Is Substitution Effective in Solving These Fractional Equations?

[siddharthmishra19]

Homework Statement

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

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

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.

 

[malawi_glenn]

have you tried the "hard way" ?

 

[symbolipoint]

Homework Statement

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

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

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.

 

[Integral]

Try multiplying each componet of the first expression by

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

and the second by

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

 

[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
$$\[<br /> - x^3 - 2x^3 + 2x^2 y - 4xy^2 + 8y^3 = 0<br /> \]$$I am not sure if this fits any useful form; it seems not to fit any.

 

[f(x)]

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.