Would $x = x+1$ have infinite solutions? If $x = x+1$ and we have the equation $x = x+1$ but since $x = x+1$ would it equal $x+1 = x+1$?
It is a dumb question but someone told me that and I want to make sure it's not false.
 A: For an equation to have a solution, you need to be able to substitute all variables in the equation with your solution and obtain a true equality. For example, $x + 3 = 5$ has the one solution of $x = 2$, since $2 + 3 = 5$.
For another example, $x + x = 2x$ has the solutions $x = 5$ and $x = 6$, since $5 + 5 = 10 = 2(5)$ and $6 + 6 = 12 = 2(6)$. In fact, this equation has infinitely many solutions, since this equation is an identity.
In your case, $x = x + 1$ does not have any solutions. For example, $x = 4$ is not a solution, because $4 = 4 + 1$ implies that $4 = 5$, which is not true.
A: You are starting from an inconsistent (or false) proposition, sating that:

if $x=x+1$

As a result, any following steps could lead to false result(s) or inconsistent result(s).
To answer your question, $x=x+1$ has no solution, that is no such $x$ may exist.
Start with a consistent assumption, proceed according to the rules and you will get a consistent result. This is the promise of Mathematics (so, I say :))
A: I love your question because it is a very common mistake to either get extra solutions or lose some of the solutions on the way of simplifying an equation.
This is what happens. You start from $$x=x+1.$$ We know that it has no solutions because by assuming that there exists a solution $x\in \mathbb{R}$ and substracting $x$ from both sides we get $0=1$ which is a contradiction to the fact there exists a solution $x$.
Nevertherless(!), we proceed. Logically it is correct to plug in as you did and get:
$$x=x+1 \implies x+1=x+1.$$
What this actually means is: If some $x\in \mathbb{R}$ is a solution of $x=x+1$, then it is also a solution of $x+1=x+1$. And the solutions for the second equation is any real number, i.e. any $x\in \mathbb{R}$. In other words, set of solutions of a first equation is a subset of solutions of a second equation:
$$\varnothing \subset \mathbb{R}.$$
Think about similar implications $$x=-x \implies x^2=x^2$$ by taking squares of both sides. Or $$x=\sqrt{5} \implies x^2 = 5$$ extends the set of solutions from $\{\sqrt{5}\}$ to $\{\sqrt{5} , - \sqrt{5}\}$
A: Is this what you are trying to ask?:

If $x = x+1,$ then, by substitution, we have $x+1 = x+1,$ so it has infinitely many solutions, right?

If so, then your argument is invalid, because it is not logical to selectively apply the substitution: the equation transformation preserves meaning only when the substituent replaces every occurrence of its placeholder.
So, $$x=x+1\iff (x+1=x+1)+1,$$ and both the LHS and RHS equations are inconsistent, which means that neither has a solution.
A: Ok let's assume what you said is true. that is if x=x+1 then by subtracting x from both sides we get 1=0, Absurdity!! isn't it? so the equation is inconsistent.
