Question

If x, y are positive real numbers such that x + y +xy = 10 and x² + y² = 29, then the approximate value of x + y is :

Answer 1

$(x+y)^2=x^2+y^2+2xy=(10-xy)^2$
$29+2xy=100+x^2y^2-20xy$
$x^2y^2-22xy+71=0$
take xy=a
$a^2-22a+71=0$
solve it
u can find that xy=3.9289  (the other value is grater than hence it is not possible)
therefore x+y=10-3.9289=6.071