The algebraic approach is OK, but if you have trouble understanding it, it may be useful to find a more straightforward solution first. I'll give you two.
Approach 1: direct simulation. This is very easy. Look at the day when the boy was born. He was $0$ years old, his father was $0+25=25$ years old. Now simulate what happens with each passing year:
$$
\begin{array}{lll}
\text{Event} & \text{Boy's age} & \text{Father's age} & \text{Sum of ages}\\
\text{Boy is born} & 0 & 25 & 25\\
\text{1 year later} & 1 & 26 & 27 \\
\text{2 years later} & 2 & 27 & 29 \\
\text{3 years later} & 3 & 28 & 31 \\
\text{4 years later} & 4 & 29 & 33
\end{array}
$$
We see immediately that when the sum was $31$, the boy was $3$ and the father was $28$, so the answer is $3$.
Approach 2: common reasoning. The first approach is good, but it would take too long to do this if instead of $31$ the sum was, say, $71$ years old. But we can look at the table above and "catch the trend". Namely, with each passing year the sum of ages increases by $2$. So, to go from $25$ to $31$, it would take $(31-25)/2 = 3$ years, hence the answer.
Approach 3: once you've solved a couple of problems like this using approaches 1 and 2, you become too lazy to internally repeat the same mantra: "with each passing year such-and-such happens, the total change was such-and-such, so the answer is such-and-such". Then you make things even more efficient in your mind. You introduce a variable, say $y$. Let's say the boy was born $y$ years ago. Then his age is $y$, his father's is $y+25$, the sum is $2y+25 = 31$, then you solve the equation by standard techniques. You subtract $25$ from both parts to get $2y=6$, then you divide both parts by $2$ to find $y=3$, and that is it.
Moral of the story: if you can't solve a problem using one method, try another. Cheat (in a good sense of the word, not literally). If you've written the equation, as in approach 3, but cannot solve it, then you can solve the problem first using common sense (approach 1 or 2). Then at least you will know the answer. And it is so much easier to provide the solution that the teacher wants if you already know the answer!
PS: to tell you the truth, my favorite solution is number 2 (common sense), not number 3 (algebra). But, as you look at more complicated situations, the "trend" becomes harder to "catch", and then you have no choice but to use algebra (and when that doesn't work, analysis, and when that doesn't work too, you come back to approach 1 (direct simulation), but on a computer/supercomputer).
Best Answer
$$P + M + B + Y = 83$$ $$6P = 7M$$ $$M = 3Y$$
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.