Let's call:
\(first\#=x\)
\(second\#=y\)
then
when we have a SUM, we have PLUS and when we have a DIFFERENCE, we have MINUS... Let's go then...
\(\begin{Bmatrix}x+y&=&95\\x-y&=&61\end{matrix}\)
now we can sum all the rows then we got it... (This is the other way to solved this question)
\(x+y+(x-y)=95+61\)
\(x+y+x-y=156\)
\(2x=156\)
\(\boxed{x=78}\)
now we can replace this value at first or at second row, you just need to pick up one...
I'll choose the second one
\(x-y=61\)
\(78-y=61\)
\(y=78-61\)
\(\boxed{y=17}\)
\(\boxed{\boxed{\begin{Bmatrix}x&=&78\\y&=&17\end{matrix}}}\)
