Unable to solve equation

siddharthmishra19

The problem statement, all variables and given/known data

[tex] \frac {xy}{x+2y} + \frac {x+2y}{xy} = 2 [/tex]

[tex] \frac {xy} {x-2y} + \frac {x-2y} {xy} = 4 [/tex]

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

say is getting nowhere as i can't find a relationship between the two.

malawi_glenn

have you tried the "hard way" ?

symbolipoint

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.

Integral

Try multiplying each componet of the first expression by

[tex] \frac {x - 2y} {x-2y} [/tex]

and the second by

[tex] \frac {x+ 2y} {x+2y} [/tex]

Dick

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.

symbolipoint

Through a large number of steps, I obtained
[tex] \[
- x^3 - 2x^3 + 2x^2 y - 4xy^2 + 8y^3 = 0
\]
[/tex]I am not sure if this fits any useful form; it seems not to fit any.

f(x)

Rather, try using the substitution as :[tex]\ \ \frac {x+2y}{xy} = p [/tex]
and [tex]\ \ \frac {x-2y}{xy} = q [/tex] and then simplify. You should notice that now p+q and p-q is much simpler to substitute in terms of x and y.