Age matters

Let Roy's age be $x$ years , So, Ben's age will be $x+6$ years (according to first condition)

According to second condition,
$x+6$ = $3x$
$6=3x-x$
$2x = 6$
$x = \dfrac {6}{2} = 3$

So, Roy's age = $x$ years = $3$ years.
And, Ben's age = $x+6$ years = $3+6$ years = $9$ years.

$\therefore$ The sum of their ages = $3+9 = 12$ years. $_\square$

Let Roy's age be $R$ . While Ben's age be $B$ . Then:

$\large R + 6 = 3B \\ \large 3R = B$

Then $R + 6 = 3R$ . Solving it gives us $R = 3$ .

By substituting in Eq.2 we get $B = 9$ .

Therefore $R + B = 3 + 9 \\ = 12$

Roy: 1 2 3 4 ...

Ben: 7 8 9 10 ... $1 * 3 = \not 7$ $2 * 3 = \not 8$ $3 * 3 = 9$ $4 * 4 = \not 10$ $3 + 9 = \boxed{12}$