Let's call:
f i rs t # = x
seco n d # = 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
2 x = 156
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
y = 17
\boxed{\boxed{\begin{Bmatrix}x&=&78\\y&=&17\end{matrix}}}
\left \{ {x+y=95} \atop {x-y=61}} \right. \\
\\2x=156\\x= \frac{156}{2} \\
\\x=78\\
\\y=95-y\\y=95-78\\y=17
First number 78, and the second is 17 .
The two numbers are 78 and 17. The sum of these numbers is 95, and their difference is 61. By setting up and solving the system of equations, we found these values systematically.
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