jmc

Let $y$ be the larger of the numbers. Since the numbers sum to 15, the other number is $15-y$. Since four times the smaller number is 60 less than twice the larger, we have: $$\begin{array}{rl}4(15-y)& =2y-60\Rightarrow \\ 60-4y& =2y-60\Rightarrow \\ 120& =6y\Rightarrow \\ 20& =y.\end{array}$$ The larger number is $\boxed{20}$, which makes the smaller number -5. We can check that our answer is correct by plugging the values into the original problem. We get $-5+20=15$ and $4(-5)=2(20)-60$, which result in $15=15$ and $-20=-20$.