Let's take specific numbers instead of symbols.
Suppose that the signal $x(t)$ being sampled is a pure sinusoid (extending from $-\infty$ to $\infty$) at $1200$ Hz, say $\cos(2\pi\cdot 1200 t)$, $-\infty < t < \infty$. (For the record, the Fourier transform of this signal is a pair of impulses at $\pm 1200$ Hz but we won't be needing to think about this very much). Suppose that we are sampling at a rate of $1000$ Hz, one sample every millisecond. Then,
\begin{alignat}{6} x[0] &= x(0) &&= \cos(2\pi\cdot 1200 \cdot 0) &&= \cos(0) &&= \cos(0)\\
x[1] &= x(0.001) &&= \cos(2\pi\cdot 1200 \cdot 0.001) &&= \cos(2\pi\cdot 1.2) &&= \cos(2\pi \cdot 0.2)\\
x[2] &= x(0.002) &&= \cos(2\pi\cdot 1200 \cdot 0.002) &&= \cos(2\pi\cdot 2.4) &&= \cos(2\pi \cdot 0.4)\\
x[3] &= x(0.003) &&= \cos(2\pi\cdot 1200 \cdot 0.003) &&= \cos(2\pi\cdot 3.6) &&= \cos(2\pi \cdot 0.6)\\
\end{alignat}
and so on. More generally, we have that $$x[n] = \cos\left(2\pi \cdot 1200\frac{n}{1000}\right) =
cos(2\pi\cdot 1.2n) = \cos(2\pi \cdot 0.2n).$$ But notice that if instead we had been sampling $\cos(2\pi \cdot 200t)$ at $1000 Hz$, we would have gotten the same set of sample values as shown above : the $n$-th sample (at $1000 Hz$ sampling rate) of $\cos(2\pi \cdot 200t)$ is precisely
$$x[n] = \cos\left(2\pi \cdot 200\frac{n}{1000}\right) =
\cos(2\pi \cdot 0.2n).$$
In short, from the samples alone, we cannot tell whether we are sampling $\cos(2\pi\cdot 1200 t)$ at $1000$ Hz or sampling $\cos(2\pi \cdot 200t)$ at $1000 Hz$. Thus, the sinusoid at $1200$ Hz has been aliased into the sinusoid at $200$ Hz. More generally, all sinusoids at frequencies $1000m + 200$ Hz will alias into the sinusoid at $200$ Hz. For the record, I note that $0 \leq 200\leq 500$ where $500$ is half the sampling frequency $F_s$ of $1000$ Hz.
But what about a sinusoid at frequency $800$ Hz that we sample at $1000$ Hz? (Note that this frequency is between $F_s/2$ and $F_s$.) Well, proceeding as above, we have that
\begin{alignat}{6} x[0] &= x(0) &&= \cos(0) &&= \cos(0)\\
x[1] &= x(0.001) &&= \cos(2\pi\cdot 0.8) &&= \cos(2\pi \cdot 0.8) &&= \cos(2\pi \cdot(-0.2))\\
x[2] &= x(0.002) &&= \cos(2\pi\cdot 1.6) &&= \cos(2\pi \cdot 0.6)&&= \cos(2\pi \cdot(-0.4))\\
x[3] &= x(0.003) &&= \cos(2\pi\cdot 2.4) &&= \cos(2\pi \cdot 0.4) &&= \cos(2\pi \cdot(-0.6))\\
\end{alignat}
and so on. More generally, we have that $$x[n] = \cos\left(2\pi \cdot 800\frac{n}{1000}\right) =
cos(2\pi\cdot 0.8n) = \cos(2\pi \cdot (-0.2)n).$$
In short, we cannot tell from the samples if we are sampling $\cos(2\pi\cdot 800 t)$ at $1000$ Hz or sampling $\cos(2\pi \cdot (-200)t)$ at $1000 Hz$. The sinusoid at $800$ Hz has been aliased to a sinusoid at $-200$ Hz.
"But, but, but," you exclaim, "I don't believe in negative frequencies, and neither should you. Do you have an explanation that doesn't involve negative frequencies at all?" Not to worry. Since $\cos(-x) = \cos(x)$, that sinusoid at $-200$ Hz is indistinguishable from a sinusoid at $200$ Hz. That is,
frequencies between $F_s/2$ and $F_s$ also alias into frequencies between $0$ and $F_s/2$.
The straightforward way of remembering what happens is to note that
Aliased frequency is the absolute difference between the actual signal frequency
and the nearest integer multiple of the sampling frequency.
For a $1200$ Hz sinusoid, the frequency difference between $1200$ and $1000$ is $200$; for a $3200$ Hz sinusoid, the frequency difference between $3200$ and $3\times 1000$ is $200$; for the $800$ Hz sinusoid, the frequency difference is $-200$. All three frequencies alias to $200$ Hz (for those who refuse to believe in negative frequencies). But more generally, if you are broad-minded enough to accept negative frequencies as a concept, then note that for each nonzero integer $k$, the frequency band $\left[kF_s - \frac 12 F_s, kF_s + \frac 12 F_s\right]$ (centered at $kF_s$, an integer multiple of the sampling frequency) aliases to the low-pass frequency band $\left[- \frac 12 F_s, \frac 12 F_s\right]$.