Question

learning goals from lesson 1.4 \thow i did (circle
✓ i can solve absolute value equations including those involving two absolute values. (standards: a.ced.1, a.rei.3) \ti got it! \ti’m still

1. \tsolve |x - 7| = 2.

\ta. x = -9, x = -5 \t\tc. x = 5, x = 9
\tb. x = 9 \t\t\td. no solution

1. \tsolve |20x| = |4x + 16|.

\ta. x = \frac{2}{3} \t\tb. x = -\frac{2}{3}, x = 1
\tc. x = 1 \t\t\td. no solution

Explanation:

Problem 6:

Step1: Recall absolute value definition

For \(|A| = B\) (\(B\geq0\)), \(A = B\) or \(A=-B\). So for \(|x - 7|=2\), we have two cases.

Step2: Case 1: \(x - 7 = 2\)

Solve for \(x\): \(x=2 + 7=9\).

Step3: Case 2: \(x - 7=-2\)

Solve for \(x\): \(x=-2 + 7 = 5\).
So the solutions are \(x = 5\) and \(x = 9\).

Step1: Recall absolute value equation rule

For \(|A|=|B|\), \(A = B\) or \(A=-B\). So for \(|20x|=|4x + 16|\), we have two cases.

Step2: Case 1: \(20x=4x + 16\)

Subtract \(4x\) from both sides: \(16x=16\), then \(x = 1\).

Step3: Case 2: \(20x=-(4x + 16)\)

Simplify: \(20x=-4x-16\), add \(4x\) to both sides: \(24x=-16\), divide by 24: \(x=\frac{-16}{24}=-\frac{2}{3}\).
So the solutions are \(x = -\frac{2}{3}\) and \(x = 1\).

Answer:

c. \(x = 5, x = 9\)

Problem 7: