Does the equation have a unique solution, if..? I have an statement thay says:

The equation: $\frac{bx}{a} = \frac{ax + b^2}{a} - b, a \neq 0$, have
  unique solution if:
I) $a = b$
II) $a > b$
III) $a \neq b$

I know the properties to analyze the solutions of a first degree equation,
$ q \neq 0 $ have unique solutions
$ A = 0, B = 0 $ have infinite solutions
$ A = 0, B \neq 0 $ not have solutions,
I also know formula, $ X = - \frac {B} {A} $,
but I can not find a way to address this problem, could you help me? The correct answer is $ D) $, but that does not interest me, I'm interested in learning it.
PD: I get that $(x -b)(b-a) = 0$, but i not know how it could help me.
 A: Multiplying by $a\ne 0$ we get
$$x(b-a)=b^2$$
If $a=b$  we get $$0=b^2$$ So if $b=0$ then we get as the solution all real numbers. If $b\ne 0$ so we get no solution.
If $$a\ne b$$ we get $$x=\frac{b^2}{b-a}$$
With the new (corrected equation) you will get
$$x(b-a)=b(b-a)$$ or
$$(x-b)(b-a)=0$$
Now we have if $b=a$ then the solutions are all real numbers, and if $b\ne a$ then we get $x=b$
A: The solution will be $a \neq b$, let me walk you through the whole reasoning step by step. Firstly, we can reorganise the equation a little bit to get a better idea of what is going on:
Your equation: $\frac{bx}{a} = \frac{ax+b^2}{a}-b$, by transferring everything to the right hand side, can be written as
$0 = \frac{(a-b)x+b^2}{a} - b = \frac{a-b}{a}x+\frac{b^2}{a}-b$, which is a linear function of x with two parameters, where $a \neq 0$ and $b \in \mathbb{R}$.
Now we have three cases: 
1) A linear equation will have 0 solutions if its slope is zero and it is not identically equal to zero, as shown in the graph below:
This image depicts the case when the slope is zero and the function has zero solutions
The slope will be zero if and only if $\frac{a-b}{a}=0$, so $a = b$. Moreover, the function will not be identically equal to zero if we additionally have that $\frac{b^2}{a}-b \neq 0$, so $b(\frac{b}{a}-1) \neq 0$. This is equivalent to $b \neq 0$ or $a \neq b$, which in our case leaves only the first condition as valid.
2) The linear equation will have infinitely many solutions if it is identically equal to zero, meaning that its slope will be zero and it will cross the y-axis at zero. The graph below depicts this case:
This image depicts the case when the function has infinitely many solutions
In our case, the conditions above translate to: $\frac{(a-b)}{a}=0$, so $a=b$ and $\frac{b^2}{a}-b=0$, so $b=0$ or $a=b$. Hence, if $a=b$ the equation always has infinitely many solutions.
3) Finally, the linear equation will have one, unique, solution when it crosses the x axis in one point. This is depicted in the graph below.
This image depicts the case when the function has one (unique) solution
This case happens whenever the slope of the function is not equal to zero. Just imagine, when the slope of a linear function is non-zero the function steadily increases or decreases. In such a case, at some point, it is doomed to hit zero.
In our case, this translates to $\frac{(a-b)}{a} \neq 0$, so $a\neq b$, which is the stated solution.
