Simultaneous equations, $\frac{1}{x}+\frac{1}{y}=1$,$x+y=a$,$\frac{y}{x}=m$

Question

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

Answer 1

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.

Answer 2

Your system has three equations and only two unknowns, so it is probably incompatible. Nevertheless, it still can have solution for certain values of $a$ and $m$.

From the last equation, we get $y=mx$. With the second eq.: $$(m+1)x=a$$ We see that the system is incompatible if $m=-1$ and $a\ne 0$. Also, if $m=-1$ and $a=0$ then $x=-y$ and the first equation is impossible.

So assume $m\neq-1$. Then $$x=\frac{a}{m+1}$$ $$y=\frac{am}{m+1}$$ then $$\frac1x+\frac1y=\frac{(m+1)^2}{am}$$ So the system has a solution only if $$a=\frac{(m+1)^2}m$$

Answer 3

Since you need to find the relation between $a$ and $m$, you can consider that you face a problem of three equations for three unknowns $x,y,a$ for which the solution is unique $$\left\{x= \frac{m+1}{m},y= m+1,a= \frac{(m+1)^2}{m}\right\}$$