Hiroto solved the equation [tex]\(6 - 4|2x - 8| = -10\)[/tex] for one solution. His work is shown below.

[tex]\[
\begin{array}{l}
6 - 4|2x - 8| = -10 \\
-4|2x - 8| = -16 \\
|2x - 8| = 4 \\
2x - 8 = 4 \\
2x = 12 \\
x = 6
\end{array}
\][/tex]

What is the other solution?

A. -6
B. -4
C. 2
D. 10



Answer :

To find the other solution of the equation [tex]\(6 - 4|2x - 8| = -10\)[/tex], let's follow a step-by-step approach similar to Hiroto's:

1. Start with the given equation and simplify:

[tex]\[ 6 - 4|2x - 8| = -10 \][/tex]

Subtract 6 from both sides:

[tex]\[ 6 - 4|2x - 8| - 6 = -10 - 6 \][/tex]

Which simplifies to:

[tex]\[ -4|2x - 8| = -16 \][/tex]

2. Isolate the absolute value:

Divide both sides by -4:

[tex]\[ |2x - 8| = 4 \][/tex]

3. Consider the definition of absolute value:

The equation [tex]\( |2x - 8| = 4 \)[/tex] gives us two separate cases to consider:

[tex]\[ 2x - 8 = 4 \quad \text{or} \quad 2x - 8 = -4 \][/tex]

4. Solve each equation separately for [tex]\(x\)[/tex]:

- Case 1:

[tex]\[ 2x - 8 = 4 \][/tex]

Add 8 to both sides:

[tex]\[ 2x = 12 \][/tex]

Divide by 2:

[tex]\[ x = 6 \][/tex]

- Case 2:

[tex]\[ 2x - 8 = -4 \][/tex]

Add 8 to both sides:

[tex]\[ 2x = 4 \][/tex]

Divide by 2:

[tex]\[ x = 2 \][/tex]

Since Hiroto already found the solution [tex]\( x = 6 \)[/tex], the other solution is:

[tex]\[ x = 2 \][/tex]

Thus, the other solution to the equation is:

[tex]\[ \boxed{2} \][/tex]