Question

what are the solutions of the following equation?
$6|x - 2| - 15 = -9|x - 2| + 15$
choose 1 answer:
a only $x = 4$
b $x = -4$ or $x = 2$
c $x = 4$ or $x = 0$
d only $x = 0$
e there are no solutions

Explanation:

Step1: Let \( y = |x - 2| \)

The equation \( 6|x - 2| - 15 = -9|x - 2| + 15 \) becomes \( 6y - 15 = -9y + 15 \).

Step2: Solve for \( y \)

Add \( 9y \) to both sides: \( 6y + 9y - 15 = 15 \), which simplifies to \( 15y - 15 = 15 \).
Add 15 to both sides: \( 15y = 15 + 15 = 30 \).
Divide by 15: \( y = \frac{30}{15} = 2 \).

Step3: Substitute back \( y = |x - 2| \)

We have \( |x - 2| = 2 \).
This gives two cases:
Case 1: \( x - 2 = 2 \) → \( x = 4 \).
Case 2: \( x - 2 = -2 \) → \( x = 0 \).

Wait, but let's check the options. Option C is \( x = 4 \) or \( x = 0 \), which matches our solution. Wait, but let's re-examine the original equation. Wait, maybe I made a mistake. Wait, original equation: \( 6|x - 2| - 15 = -9|x - 2| + 15 \). Let's verify with \( x = 4 \):
Left: \( 6|4 - 2| - 15 = 6*2 - 15 = 12 - 15 = -3 \).
Right: \( -9|4 - 2| + 15 = -9*2 + 15 = -18 + 15 = -3 \). So \( x = 4 \) works.
For \( x = 0 \):
Left: \( 6|0 - 2| - 15 = 6*2 - 15 = 12 - 15 = -3 \).
Right: \( -9|0 - 2| + 15 = -9*2 + 15 = -18 + 15 = -3 \). So \( x = 0 \) works.
So the solutions are \( x = 4 \) or \( x = 0 \), which is option C.

Answer:

C. \( x = 4 \) or \( x = 0 \)