Find the ages of three people, given their sum and two relations between the ages 
Solve this riddle:  
“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 by setting up three equations from the problem 
$$A+B+C+D=83$$ where I took $A$ to be Ma's age, $B$ to be Pa's age, $C$ to be my brother's age and $D$ to be my age.
I then came up with two other equations $6B=7A$ and $A=3D$. I later solve these and came up with $15D+2C=166$ but I am lost and need help.
 A: *

*$P$ Pa

*$M$ Ma

*$B$ Brother

*$Y$ You


$$P + M + B + Y = 83$$
$$6P = 7M$$
$$M = 3Y$$
Combine and write everything in terms of $B$ and $Y$.
$$21/6 Y + 3Y + B + Y = 83$$
$$45 Y + 6B = 498$$
Sum of even numbers is even, so $Y$ must be even, call it $Y=2n$:
$$90n + 6Y = 498$$
$$15n + Y = 83$$
$$\begin{bmatrix} n \\ B \end{bmatrix} = \begin{bmatrix} n \\ 83 - 15n \end{bmatrix}$$
$$\begin{bmatrix} Y \\ B \end{bmatrix} = \begin{bmatrix} 2n \\ 83 - 15n \end{bmatrix}$$
Insert back in the parents:
$$\begin{bmatrix} P \\ M \\ Y \\ B \end{bmatrix} = \begin{bmatrix} 7n \\ 6n \\ 2n \\ 83 - 15n \end{bmatrix}$$
I think we can assume $B \ge 0$, so $83 - 15n \ge 0$, so $n \le 5$.
Hopefully $B < M$, so $83 - 15n < 6n$, so $n \ge 4$.
So your choices are $n=4$ or $n=5$:
$$\begin{bmatrix} P \\ M \\ Y \\ B \end{bmatrix} = \begin{bmatrix} 28 \\ 24 \\ 8 \\ 23 \end{bmatrix} \text{ or } \begin{bmatrix} 35 \\ 30 \\ 10 \\ 8 \end{bmatrix}$$
Assuming you are humans and your brother isn't adopted, it's probably the second one.
A: From$$15D+2C=166$$You can conclude that $D$ must be a multiple of $2$ and its maximum value would be when $C=1$ which means max $D=10$.
So then try and see what values you get for $A,B,C$ when $D=2,4,6,8,10$.
Only one of these combinations will make sense.
A: $6B=7A$ means that $B$ is a multiple of $7.$ Say, $B=7x.$ Now, it is $7A=6B=42x\implies A=6x.$ Since $6x=A=3D$ we have $D=2x.$ Thus,
$$A+B+C+D=6x+7x+C+2x=15x+C=83,$$ or equivalently $$15x=83-C.$$ The LHS is a multiple of $15$ (that is, $15,30,45,60$ or $75$). So, $C=8,23,38,53,68$ or $83.$ Since $C\le A$ and $C\le B$ it is $3C\le 83,$ that is,  $C=8$ or $C=23.$ 
If $C=23$ then $x=4$ and $A=24, B=28$ and $D=8.$  If $C=8$ then $x=5$ and $A=30, B=35$ and $D=10.$ The first solution doesn't make sense because the difference between the age of the father and the son is one year. So $A=30, B=35,$ $C=8$ and $D=10.$
