L = Lee's class amount
F = Fred's class amount
we know the total of students for both classrooms is 60,
thus, L, whatever that is, plus F, whatever that may be, is 60,
or
L + F = 60
now, we also now, whatever Fred's class is, Lee's class is 1.4
times bigger
since Fred's class is F in size, Lee's is 1.4 * F then, or 1.4F
\(\bf \begin{cases}
L + F =60\\
\boxed{L}=1.4F\\
---------\\
\boxed{1.4F} + F =60
\end{cases}\)
solve for F
what's Lee's class? well, 1.4F
