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.