Consider the following inequality: $\sqrt { 8x-7 } + \sqrt {2x+2} \lt 3$
Attempt a solution: after doing all the calculations I got: $x \gt 7$, or, $\frac 78 \le x \lt 1$. On checking this result, however applying greater than $7$ values to the inequality yielded absurdities. Can I cancel an entire part of the solution based on one result? Or were my calculations wrong?
Clearly if $x$ is very large, the left side will be very large and greater than $3$, so the range cannot be unbounded above. You probably squared something to get there and introduced an extraneous solution, but we can't be sure unless you show your work. We must have $x \ge \frac 78$ for the first square root to make sense.
In order to avoid negative square roots, we need $8x-7 \geq 0$ and $2x+2 \geq 0.$ This gives the conditions $$x \geq \tfrac{7}{8}$$ and $x \geq -1$ (this second inequality being redundant).
We also observe that $f:[\tfrac{7}{8},\infty) \rightarrow [\tfrac{7}{8},\infty)$, defined by $f(x)=8x-7$, is a non-negative, increasing and continuous function. The same is true for $g:[\tfrac{7}{8},\infty) \rightarrow [\tfrac{7}{8},\infty)$ defined be $g(x)=2x+2$. We conclude that $h:[\tfrac{7}{8},\infty) \rightarrow [\tfrac{7}{8},\infty)$ defined by $h(x)=\sqrt{f(x)}+\sqrt{g(x)}$ is a well defined function, and is increasing and continuous.
We observe that $h(1)=\sqrt{1}+\sqrt{4}=3$. Hence $h(x) \geq 3$ whenever $x \geq 3$ and $h(x)<3$ whenever $x \in [\tfrac{7}{8},1)$.
When solving an inequality with square roots like this, extraneous solutions should be expected. You have already clearly solved this, so I will not recreate all of that, but I know exactly what you did, so I will sketch that and the proper logic for the record.
Before solving this you set the bounds for the radicals themselves:
$$8x-7 \geq 0 \wedge 2x+2 \geq 0 \Rightarrow x \geq \frac{7}{8} \wedge x \geq -1 \Rightarrow x \geq \frac{7}{8}.$$
Ok good, one lower boundary. Next you solved the equation, and I am quite sure you squared both sides of an inequality twice. In the end you had
$$(x-7)(x-1)>0.$$
Your thinking here should have been that the product of two binomials being positive implies that they are either both positive, or they are both negative.
$$(x-7)(x-1)>0 \Rightarrow \left((x-7)>0 \wedge (x-1)>0\right) \vee \left((x-7)<0 \wedge (x-1)<0\right).$$
So if both binomials are positive you conclude
$$(x-7)>0 \wedge (x-1)>0 \Rightarrow x>7 \wedge x>1 \Rightarrow x>7.$$
If both are negative you conclude
$$(x-7)<0 \wedge (x-1)<0 \Rightarrow x<7 \wedge x<1 \Rightarrow x<1.$$
So you not only need to test the conditions $x>7$ and $x<1$, but you should expect extraneous solutions as a result of squaring square roots twice. Clearly as you noted $x \ngtr 7$, but $x <1$ works. You combine this with your initial lower bound $x \geq \frac{7}{8}$ to conclude as you have that
$$\frac{7}{8} \leq x < 1.$$
So to precisely answer your question, knowing precisely what you experienced, your calculations were correct, and yes you can and should cancel a part of your solution as it is not actually a solution. It is just an extraneous result of repeated squaring on square roots.
