Solve the following equations.
(C) $0.4\mid 3y+2\mid =3.6$

• A. $y=\{14,-\frac{38}{3}\}$
• B. $y=\{\frac{7}{3},-\frac{11}{3}\}$
• C. $y=\{-\frac{7}{3},\frac{11}{3}\}$
• D. $y=\{-14,\frac{38}{3}\}$

Answer: (B)

Verified step by step guidance

1. Start with the given equation: \(0.4 \left|3y + 2\right| = 3.6\).
2. Isolate the absolute value expression by dividing both sides of the equation by 0.4: \(\left|3y + 2\right| = \frac{3.6}{0.4}\).
3. Simplify the right side to find the value inside the absolute value equals a positive number, say \(k\): \(\left|3y + 2\right| = k\).
4. Recall that if \(\left|A\right| = k\), then \(A = k\) or \(A = -k\). Apply this to get two equations: \$3y + 2 = k\( and \)3y + 2 = -k$.
5. Solve each linear equation separately for \(y\) by isolating \(y\): subtract 2 from both sides and then divide by 3.