Question

1. \\(\frac{x}{15} = -10\\)\
2. \\(12 + x = 22\\)\
3. \\(60 = -10x\\)

Explanation:

To solve each equation, we'll use basic algebraic operations to isolate \( x \).

Equation 1: \( \frac{x}{15} = -10 \)

Step 1: Multiply both sides by 15

To isolate \( x \), multiply both sides of the equation by 15. This cancels out the denominator on the left side.
\[
\frac{x}{15} \times 15 = -10 \times 15
\]

Step 2: Simplify both sides

Simplifying the left side, the 15s cancel out, leaving \( x \). On the right side, \( -10 \times 15 = -150 \).
\[
x = -150
\]

Equation 2: \( 12 + x = 22 \)

Step 1: Subtract 12 from both sides

To isolate \( x \), subtract 12 from both sides of the equation. This cancels out the 12 on the left side.
\[
12 + x - 12 = 22 - 12
\]

Step 2: Simplify both sides

Simplifying the left side, \( 12 - 12 = 0 \), leaving \( x \). On the right side, \( 22 - 12 = 10 \).
\[
x = 10
\]

Equation 3: \( 60 = -10x \)

Step 1: Divide both sides by -10

To isolate \( x \), divide both sides of the equation by -10. This cancels out the coefficient of \( x \) on the right side.
\[
\frac{60}{-10} = \frac{-10x}{-10}
\]

Step 2: Simplify both sides

Simplifying the left side, \( \frac{60}{-10} = -6 \). On the right side, the -10s cancel out, leaving \( x \).
\[
x = -6
\]

Answer:

s:

• For \( \frac{x}{15} = -10 \), \( x = -150 \)
• For \( 12 + x = 22 \), \( x = 10 \)
• For \( 60 = -10x \), \( x = -6 \)