OG 11 #245 Algebra

[Mr_T]

Hi all,

Here's the question:

If x + y = a and x-y = b then 2xy = ?

a) (a^2 - b^2) / 2
b) (b^2 - a^2) / 2
c) (a - b) / 2
d) ab / 2
e) (a^2 + b^2) / 2

Now...I was able to solve it, but it took some time because my first approach was wrong. My first approach was to isolate x and y. => x = a - y and y = x - b => 2xy = 2(a-y)(x-b). This obviously brought me nowhere...

So I then used substitution => x + y = a
- ( x - y ) = b which gave 2y = a - b
then x+ y = a
- (- x + y)) which gave 2x = b - a

Therefore 2y * 2x = (a-b) (b-a) => 2xy = (a^2 - b^2) / 2

My question is how from the start should I have known that the first approach was wrong? By the way, I don't like the way the OG 11 solves it. I found my approach quicker.

Thanks,

Mr T

[regor60]

I just saw that by adding the two eqs you can see 2X=a+b.

Y is obviously a-X and X is (a+b)/2 per above

so 2XY = (a+b) x (a- (a+b)/2)

or (a+b) x ((a-b)/2)

etcetera

[Gmatter2.0]

Squaring two eqns

x^2+y^2+2xy=a^2 (1)
x^2+y^2-2xy=b^2 (2)

(1)-(2)

4xy=a^2-b^2

2xy=a^2-b^2/2