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

For the corresponding x expressions you should have $\mp$ not $\pm$ — David Quinn, Jul 26 '15 at 11:46

score 3 · Accepted Answer · answered Jul 26 '15 at 11:45

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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.

answered Jul 26 '15 at 11:45

Eclipse Sun

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At what step does my "solution" break? – Naz Jul 26 '15 at 12:24
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@isquared-KeepitReal When you get $y_{1,2}=\cdots$, you should solve for $x$ for each $y$. For example, your $x_2$ corresponds to $y_1$ and $m=y_1/x_2$ is a possible answer. You cannot say $m=y_1/x_1$. – Eclipse Sun Jul 26 '15 at 12:48
How come $y_1$ corresponds to $x_2$? – Naz Jul 26 '15 at 15:19
See @David Quinn's comment. – Eclipse Sun Jul 26 '15 at 15:21
But then that means that answers for second and third cases are correct? – Naz Jul 26 '15 at 15:56
No, the denominator should be $4a$. – Eclipse Sun Jul 26 '15 at 16:00
@isquared-KeepitReal: $y_1$ corresponds to $x_2$ because your first two equations are symmetrical in $x$ and $y$. It makes sense that, when you find the two roots of the quadratic, one is $x_i$ and the other is $y_i$. – Steven Alexis Gregory Jul 27 '15 at 01:33

score 1 · Answer 2 · answered Jul 26 '15 at 12:01

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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 $$\left\{x= \frac{m+1}{m},y= m+1,a= \frac{(m+1)^2}{m}\right\}$$

answered Jul 26 '15 at 12:01

Claude Leibovici

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ajotatxe · Answer 3 · 2015-07-27T01:05:00.597

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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 $$a=\frac{(m+1)^2}m$$

edited Jul 27 '15 at 01:05

answered Jul 26 '15 at 11:49

ajotatxe

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Shouldn't it be $x=\frac{a}{m+1}$? – entrelac Jul 26 '15 at 11:58
Yes, fixed now. – ajotatxe Jul 27 '15 at 01:05

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

3 Answers3