The age difference between two brothers is three years. One year ago, the father's age of these brothers was double the sum of the brothers' ages, and in twenty years, the father's age will be the sum of the ages of these two sons. What is the age of each of the brothers?
Let $x$ and $y$ be the current ages of the two siblings and $z$ the age of the father. We have

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-112.jpg?height=168&width=882&top_left_y=1983&top_left_x=610)

A simple way to find $z$ is to multiply the third equation by 2 and subtract the second equation from the result, obtaining $2z + 40 - (z - 1) = 80 - (-4)$, which implies $z = 43$. Using the first two equations, we can now calculate the ages of the two children. From the first equation, $x = y + 3$, and from the second, $43 - 1 = 2x + 2y - 4$, or $x = 23 - y$. Adding these two equations, we get $2x = 26$, so $x = 13$ and, therefore, $y = 10$.