Question

Find two numbers whose sum is 64 and whose difference is 42

Answer 1

These numbers are "a" and "b".
Sum is 64, so a+b=64
Difference is 42 so a-b=42
So you've got:

`+\begin{cases} a+b=64 \\ a-b=42\end{cases} \\ \overline{, \quad2a \quad \ = \ 106} \qquad /:2 \\ a=53`
When a+b=64 and a=53 you've got:
53+b=64
b=11

So these numbers is 53 and 11.