If (x 2y)(x + 2y) = 5 and (2x y)(2x + y) = 35, then (x^2 - y^2)

Question

If  (x – 2y)(x + 2y) = 5  and  (2x – y)(2x + y) = 35 , then  \frac{x^2 - y^2}{x^2+ y^2} =

(A) –8/5
(B) –4/5
(C) 0
(D) 4/5
(E) 7/5

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nikaro · Answer

Simplifying the two equations and solving will give the values of x^2 as 9 and y^2 as 1. Substituting the same will give the answer as 4/5 (D)

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Oppenheimer1945 · Answer

(x-2y)(x+2y)=1*5
x-2y=1; x+2y=5
=>x=3,y=1

check in (2x-y)(2x+y)=5*7 (correct)

(x^2-y^2)/(x^2+y^2)=(9-1)/(9+1)=8/10=4/5

Nidzo · Answer

First equation:

(x-2y)(x+2y) = 5

x^2 - 4y^2 = 5   (1)

Second equation:

(2x - y)(2x + y) = 35

4x^2 - y^2 = 35   (2)

Solving:

Looking at  x^2 - y^2 , we can solve for this by adding  (1)  and  (2) .

(1)  +  (2) :  5x^2 - 5y^2 = 40

x^2 - y^2 = 8

And, to find the value for  x^2 + y^2 , we can subtract  (1)  from  (2) .

(1)  +  (2):   3x^2 - 3y^2 = 30

x^2 + y^2 = 10

Therefore,  \frac{x^2 - y^2}{x^2+y^2} = \frac{8}{10}  which can be simplified to  \frac{4}{5}

ANSWER D

unraveled · Answer

If  (x – 2y)(x + 2y) = 5  and  (2x – y)(2x + y) = 35 , then  \frac{x^2 - y^2}{x^2+ y^2} =

(A) –8/5
(B) –4/5
(C) 0
(D) 4/5
(E) 7/5

Solution with a different approach: 
In  \frac{x^2 - y^2}{x^2+ y^2}  the numerator must be less than denominator since squares are positive. Hence any choice with numberator > denominator is out. E eliminated.
 x^2 - y^2  ≠ 0 since in that case x = y whether negative or positive is not an issue. 
If that's the case 
 (x – 2y)(x + 2y) = 5  does not make sense as it's not possible(take any value).

Hence C is out too.

Now we are left with A , B and D in which caase we need to find whether x > y.
Writing  (x – 2y)(x + 2y) = 5  in the form as
A. (+)*(+) = (+)
OR 
B. (-)*(-) = (+)

Taking A(B too would lead to same result) we have 
x -2y > 0
implying that 
x > y

Thus,  x^2 - y^2 > 0

Therefore only D is satisfied.

Answer D.

If (x 2y)(x + 2y) = 5 and (2x y)(2x + y) = 35, then (x^2 - y^2)

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GMAT Problem Solving (PS) Questions

If (x 2y)(x + 2y) = 5 and (2x y)(2x + y) = 35, then (x^2 - y^2)

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