How to find a pair of possible solutions (x, y) of this system of equations?

Question

Let $(x, y)$ be a pair of possible solutions of this system of equations: $x + y = 7$; $x^2 + y^2 = 25$.

This exercise is all about finding $xy$ product. there are four possible solutions,

(a) 15

(b) 12

(c) 8

(d) 6

Here's how I've solved it: I've rewritten the second equation as "$x + y = 5$", since each member was squared.
Now, I've rewritten again this equation as "$y = 5 - x$". I've substituted and I've got: "$x + 5 - x = 7$", but x cancels out. Therefore I can say this is not the way to solve it.

I try another one: I've rewritten the first equation as "$y = 7 - x$". Then, I've substituted, and I've got: "$x^2 + (7-x)^2 = 25$" The last step is: $x^2 = -12$, which isn't solvable using real numbers. So, it's not solvable using this method.

If you know an alternative method, let me know.

Answer 1

$x^2 + (7 - x)^2 = 25 \implies$

$x^2 + (49 - 14x + x^2) = 25 \implies$

$2x^2 - 14x + 24 = 0 \implies$

$x^2 - 7x + 12 = 0 \implies$

$(x - 3)(x - 4) = 0.$

$x = 3$ or $x = 4$.

$y = 4$ or $y = 3.$

Answer 2

Since you asked for alternative methods, here is another one:

$$(x+y)^2 = x^2+y^2+2xy\Leftrightarrow $$ $$xy=\frac 12\left(\underbrace{(x+y)^2}_{=7^2}-\underbrace{(x^2+y^2)}_{=25}\right)=12$$