## Unable to solve equation

The problem statement, all variables and given/known data

$$\frac {xy}{x+2y} + \frac {x+2y}{xy} = 2$$

$$\frac {xy} {x-2y} + \frac {x-2y} {xy} = 4$$

The attempt at a solution
I tried replacing
($$\frac {xy} {x+2y} = p$$, say and
$$\frac {xy} {x-2y} = q$$,

say is getting nowhere as i can't find a relationship between the two.
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 Blog Entries: 9 Recognitions: Homework Help Science Advisor have you tried the "hard way" ?

Recognitions:
Homework Help
 Quote by siddharthmishra19 The problem statement, all variables and given/known data $$\frac {xy}{ x+2y} + \frac { x+2y }{xy} = 2$$ $$\frac {xy} {x-2y} + \frac {x-2y} {xy} = 4$$ The attempt at a solution I tried replacing ($$\frac {xy} {x+2y} = p$$, say and $$(xy/(x-2y)) = q$$, say is getting nowhere as i can't find a relationship between the two.
If I had the ambition and initiative, I would try changing x and y into polar coordinates and attempt solving ... but I just do not have that much energy to try. What kinds of methods are used in that kind of equation to solve? That must be beyond the intermediate, even beyond the "College Algebra" level.

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## Unable to solve equation

Try multiplying each componet of the first expression by

$$\frac {x - 2y} {x-2y}$$

and the second by

$$\frac {x+ 2y} {x+2y}$$
 Recognitions: Homework Help Science Advisor Your substitutions do give a short cut. Use the first equation to show p=1. Use the second to show q=two roots of a simple quadratic. Now notice 1/p+1/q=2/y. Now it's just slinging square roots around.
 Recognitions: Homework Help Through a large number of steps, I obtained $$$- x^3 - 2x^3 + 2x^2 y - 4xy^2 + 8y^3 = 0$$$I am not sure if this fits any useful form; it seems not to fit any.

 Quote by siddharthmishra19 The attempt at a solution I tried replacing ($$\frac {xy} {x+2y} = p$$, say and $$\frac {xy} {x-2y} = q$$, say is getting nowhere as i can't find a relationship between the two.
Rather, try using the substitution as :$$\ \ \frac {x+2y}{xy} = p$$
and $$\ \ \frac {x-2y}{xy} = q$$ and then simplify. You should notice that now p+q and p-q is much simpler to substitute in terms of x and y.