Simultaneous equations, $\frac{1}{x}+\frac{1}{y}=1$,$x+y=a$,$\frac{y}{x}=m$ By eliminating  $x$ and $y$ from the following equations, I need to find the relation between $m$ and $a$.
\begin{align*}
\frac{1}{x}+\frac{1}{y}=1 \\
x+y=a \\
\frac{y}{x}=m
\end{align*}
I tried different ways, but cannot arrive at the answer. I end up with a quadratic.
This is what I have.
\begin{align*}
\frac{1}{x}=1-\frac{1}{y} \\ 
x(y-1)=y \\
x=\frac{y}{y-1}
\end{align*}
Now using second equation:
\begin{align*}
\frac{y}{y-1}+\frac{y(y-1)}{y-1}=a \\
\frac{y^2}{y-1}=a \\
y^2-ay+a=0 
\end{align*}
Now roots are: $y_{1,2}=\displaystyle{\frac{a\pm\sqrt{a^2-4a}}{2}}$
Then it follows that: $x_{1,2}=\displaystyle{\frac{a\pm\sqrt{a^2-4a}}{2}}$
Hence there are three scenarios for the relation between $a$ and $m$.
First: $\frac{4a}{4}=m$, $a=m$
Second: $\displaystyle{\frac{(a+\sqrt{a^2-4a})^2}{4}=m}$
Third: $\displaystyle{\frac{(a-\sqrt{a^2-4a})^2}{4}=m}$
Is this correct? Thank you

No, since for each value of $y$ there is at most one $x$ satisfying $x+y=a$. You can now reexamine your solution.  
Here is a different approach:
From the last two equations, we can derive that $x=\frac{a}{1+m}$ and $y=\frac{ma}{1+m}$. Substitute into the first equation, and we have $(1+m)^2=ma$. It is clear $m\ne0$. Hence, $$a=\frac{(1+m)^2}{m}$$ is what you want.
