1
$\begingroup$

Problem: $(x - 1)(x - 2)(x - 3) = (x - 2)(x - 3)(x - 4)$

Heres my question with this problem: why do I end up with a wrong answer when I divide both sides by $(x-2)(x-3)$ to cancel out the $(x-2)(x-3)$ on both sides. Is this not allowed and why? Please provide the explanation to this question. You do not have to solve the problem.

Thank you very much.

$\endgroup$
3
  • 1
    $\begingroup$ Do you mean "wrong answer" or "a contradiction"? $\endgroup$
    – Sávio
    Feb 8, 2023 at 2:32
  • 3
    $\begingroup$ Subtract the RHS from the LHS and pull out the factor $(x - 2)(x - 3)$ in common. $\endgroup$ Feb 8, 2023 at 2:33
  • 6
    $\begingroup$ If you want to divide an equation by an expression, you must first assume it is not zero. The cases where the expression is zero (if there are any) must then be treated separately. $\endgroup$
    – Ian
    Feb 8, 2023 at 2:36

3 Answers 3

7
$\begingroup$

When you divide by $x-2$ you assume that $x-2\ne 0$. If $x-2=0$ then you can’t divide by zero. Before dividing check if $x=2$ was a solution.

Again, before dividing by $x-3$ check if $x=3$ solves the equation.

After that, if $x\ne2$ and $x\ne3$ then divide-as you did and see if you can solve for x.

If you can’t then it means that the only solutions were the ones you found when $x-2=0$ or $x-3=0$.

$\endgroup$
5
$\begingroup$

You cannot divide both sides by $(x-2)(x-3)$ if it equals $0$. When you divide both sides by it, you are ignoring the solutions $x=2,3$ $$(x-1)(x-2)(x-3)=(x-2)(x-3)(x-4)$$ A proper way to solve this is as follows $$(x-1)(x-2)(x-3)-(x-4)(x-2)(x-3)=0$$$$[(x-1)-(x-4)](x-2)(x-3)=0$$ $$3(x-2)(x-3)=0$$

$\endgroup$
1
  • $\begingroup$ That's right: always resit the temptation to divide by a variable but factorise instead. $\endgroup$ Feb 8, 2023 at 10:32
2
$\begingroup$

Obviously, both $x=2$ and $x=3$ are solutions.

If $x$ is neither above, then we can cancel $(x-2)(x-3)$ on both sides of the equation, so we get $$ x-1 = x-4, \text{or} -1=-3, $$ which is a contradiction.

Therefore we only have two solutions: $x=2$ and $x=3$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .