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

[guest2]

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


I started out by adding the two equation together to get

2x = a+b

Then I solved for y using equation x+y=a to get y = a-x, and then plugged this into the equation above.

(a+b)(a-x), but this wasn't one of the answer choices...any help would be appreciated. Thanks

[Guest]

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

you can use the formula:

(x+y) ^ 2 = x^2 + y^2 + 2xy -- > 1
(x-y)^2 = x^2 + y^2 - 2xy --> 2

Subtracting equation 2 from 1
(x+y) ^ 2 - (x-y)^2 = 4xy
a^2 - b^2 = 4xy

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

[divya]

please confirm if that is one of the answer choices.

Also alternatively:
2xy = (a^2 - b^2)/2

can also be written as

2xy = (a-b)(a+b)/2

[guest 2]

answer choice is correct.

Thanks.

[RonPurewal]

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


I started out by adding the two equation together to get

2x = a+b

Then I solved for y using equation x+y=a to get y = a-x, and then plugged this into the equation above.

(a+b)(a-x), but this wasn't one of the answer choices...any help would be appreciated. Thanks

if you're going to do this, you may as well solve for y using essentially the same method:
if you want x, ADD
if you want y, SUBTRACT

if you ADD the equations, you'll get 2x = a + b --> x = (a + b)/2
if you SUBTRACT the equations, you'll get 2y = a - b --> y = (a - b)/2

therefore,
2xy = ((a + b)/2)((a - b)/2)(2)
= (a^2 - b^2)/2 using the difference of squares
done.

--

TAKEAWAY: if you have "this + that" and "this - that", then you should ADD to find "this" and SUBTRACT to find "that".

--

from the other poster's stuff:
takeaway: you should know what cancels and what doesn't when you combine (x + y)^2 and (x - y)^2 by addition or subtraction.
when you add them, the xy terms cancel, and all that's left are the squared terms (2x^2 + 2y^2). when you subtract them, the squared terms cancel, and all that's left is 4xy.

[amaliachatzi]

x+y = a => y = a-x
x-y = b => x = y+b
-------------
(x+y)+(x-y)=a+b=> 2x =a+b(1)

xy = (a-x)(y+b)
xy = ay+ab-xy-xb
2xy = ay+ab-xb
2xy = a(y+b)-xb
2xy = ax-xb
2xy = x(a-b)
2y = a-b(2)
[editor: this is correct, but it takes seven lines, along with some very weird steps, to do what i did in one step in the post above.
if you just subtract the two original equations, you'll get this equation instantly.]

(1)*(2)=> 2x2y = (a+b)(a-b)
4xy = a^2-ab+ab-b^2
4xy = a^2 - b^2
2xy = (a^2 - b^2)/2

[RonPurewal]

by the way, the easiest way to do this problem -- by far -- is to plug in your own numbers for the variables. since the variables are still there at the end of the problem, it follows that you can't get actual values for them -- so you can pick any values you want, as long as they obey the initial conditions (i.e., x + y = a and x - y = b).

so let's pick
x = 3
y = 2
therefore
a = 5
b = 1

the question becomes
2xy = 2(3)(2) = 12

(A) is 12
(B) is not 12
(C) is not 12
(D) is not 12
(E) is not 12

done.
(A)