if x+y=5 and x-y=3 then xy=
if x+y=5 and x-y=3 then xy= ?
$$\boxed{\\(x+y)^2=x^2+2xy+y^2 \qquad (x-y)^2=x^2-2xy+y^2} \\\\\dfrac{(x+y)^2-(x-y)^2}{2}=\dfrac{2xy+2xy}{2}=2xy\\\\2xy=\dfrac{5^2-3^2}{2}\\\\xy=\dfrac{5^2-3^2}{4} = \dfrac{25-9}{4}=\dfrac{16}{4}=4\\\\$$
x + y + x - y = 2x = 5 + 3 = 8 => x = 4
4 - y = 3 => y = 1
xy = 4*1 = 4
x-y=3 => x=3+y. 3+y+y=5 => 2y=2 => y=1
x+1=5 => x=4
xy=4*1=4