Find all real numbers x and y such that
\(x^2y^2+x^2+y^2+2xy=40\) and \(xy + x+y=8\)
Please explain your answer!
\(x^2+y^2+2xy=(x+y)^2\\ and\\ x+y=8-xy\)
After I continued on a little further I found it easier to substitute z=xy and solve for z
then I solved for x and y after that.