# Eliminating x and y

• Nov 14th 2009, 07:17 PM
scrible
Eliminating x and y
Hi there I have a problem that I am trying to figure out. It said that by eliminating $x$ and $y$ from the equations $\frac {1}{y}+\frac {1}{y}=1$, $x +y = a$, $\frac {y}{x}=m$ where $a =/0$. obtain a relation between $m$adn $a$. Given that a is real, determine the ranges of values of $a$ for which $m$is real. Can some one help me with this problem. Thank you in advance.
• Nov 14th 2009, 07:24 PM
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.
• Nov 14th 2009, 07:40 PM
scrible
Quote:

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 $x=a-y$ but when I substitute that value for the other two eqaution every thing just seem confusing from there.
• Nov 14th 2009, 08:01 PM
topsquark
Quote:

Originally Posted by scrible
I tried that and I am just not getting the problem solved at all.for the second I got $x=a-y$ 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:
$\frac{1}{x} + \frac{1}{y} = 1$
and
$x + y = a$

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

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

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

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

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

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

-Dan
• Nov 14th 2009, 09:41 PM
scrible
Quote:

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:
$\frac{1}{x} + \frac{1}{y} = 1$
and
$x + y = a$

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

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

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

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

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

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

-Dan

Thank you very much