A man is twice as old as his son. When his son is the same age as his father was when he was twice as old as his son, both of their ages add to 180 years old. What are the ages of the man and his son through both of these time periods?
My attempt: $$m_1=2s_1=s_2$$ $$m_2+s_2=180$$
I really can't go farther than this, as I don't know what to do.
Let $x$ be the age of the son and $y$ the father. We have first off $2x=y$ right now. The second sentence says that "when the son is the age of the father...", we note that right now the son is $x$ and his father is $y=2x$, so when the son is the father's age, that is when the son is $2x$ years old, their ages add to 180. The number of years that pass from the time that the son is $x$ till the time the son is $2x$ is simply $x$. Thus $(2x) + (y+x) = 180$. Now we have a system $$2x=y \\ (2x) + (y+x) = 180$$ The solution is $x=36$ and $y=72$. Those are the ages of the two right now, while the ages later on are $72$ and $108$ respectively.
