# How do you solve the system x^2+y^2=18 and x-y=0?

Jul 26, 2016

$x = 3 , y = 3$ or $x = - 3 , y = - 3$

#### Explanation:

Multiplying by $x + y$ both sides of

$x - y = 0$

results the following system

 { (x^2+y^2=18), (x^2-y^2 = 0) :}

So $2 {x}^{2} = 18 \to x = \pm 3$ and $y = \pm 3$

The feasible solutions are

$x = 3 , y = 3$ and $x = - 3 , y = - 3$. Their feasibility was checked against the equation $x - y = 0$