# Ten years ago, a man was 3 times as old as his son. In 6 years, he will be twice as old as his son. How old is each now?

May 20, 2018

The son is $26$ and the man is $58$.

#### Explanation:

Consider their ages $10$ years ago, now,and in $6$ years time.

Let the son's age $10$ years ago be $x$ years.
Then the man's age was $3 x$

It is useful to draw a table for this

$\underline{\textcolor{w h i t e}{\times \times \times x} \text{past" color(white)(xxxxxxx)"present"color(white)(xxxxxxx)"future}}$
SON:$\textcolor{w h i t e}{\times \times x} x \textcolor{w h i t e}{\times \times \times x} \left(x + 10\right) \textcolor{w h i t e}{\times \times \times} \left(x + 16\right)$

MAN:$\textcolor{w h i t e}{\times \times} 3 x \textcolor{w h i t e}{\times \times \times x} \left(3 x + 10\right) \textcolor{w h i t e}{\times \times x} \left(3 x + 16\right)$

In $6$ years time, the man's age will be twice his son's age.
Write an equation to show this.

$2 \left(x + 16\right) = 3 x + 16$

$2 x + 32 = 3 x + 16$

$32 - 16 = 3 x - 2 x$

$16 = x$

Ten years ago, the son was $16$ years old.

Use this value for $x$ to find the ages in the table.

$\underline{\textcolor{w h i t e}{\times \times \times x} \text{past" color(white)(xxxxxxx)"present"color(white)(xxxxxxx)"future}}$
SON:$\textcolor{w h i t e}{\times \times x} 16 \textcolor{w h i t e}{\times \times \times x} \left(26\right) \textcolor{w h i t e}{\times \times \times \times} \left(32\right)$

MAN:$\textcolor{w h i t e}{\times x . x} 48 \textcolor{w h i t e}{\times \times \times x} \left(58\right) \textcolor{w h i t e}{\times \times x . . \times} \left(64\right)$

We see that $2 \times 32 = 64$ so the ages are correct.

The son is $26$ and the man is $58$.