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I came across this question while answering an intermediate algebra textbook:

A student tried to solve the equation $8x = 7x$ by dividing both sides by x, obtaining $8 = 7$. He gave the solution set to $\emptyset$. Why is this incorrect?

As we know it the correct answer is x = $0$. Can you give any possible explanations to the student why is this so?

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    $\begingroup$ Anything can happen when you divide by zero. So don't divide by zero. $\endgroup$ – Jacky Chong Oct 1 '16 at 4:49
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    $\begingroup$ Dividing by "$x$" is only valid if $x$ is not zero. $\endgroup$ – Math1000 Oct 1 '16 at 4:50
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    $\begingroup$ As others have pointed out, the division by $x$ is unjustified. We can still subtract $7x$ from both sides of the equation though. This directly shows $x=0$. $\endgroup$ – Glare Oct 1 '16 at 4:52
  • $\begingroup$ You can post you answers as answers instead of comments. $\endgroup$ – Ralf Rafael Frix Oct 1 '16 at 5:28
  • $\begingroup$ $Ax=Bx\iff [(x\ne 0 \land A=B)\lor (x=0)]. $ If $A\ne B$ then $[(x\ne 0\land A=B)\lor (x=0)] \iff x=0.$ $\endgroup$ – DanielWainfleet Oct 1 '16 at 20:34
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We can solve the equation $$8x = 7x$$ by subtracting $7x$ from each side of the equation to obtain $$x = 0$$ Substituting $x = 0$ into the original equation reveals that it is a valid solution, so the solution set is $S = \{0\}$, which is not empty.

Division by zero is undefined. When you divide by a variable, you are implicitly assuming that the variable does not equal zero. That is not a valid assumption in this problem since zero is the only real number that satisfies the equation $8x = 7x$.

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In my recent reply it is shown $x=0,$ and division by $0$ is not allowed, and if done we can see the differences between big quantities compared to $\infty$ are not sensitive and would be wrong in comparision..

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I guess one way to think of this is that the solution set is the union of all sets of solutions to the equation. So by diving by zero your student has found a solution for x different from 0, which is the empty set. However, x may be 0, thus the solution set is {0}. Since the union of a set with the empty set is itself, this is not a contradiction. Your student simply did not consider all posssible solutions.

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