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Solve $\dfrac{x+25}{x-5}=\dfrac{2x+75}{2x-15}$

$\Rightarrow \dfrac{(x-5)+30}{x-5}=\dfrac{(2x-15)+90}{2x-15} \ \ \ ...(1)$

$\Rightarrow 1+\dfrac{30}{x-5}=1+\dfrac{90}{2x-15}\ \ \ ...(2)$

$\Rightarrow \dfrac{30}{x-5}=\dfrac{90}{2x-15}\ \ \ ...(3)$

$\Rightarrow 3(2x-15)=9(x-5)\ \ \ ...(4)$

$\Rightarrow 6x-45=9x-45\ \ \ ...(5)$

$\Rightarrow 6x=9x\ \ \ ...(6)$

$\Rightarrow 2x=3x\ \ \ ...(7)$

So the answer becomes $3x-2x=0 \Rightarrow x=0$ which works in the original equation. My question is if I divide $2x=3x$ by $x$, I get $2=3$ which is not valid. Why can't I divide by $x$? Is it because for $x=0$, $\frac{x}{x}$ is not defined and hence it gives $2=3$?

And if that is the case, then why is it in $(1)$, $\frac{x-5}{x-5}=1$ is valid? What if $x=5$, then wouldn't this fraction become undefined as well? I'm confused why in this case the variable $x$ can be divided in the beginning but at $(7)$ it doesn't work. Thanks for helping.

Martin Sleziak
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ronald christenkkson
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    If $x=0$, you can't divide by $x.;$ If $x\ne5$, you can divide by $x-5$ – J. W. Tanner Feb 07 '23 at 03:01
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    I just want to point out that you could have cross-multiplied at the start. The quadratic terms would have cancelled and you'd have been left with a linear equation: $35x-375=65x-375 \Rightarrow 35x=65x \Rightarrow x=0$. The cross-multiplication is valid because neither denominator can be zero (or that expression is undefined), so you're not multiplying by zero. – Robert Shore Feb 07 '23 at 03:14

1 Answers1

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When you want to divide by something that involves a variable you need to make sure it is not $0$. Your point about $2x=3x$ is a good example. You can then say "either $x=0$ or $x \neq 0$". The first one is easy to try and you see it works. In the second case you can divide by $x$ and you reach a contradiction. The conclusion is that $x=0$ is the only solution.

The same applies for dividing by $x-5$ at the start. You could say "either $x=5$ or $x \neq 5$", try $x=5$ and then go ahead and divide by $x-5$. In this case the problem statement divides by $x-5$ so you are assured by the problem setter that $x \neq 5$. Some teachers would expect you to comment about this, others would not.

Ross Millikan
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