How solve the nedeed value of $t$ in this statement

Question

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.

Answer 1

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 $x*0 = 24$.

Answer 2

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.