A brother $G$ is 6 years older than his younger brother $L$. The first digit in G's age is twice the second digit in L's age. The second digit in G's age is twice the first digit in L's age. Find their ages.
I assumed that: $G = L + 6$
and G's age is compose by two digits: $G = XY $ and L's age is composed by two digits as well: $L = KT $
and we know that: $X = 2T$ and $Y = 2K$
How can I proceed to find their ages?
Let $G = 10a + b$ and $L = 10c + d$, their ages.
We are given that $G = 6 + L$, $a = 2d$, and $b = 2c$.
Solving this system (and keeping $a,b,c,d$ bound to the $0-9$ range), we have $a=4, b=8, c=4, d=2$, so $G=48$, and $L=42$