Question

algebra 2a semester online practice
complete this assessment to review what you’ve learned. it will not a
solve the absolute value equation $6 - 2|3x + 2| - 9 = -11$. (1 point)
\\(\boldsymbol{x = -2}\\) and \\(\boldsymbol{x = \frac{2}{3}}\\)
\\(\boldsymbol{x = \frac{2}{3}}\\)
\\(\boldsymbol{x = -5}\\) and \\(\boldsymbol{x = \frac{11}{3}}\\)
no solution

Explanation:

Step1: Simplify the equation

First, combine the constant terms on the left side. The equation is \(6 - 2|3x + 2| - 9=-11\). Combine \(6\) and \(-9\): \(6 - 9=-3\), so the equation becomes \(-3 - 2|3x + 2|=-11\).

Step2: Isolate the absolute value term

Add \(3\) to both sides of the equation: \(-2|3x + 2|=-11 + 3\). Calculate the right side: \(-11 + 3=-8\), so we have \(-2|3x + 2|=-8\). Then divide both sides by \(-2\): \(|3x + 2|=\frac{-8}{-2}=4\).

Step3: Solve the absolute value equation

The absolute value equation \(|3x + 2| = 4\) means two cases:

• Case 1: \(3x + 2 = 4\)

Subtract \(2\) from both sides: \(3x=4 - 2 = 2\). Divide by \(3\): \(x=\frac{2}{3}\).

• Case 2: \(3x + 2=-4\)

Subtract \(2\) from both sides: \(3x=-4 - 2=-6\). Divide by \(3\): \(x=\frac{-6}{3}=-2\).

Answer:

\(x = -2\) and \(x=\frac{2}{3}\) (the first option: \(x = -2\) and \(x=\frac{2}{3}\))