Firstly I believe there is probably a very simple answer here but I don't quite understand.
Let's say we have the equation $77x-x=0$. Obviously this is equal to $76x=0$ so $x=0$. But why doesn't the following also work:
$$77x-x=0$$ $$77x=x$$ $$77=1$$
I'm pretty sure that $77$ does not equal $1$ so can someone please explain to me why the second method does not work. Thanks.
On
$77x=x$ implies nothing more nor less than $x (77-1) = 0$, which implies nothing more nor less than $76x=0$, which implies nothing more nor less than "$x = 0$ or $76 = 0$". From the fact that "$x=0$ or $76=0$", you can't deduce that $76=0$ (and hence that $77=1$).
On
When you have $77x = x$, it seems that your next step was to divide both sides by $x$.
But, dividing both sides by $0$ is problematic; and, indeed, here you have $x = 0$.
In a similar vein, if $x = 0$, then $10x = 0$. Adding $x$ to both sides yields $11x = x$; again, it would be unwise to divide both sides by $x$ and conclude that $11 = 1$. You could continue in this way to "conclude" that any two numbers are equal!
This is one of the most common mistake made by beginning algebra students (and on occasion, even by experts when its sufficiently disguised):
This is not a legal operation! What is legal is
Passing from $77x = x$ to $77 = 1$ can only be done in the case that $x$ is a nonzero number. But you haven't established that, so you can't do it!
What you can do is split the problem into two cases: one where $x=0$ and one where $x$ is nonzero. In the latter case, you can divide by $x$ to conclude $77=1$.
So, what you can actually conclude here is, given the equation $77x - x = 0$, that
Of course, you had already determined that $x=0$; you could skip all of the above since you know the "$x$ is nonzero" case doesn't happen.
Incidentally, if you were able to correctly deduce $77=1$, the result is known to be a contradiction, and so you would conclude the premise is false: that $77x - x \neq 0$.