Problem Solving

If (x - y)^2 = 10, and if x^2 - y^2=2, then what is the value of (5x+5y)/(6x-6y) ?

Answer Choices:
A. 1/6
B. 1/5
C. 1/4
D. 1/3
E. 1/2
My approach is somehow wrong i guess because I don't get the right answer. I started with calculating x-y= +5, -5 (is this wrong?) and then plugging the value in the other equation! I feel so dumb right now because I can't see where I am wrong! Please explain!

ilovemgmat wrote:If (x - y)^2 = 10, and if x^2 - y^2=2, then what is the value of (5x+5y)/(6x-6y) ?

Answer Choices:
A. 1/6
B. 1/5
C. 1/4
D. 1/3
E. 1/2
My approach is somehow wrong i guess because I don't get the right answer. I started with calculating x-y= +5, -5 (is this wrong?) and then plugging the value in the other equation! I feel so dumb right now because I can't see where I am wrong! Please explain!

(x^2-y^2)/(x-y)(x-y) = 2/10
=> (x-y)(x+y)/(x-y)(x-y) = 1/5
=> (x+y)/(x-y) = 1/5

hence, (5x+5y)/(6x+6y) = 5(x+y)/6(x-y)
= (5/6)(1/5)
= 1/6

Yes, that is wrong. It looks like you're trying to solve it by taking the square root of both sides, but 5 is not the square root of 10.

In any case, that is not the best way to solve it. Start by everything you can out of the expression (5x+5y)/(6x-6y). Then factor both equations and see if you can combine them in a way that gives you the value of your target expression.

Yeah, like that.

Yeah the square root of 10 is not 5. I feel like the dumbest person alive! Thanks!

GmatMathPro wrote:Yes, that is wrong. It looks like you're trying to solve it by taking the square root of both sides, but 5 is not the square root of 10.

In any case, that is not the best way to solve it. Start by everything you can out of the expression (5x+5y)/(6x-6y). Then factor both equations and see if you can combine them in a way that gives you the value of your target expression.

We can easily solve by taking common

We have (x-y)2 = 10, x2-y2=2

Equation is (5x+5y)/(6x-6y), Taking 5 and 6 common in numerator and denominator we have

5/6 * (x+y)/(x-y) , Now multiply both numerator and denominator using x-y

5/6 * (x+y)(x-y)/(x-y)(x-y) = 5/6 * x2-y2/(x-y)2 = 5/6 * 2/10 = 1/6

Lot of practice problems has this equation used most of the time (x+y)(x-y) = x2-y2

Cheers,
SVD