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I am trying to show a proof of the uniqueness of two equations. If $a,b,c,d\in\mathbb R$ and $ad-bc\neq 0$, then or any $\alpha,\beta \in \mathbb R$ the pair of equations: $$ax + by = \alpha\\cx + dy = \beta$$ have a unique solution where $x = x_0$ and $y=y_0$ that depends on $a,b,c,d,\alpha,$ and $\beta$.

I am going to find a solution first by setting $x=x_0$ and $y=y_0$, then divide the first equation by $a$ and the seccond equation by $c$ I have:$$x_0+\frac{by_0}{a}=\frac{\alpha}{a}\\x_0 + \frac{dy_0}{c}=\frac{\beta}{c}$$ Subtract the two equations from each other: $$\left(\frac{b}{a}-\frac{d}{c}\right)y_0=\left(\frac{\alpha}{a}-\frac{\beta}{c}\right)$$ Multiply both sides by $ac$: $$(cb-ad)y_0=(c\alpha-a\beta)$$ Isolate $y_0$ by divining by the coefficients: $$y_0=\frac{c\alpha-a\beta}{cb-ad}$$ I isolate $x_0$ by doing the same as $y_0$ I divide the original top equation by $b$ and the bottom equation by $d$: $$\frac{ax_0}{b}+y_0=\frac{\alpha}{b}\\\frac{cx_0}{d}+y_0=\frac{\beta}{d}$$ Subtract the two equations from each other: $$\left(\frac{a}{b}-\frac{c}{d}\right)x_0=\left(\frac{\alpha}{b}-\frac{\beta}{d}\right)$$ Multiply both sides by $bd$: $$\left(da-bc\right)x_0=\left(d\alpha-b\beta\right)$$ Divide both sides by the coefficient of $x_0$: $$x_0=\frac{d\alpha-b\beta}{da-bc}$$ To show uniqueness I assume there is an alternative solution where $x=m_0$ and $y=n_0$. $$ax_0 + by_0 = \alpha\qquad am_0+bn_0=\alpha\\cx_0 + dy_0 = \beta\qquad cm_0+dn_0=\alpha$$

Question: How do I show that $x_0=m_0$ and $y_0=n_0$? Do I do the same process as above to the new set of equations with $m_0$ and $n_0$ to show that they equal the same thing?

Here is what I think I should do: Set the top equations equal to each other because they are both equal to constant $\alpha$, and set the bottom equations equal to each other because they are both equal to constant $\beta$. $$ax_0+by_0=am_0+bn_0\\cx_0+dy_0=cm_0+dn_0$$ Divide the top equation by $a$ and the bottom equation by $c$: $$x_0+\frac{by_0}{a}=m_0+\frac{bn_0}{a}\\x_0+\frac{dy_0}{c}=m_0+\frac{dn_0}{c}$$ Subtract the two equations: $$\left(\frac{b}{a}-\frac{d}{c}\right)y_0=\left(\frac{b}{a}-\frac{d}{c}\right)n_0$$ And divde both sides by the coefficient in front of $y_0$ or $n_0$ to show that$$y_0=n_0$$ I can then do the same for $x_0$ and $m_0$

Ultimate Question: Is this valid as proof of uniqueness?

Tsangares
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    Why do you want to prove uniqueness if you have got $x_0,y_0?$ On the other hand, what happens if $a=0?$ – mfl Oct 12 '16 at 20:08
  • The whole proof is for uniqueness. Getting $x_0, y_0$ was to show existence, an intermediary process to uniqueness. If $a=0$ I did account for, thank you. – Tsangares Oct 12 '16 at 21:43
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    Once you have shown that if $(x_0,y_0)$ is a solution it must be $x_0=\frac{d\alpha-b\beta}{da-bc},y_0=\frac{c\alpha-a\beta}{cb-ad}$ you have obtained the solution explicitly. So, it is unique. – mfl Oct 12 '16 at 22:07
  • When I stumble across something like $b/a$ or $d/c$, do I have to state that $a\neq0$ or can I conclude that this will not be a problem because of $ad-bc/neq0$ stated at the beginning? Is there a way I can come to a solution without the intermittent $b/a$? – Tsangares Oct 12 '16 at 22:40
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    If $a=0$ you have $y_0=\frac{\alpha}{b}=\frac{c\alpha}{cb}$ (note that $cb\ne 0$ in this case). Now, you can get $x_0$ from $cx_0+dy_0=\beta.$ You can work in a similar way if $b,c,d$ are zero. In any case your formula for $x_0,y_0$ is correct, because the key is that $ad-bc\ne 0.$ – mfl Oct 13 '16 at 00:06

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It is more simple and more correct to prove the unicity as a consequence of the unicity of the solution of the equations: $$(cb-ad)y_0=(c\alpha-a\beta)$$ and $$\left(da-bc\right)x_0=\left(d\alpha-b\beta\right)$$ Solving this equation you cannot say something as

''Isolate $y_0$ by divining by the coefficients:''

But you have to distinguish the cases:

$1) \quad (cb-ad) \ne 0$: in this case $(cb-ad)$ has an inverse and this inverse is unique so we have the unique solution $y_0=(c\alpha-a\beta)(cb-ad)^-1$

$2) \quad (cb-ad)=0 $ and $(c\alpha-a\beta)\ne 0$, in this case the equation has no solutions.

$2) \quad (cb-ad)=0 $ and $(c\alpha-a\beta)= 0$, in this case we have an identity with infinitely many solutions.

Emilio Novati
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