Boss is currently 8 times as old as Bew. In 8 years, Boss will be 3 times as old as Bew. How old is Boss now?

Step-by-step solution

1. 1

   Choose the smaller quantity as the variable. Let Bew's present age be $x$ years. Because Boss is $8$ times as old now, Boss's present age is $8x$. Naming the younger age keeps both expressions free of fractions at this stage.

2. 2

   Age both people by the same amount. Eight years from now everyone is exactly $8$ years older, so Bew is $x + 8$ and Boss is $8x + 8$. A very common slip is to write Boss's future age as $8(x+8)$ — that would multiply the future ages by 8, which the problem never claims.

3. 3

   Turn the future comparison into an equation. "Boss will be 3 times as old as Bew" means

   $8x + 8 = 3(x + 8)$

4. 4

   Solve the linear equation. Expand the right side and collect:

   $8x + 8 = 3x + 24 \quad\Longrightarrow\quad 5x = 16 \quad\Longrightarrow\quad x = \frac{16}{5} = 3\tfrac{1}{5}$

5. 5

   Answer the question that was asked. The problem wants Boss's age, not $x$:

   $8x = 8 \cdot \frac{16}{5} = \frac{128}{5} = 25\tfrac{3}{5}\ \text{years}$

6. 6

   Check both conditions in the original wording. Now: $25\tfrac{3}{5} = 8 \times 3\tfrac{1}{5}$, since $8 \times \tfrac{16}{5} = \tfrac{128}{5}$. In 8 years: Bew is $\tfrac{16}{5} + 8 = \tfrac{56}{5}$ and Boss is $\tfrac{128}{5} + 8 = \tfrac{168}{5}$, and $3 \times \tfrac{56}{5} = \tfrac{168}{5}$. Both statements hold. The ages are not whole numbers, which is fine algebraically — the data simply were not chosen to give integer ages.