Abdulla wrote:If x+y = a and x-y = b

What is 2xy in terms of a and b?

OA is [spoiler]a^2-b^2/2[/spoiler]

I know how to do it by picking numbers, but can someone do it algebraically??

Two algebraic alternatives to picking numbers:

Square both sides of the first equation:
x+y = a
(x+y)^2 = a^2
x^2 + 2xy + y^2 = a^2

Do the same with the second equation:
x-y = b
(x-y)^2 = b^2
x^2 - 2xy + y^2 = b^2

Now subtract the second equation from the first to get:

4xy = a^2 - b^2
2xy = (a^2 - b^2)/2


Alternatively, you could do:

x+y = a, so x = a-y
x-y = b, so x = b+y

Since these are both equal to x, they must be equal; a-y = b+y, or 2y = a-b

Similarly, solving for y instead of x in each equation, we get y = a-x, and y = x-b, so we have a-x = x-b and 2x = a+b.

So 2x = a+b, and 2y = a-b; if we multiply these equations we get 4xy = (a+b)(a-b) = a^2 - b^2, so 2xy = (a^2- b^2)/2

heshamelaziry wrote:This one is in OG 12. I got (b^2-a^2)/2. I added both equations and got x = (b+a)/2----> substituted x in x+y=b to get y, then multiplied.

That's yet another way to solve, though you've not arrived at the right answer because of a typo: the equation is x-y = b, not x+y = b.