Question

1. -6y + 4 = |4y + 12|

Explanation:

Step1: Consider two cases for absolute value

Case 1: \(4y + 12\geq0\) (i.e., \(y\geq - 3\)), then \(|4y + 12|=4y + 12\). The equation becomes \(-6y + 4=4y + 12\).

Step2: Solve the equation for Case 1

Subtract \(4y\) from both sides: \(-6y-4y + 4=4y-4y + 12\), so \(-10y + 4 = 12\).
Subtract 4 from both sides: \(-10y+4 - 4=12 - 4\), so \(-10y=8\).
Divide both sides by - 10: \(y=\frac{8}{-10}=-\frac{4}{5}=-0.8\). Since \(-0.8\geq - 3\), this is a valid solution.

Step3: Consider Case 2: \(4y + 12\lt0\) (i.e., \(y\lt - 3\)), then \(|4y + 12|=-(4y + 12)=-4y-12\). The equation becomes \(-6y + 4=-4y-12\).

Step4: Solve the equation for Case 2

Add \(6y\) to both sides: \(-6y + 6y+4=-4y + 6y-12\), so \(4 = 2y-12\).
Add 12 to both sides: \(4 + 12=2y-12 + 12\), so \(16 = 2y\).
Divide both sides by 2: \(y = 8\). But \(8\) is not less than \(-3\), so this solution is extraneous.

Answer:

\(y =-\frac{4}{5}\)