(x + y)^2 = 4 x^2 - y^2 = 0 How many distinct solution of (x,y)

Question

(x + y)^2 = 4
x^2 - y^2 = 0

How many distinct solution of (x,y) satisfy the above equations?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

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niels.v_ · Answer

How I solved it (Feel free to correct me): 
the first equation can be rewritten as x^2+2xy+y^2,

the second equation tells us that x^2=y^2 since if x^2-y^2=0 it means they are equal. this means that x= ±y. we can now subsitite this in the first equation to get x^2 + 2x^2+x^2=4, so 4x^2=4, x= ±1. Thus since x=y the only solutions that satisfy the above equation are: (1,1) and (-1,-1). So answer C

Zakmaster · Answer

I got (b)

I simplified both equations by taking the square root of both sides, so:
(x+y)^2 = 4 == x+y = 2
x^2 - y^2 = 0 == x-y = 0

If my logic above is correct, there can only be one solution (1). -1 does not work because -1+(-1) = -2

Mirko92 · Answer

If (x+y)^2 = 4, this means that x+y is either 2 or -2, since 2^2 and (-2)^2 are both equal to 4.
Second statement x^2 - y^2 = 0 can be rewritten as (x+y)(x-y) = 0. Since we know x+y is non-zero, x-y has to be equal to 0, which means x=y.

If x=y, and we need to solve for both x+y=2 and x+y=-2, the only combinations that work are (1,1) and (-1,-1).

Answer is C.

Shubhradeep · Answer

-1+(-1) = -2 that's correct
But why do you think that doesn't satisfy the above equations?
 (-2)^{2}  = 4
 (-1)^{2} - (-1)^{2}  = 0

Nothing wrong. Answer is C

Tarunw111 · Answer

We can write the second eq. as (x-y)(x+y)=0
We know that x+y=2, which is non zero, we have to take x=y.
Thus solving for 2 eq.
x+y=2
x+y=-2

There are two solutions.

(x + y)^2 = 4 x^2 - y^2 = 0 How many distinct solution of (x,y)

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GMAT Problem Solving (PS) Questions

(x + y)^2 = 4 x^2 - y^2 = 0 How many distinct solution of (x,y)

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

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From 555 to 765: 210-point Improvement

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My Rewards