + 2|=-140
4. |3 + 7x|+4 = 22

Explanation:

Step1: Solve the first - equation \(|x - 3|+2 = 18\)

Isolate the absolute - value term. Subtract 2 from both sides of the equation:
\(|x - 3|=18 - 2=16\)
Then, we have two cases:
Case 1: \(x−3 = 16\), so \(x=16 + 3=19\)
Case 2: \(x−3=-16\), so \(x=-16 + 3=-13\)

Step2: Solve the second - equation \(3|9 + x|-16 = 20\)

Isolate the absolute - value term. First, add 16 to both sides:
\(3|9 + x|=20 + 16=36\)
Then divide both sides by 3: \(|9 + x| = 12\)
We have two cases:
Case 1: \(9 + x=12\), so \(x=12 - 9 = 3\)
Case 2: \(9 + x=-12\), so \(x=-12 - 9=-21\)

Step3: Solve the third - equation \(-20|x + 2|=-140\)

Isolate the absolute - value term. Divide both sides by - 20:
\(|x + 2|=\frac{-140}{-20}=7\)
We have two cases:
Case 1: \(x + 2=7\), so \(x=7 - 2 = 5\)
Case 2: \(x + 2=-7\), so \(x=-7 - 2=-9\)

Step4: Solve the fourth - equation \(|3 + 7x|+4 = 22\)

Isolate the absolute - value term. Subtract 4 from both sides:
\(|3 + 7x|=22 - 4=18\)
We have two cases:
Case 1: \(3+7x = 18\), then \(7x=18 - 3 = 15\), so \(x=\frac{15}{7}\)
Case 2: \(3+7x=-18\), then \(7x=-18 - 3=-21\), so \(x=-3\)

Answer:

For \(|x - 3|+2 = 18\), \(x = 19\) or \(x=-13\); for \(3|9 + x|-16 = 20\), \(x = 3\) or \(x=-21\); for \(-20|x + 2|=-140\), \(x = 5\) or \(x=-9\); for \(|3 + 7x|+4 = 22\), \(x=\frac{15}{7}\) or \(x=-3\)