SAT Problem - "If ... what is the value of x?"

[Guest]

I'm stumped

If x^2 + 2xy + y^2 = (x-y)^-2 and x + y = 2, what is the value of x?

(A) 1/2
(B) 3/4
(C) 5/4
(D) 2
(E) 4

- Styles

 

[tridelta246]

If x^2 + 2xy + y^2 = (x-y)^-2 and x + y = 2, what is the value of x?

(A) 1/2
(B) 3/4
(C) 5/4
(D) 2
(E) 4

Left side of first equation is just (x + y)^2, so it's equal to 4 from the second equation. You now have 2 equations in 2 unknowns:
x - y = (+1/2) or (-1/2) ----- from first equation after solving for (x - y)
x + y = 2
so that x=(5/4) and x=(3/4) are solutions.

.

 

[Unco]

Note that if

\(\displaystyle \L 4 = \frac{1}{(x-y)^2}\)
then \(\displaystyle \L (x \, - \, y) = \pm \, \frac{1}{2}\)

There are two solutions; the question is incorrect to suggest there is one.

 

[Guest]

:shock:
Unless that's not the full question that was posted?