Question

Solve the system by the substitution method.

xy = 12
x2 + y2 = 40

Answer 1

\[x^2+y^2=40\]\[x*y=12\] so \[(x-y)^2=16\]\[(x+y)^2=64\]\[x-y=\pm4\] and, \[x+y=\pm8\]

Answer 2

so by solving the above equations that are
x+y=8
and, x-y=8 we get the following ordered pairs (x,y)=\[(6,2), (-6,-2)\]