How solve the nedeed value of $t$ in this statement What conditions must the parameter $t$ fulfill so that the equation:
$x(1+4t) - 24 = 3xt - \frac{x}{2}$, have a unique solution ?
I resolve it, but I think it was coincidence.
This was:
$x + 4xt - 24 = 3xt - \frac{x}{2}$
$3x - 48 = -2xt$
$3x + 2xt = 48$
$x(3 + 2t) = 48$
Now, i will work with $(3 + 2t)$
$3 + 2t = 0$
$t = -\frac{3}{2}$, so $t$ should be different of $-\frac{3}{2}$
Why do I think it's a coincidence?
Because I had never done, nor seen that of:
Work only with $(3 + 2t) = 0$, and forget about the factorized $x$ and $48$.
So, I'd like you to tell me if it was a simple coincidence or if it really works out that way and why.
Also how do you solve it personally?
Please, if you will have a response that is not complete, refrain from answering.
 A: For any fixed value of $t$, $x(1+4t) - 24 = 3xt - \frac{x}{2}$ has the equation of a line on the left and the equation of a line on the right.  If they have different slopes, they must intersect exactly once, yielding a unique solution.  If they have identical slopes, they intersect either zero (parallel and distinct lines) or infinitely many (identical lines) times, failing to yield a unique solution.
The slope of the line on the left is $1+4t$.  The slope of the line on the right is $3t-\frac{1}{2}$.  These are the same when $1+4t = 3t - \frac{1}{2}$, so when $t = \frac{-3}{2}$.  Consequently, for $t \in (-\infty, \frac{-3}{2}) \cup (\frac{-3}{2}, \infty)$, the equation has a unique solution.
A: Attempt to solve it:
$x(1+4t) - 24 = 3xt - \frac{x}{2}$
$x(1+4t) - 24 = x(3t - \frac 12)$
$x(1+4t - 3t + \frac 12) = 24$
$x(t + \frac 32) = 24$.  If $t+ \frac 32 \ne 0$ then we can solve this.
$x = \frac {24}{t + \frac 32}$ is the unique distinct solution.
So $t$ can be anything AS LONG AS $t + \frac 32 \ne 0$.  So $t$ can be any value except for $-\frac 32$.
If $t = -\frac 32$ we get the impossible $x*0 = 24$. 
