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Mr_T
- Senior | Next Rank: 100 Posts
- Posts: 48
- Joined: Sat Nov 28, 2009 1:01 pm
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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
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
