When you collect the variable terms, they sometimes cancel completely. Then the equation has no $x$ left, and what remains tells you how many solutions there are.
Why it works
If the $x$ terms cancel, what is left is a statement about numbers only, like $6 = 6$ or $-1 = 6$. Each step kept the two sides balanced, so this statement is true for exactly the values of $x$ that made the original true. A true statement such as $6 = 6$ does not depend on $x$, so every number works. A false one such as $-1 = 6$ can never be made true, so no number works.
Key idea
A linear equation has exactly one solution when you end with $x$ = a number, infinitely many when you end with a true statement like $6 = 6$, and no solution when you end with a false statement like $-1 = 6$.
Worked examples
Worked example: Every number works
Solve $3(x + 2) - x = 2(x + 3)$.
- Simplify each side: $2x + 6 = 2x + 6$.
- Subtract $2x$: $6 = 6$. True, whatever $x$ is.
- Check two numbers in the original: $x = 0$ gives $6 = 6$; $x = 5$ gives $16 = 16$.
Answer: infinitely many solutions (all real numbers)
Worked example: No number works
Solve $4x - 1 = 2(2x + 3)$.
- Distribute: $4x - 1 = 4x + 6$.
- Subtract $4x$: $-1 = 6$. False, whatever $x$ is.
Answer: no solution
Worked example: Exactly one
Solve $5x + 2 = 3x - 4$.
- Subtract $3x$: $2x + 2 = -4$.
- Subtract 2, then divide by 2: $x = -3$.
Answer: one solution, $x = -3$
Worked example: Tell without solving
For which number $c$ does $7x + 4 = 7x + c$ have infinitely many solutions? For which does it have none?
- Both sides have $7x$, so the $x$ terms will cancel, leaving $4 = c$.
- If $c = 4$ the sides are identical: infinitely many solutions. Any other $c$ gives a false statement: no solution.
Answer: $c = 4$ gives infinitely many; any other $c$ gives none.
Common mistake
Reading $0 = 0$ as “$x = 0$,” or thinking a false statement means you made a mistake. $0 = 0$ says nothing about $x$; it is true for every $x$. A false statement like $4 = 9$ is a real result: no solution. It is still wise to recheck your algebra once, and a quick test with two values of $x$ in the original equation settles it.