Proof of Uniqueness of Two Lines 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?

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.
