# Thread: Eliminating x and y

1. ## Eliminating x and y

Hi there I have a problem that I am trying to figure out. It said that by eliminating $\displaystyle x$ and $\displaystyle y$ from the equations $\displaystyle \frac {1}{y}+\frac {1}{y}=1$, $\displaystyle x +y = a$, $\displaystyle \frac {y}{x}=m$ where $\displaystyle a =/0$. obtain a relation between $\displaystyle m$adn $\displaystyle a$. Given that a is real, determine the ranges of values of $\displaystyle a$ for which $\displaystyle m$is real. Can some one help me with this problem. Thank you in advance.

2. I'm guessing the first term should be 1/x rather than another 1/y?

Solve the second for x and substitute this expression into the other two equations. You'll be nearly done.

3. Originally Posted by TKHunny
I'm guessing the first term should be 1/x rather than another 1/y?

Solve the second for x and substitute this expression into the other two equations. You'll be nearly done.
I tried that and I am just not getting the problem solved at all.for the second I got $\displaystyle x=a-y$ but when I substitute that value for the other two eqaution every thing just seem confusing from there.

4. Originally Posted by scrible
I tried that and I am just not getting the problem solved at all.for the second I got $\displaystyle x=a-y$ but when I substitute that value for the other two eqaution every thing just seem confusing from there.
Perhaps this is a bit sloppy (but then I tend to make problems a bit rougher than they need to be.) Start with the two equaitons:
$\displaystyle \frac{1}{x} + \frac{1}{y} = 1$
and
$\displaystyle x + y = a$

Then we know that
$\displaystyle \frac{x + y}{xy} = 1$

Put the second equation in:
$\displaystyle \frac{a}{x(a - x)} = 1$

or
$\displaystyle x^2 - ax + a = 0$

Now start from x + y = a and y = mx.
$\displaystyle x + mx = a$

giving
$\displaystyle x = \frac{a}{m + 1}$

Put this into the $\displaystyle x^2 - ax + a = 0$ equation to cancel the x and you are done.

-Dan

5. Originally Posted by topsquark
Perhaps this is a bit sloppy (but then I tend to make problems a bit rougher than they need to be.) Start with the two equaitons:
$\displaystyle \frac{1}{x} + \frac{1}{y} = 1$
and
$\displaystyle x + y = a$

Then we know that
$\displaystyle \frac{x + y}{xy} = 1$

Put the second equation in:
$\displaystyle \frac{a}{x(a - x)} = 1$

or
$\displaystyle x^2 - ax + a = 0$

Now start from x + y = a and y = mx.
$\displaystyle x + mx = a$

giving
$\displaystyle x = \frac{a}{m + 1}$

Put this into the $\displaystyle x^2 - ax + a = 0$ equation to cancel the x and you are done.

-Dan
Thank you very much