Finding x and y in a Right-Angled Triangle with Given Hypotenuse and Area

[footprints]

The shorter sides of a right-angled triangle are of length (x+y)cm, (x-y)cm respectively. Given that the length of the hypotenuse is \sqrt68cm and that the area of the triangle is 8cm^2, find x and y.

\displaystyle{(x+y)^2 + (x-y)^2 = (\sqrt68)^2}
\displaystyle{x^2+y^2=34} ----------- (1)
\displaystyle{\frac{1}{2}(x+y)(x-y)=8}
\displaystyle{x^2+y^2=16} ----------- (2)
I can't seem to solve it. Did I do something wrong?

 

[Galileo]

Equation (2) should be x^2-y^2=16.

Can you solve it now?

 

[footprints]

Why?
This is how I did it.
\displaystyle{\frac{1}{2}(x+y)(x-y)=8}
x^2-xy+xy+y^2=16
\displaystyle{x^2+y^2=16}
Does that mean that -y multiply by y equal to -y^2? Isn't it y^2?

 

[ivfdad]

-y \times y = -y^2 is correct. Maybe you're thinking of -y \times -y = y^2

 

[footprints]

-y \times y = -y^2 is correct. Maybe you're thinking of -y \times -y = y^2

I was thinking of the first one. I always thought that because there is a square, so it would be positive.

 

[Zaimeen]

No, it's negative!

 

[hypermorphism]

I was thinking of the first one. I always thought that because there is a square, so it would be positive.

-y*y = -1*y*y = -1*(y*y) = -1*y^2 = -y^2