System with homogeneous equation in denominator help

[Hivoyer]

Homework Statement

The system is declared as follows:

8/(2*x - y) - 7/(x + 2*y) = 1
4/((2*x - y)^2) - 7/((x + 2*y)^2) = 3/28

Homework Equations

The Attempt at a Solution

I define 'x' to equal k*y and I replace it inside the equation:

8/(2*k*y^2) - 7(k*y + 2*y) = 1
4(4*k^2*y^2 - 4*k*y^2 + y*2) - 7(k^2*y^2 + 4*k*y^2 + 4*y^2) = 3/28

I know I have to divide the top one by the bottom one and take y^2 out of the brackets, however the way it is now, it would become impossibly complex if I divide them.How can I simplify them before I divide?Not sure how to proceed.Using least common multiple also results in a monstrosity.

 

[mfb]

How do you get 2ky^2 if you replace x with ky in 2x - y?
What happens to the fractions in equation 2?

I know I have to divide the top one by the bottom one

I doubt that this will work.

I would substitute 2x-y and x+2y with new parameters, and get rid of the denominators quickly.

 

[Hivoyer]

How do you get 2ky^2 if you replace x with ky in 2x - y?
What happens to the fractions in equation 2?

I doubt that this will work.

I would substitute 2x-y and x+2y with new parameters, and get rid of the denominators quickly.

By doing that I get:
8*v - 7*u = u*v
12*v^2 - 196*u^2 = 3*u^2*v^2

I can't use any of them to express u = # or v = # :(

 

[mfb]

You can solve the first equation for v or u and plug it into the second equation.

 

[YawningDog27]

By doing that I get:
8*v - 7*u = u*v
12*v^2 - 196*u^2 = 3*u^2*v^2

I can't use any of them to express u = # or v = # :(

First, shouldn't the $12v^2$ in your second equation be $112v^2$? (Probably just a typo)

Second, I suggest manipulating the first equation to get $u = \frac{8v}{v+7}$(assuming that $v \neq -7$, you should check that this cannot be true by substituting $v = -7$ into the system). Then, as mentioned above, substitute this expression into the second equation and solve for v.