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