How do you solve "sum of ages" puzzles?

Question

Ma and Pa and brother and me. The sum of our ages is eighty-three. Six times Pa’s age is seven times Ma’s age, and Ma’s age is three times my age.

What is Pa’s age? What is Ma’s age? What is my brother’s age? What is my age?

I try setting up the first equation in four variables letting $A$ be ma's age, $B$ be pa's age, $C$ be my brother's age and $D$ be my age.

$$A+B+C+D=83$$

The problem gives

\begin{align} 7B&=6A\\ A&=3D \end{align}

but I am still lost observing the fact that there was not any fact on my brother.

Answer 1

If Pa’s age $=7a,$ Ma’s age $=6a,$ your age$=2a$

If your brother’s age $=b,$ we have $7a+6a+2a+b=83\iff b=83-15a$

If ages are integers, $a\le5$

If $a=5,b=8$

If $a=4,b=23$ and Ma’s age $=6a=24$ which is impossible

Answer 2

You got the first equation backwards. It should be $6B=7A$. (That is, six times Pa's age, which you've called $B$, is seven times Ma's age, which you've called $A$.) So we have

$$\begin{align} A+B+C+D&=83\\ 6B&=7A\\ A&=3D\\ \end{align}$$

Can you take it from here? (Hint: Ages are typically positive whole numbers, and parents should be a good deal older than their children.)