# Solving for $x$ when $x$ is the denominator

How do you solve for $x$ when $x$ is in the denominator?

E.g.

$$10 = \frac{g-1}{x}$$

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(g-1)/10 = x.... –  user16697 Nov 23 '11 at 21:10
If there is no $x$ in the numerator, just flip both sides over so that $x$ is where you like it to be: $\frac{1}{10}=\frac{x}{g-1}$. –  André Nicolas Nov 23 '11 at 21:41

## 2 Answers

You can use the technique of "clearing denominators". To do so, just multiply both sides of the equation by whatever denominator you wish to get rid of. \begin{align*} 10 &= \frac{g-1}{x}\\ 10 \cdot x &= \frac{g-1}{x} \cdot x\\ 10x &= g-1. \end{align*} No matter how many denominators there are, you can use this trick to clear them all. Now you have a much simpler equation to work with.

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However, note that this technique can introduce spurious solutions. For example, if $g=1$, the original equation has no solution, even though the rewritten version has the solution $x=0$. Always check that the solutions you get actually satisfy the original equation. –  Ilmari Karonen Nov 23 '11 at 21:28
Y'all make this look so easy!!!! I think my head id going to fall off! –  user54720 Dec 30 '12 at 18:54

Zillions of undergraduates with no interest in mathematics but a fair amount of interest in "surviving" required math courses will reflexively say "cross-multiply!". And in this case, they'd be right. But maybe they'd miss it because they don't quite see two fractions: \begin{align} \frac{10}{1} & = \frac{g-1}{x} \\ \\ \\ 10x & = 1(g-1) \end{align} then divide both sides by $10$.

(Then you have to talk them out of "cross-multiplying" when they see $\dfrac ab - \dfrac cd$ and claim that it's of course equal to $ad - bc$, etc.)

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How about when we have x-3/1 = y/x+2? –  methuselah Nov 23 '11 at 22:36
@methusaleh : In standard usage, $x-3/1=y/x-2$ means $x-(3/1)=(y/x)-2$, but I've been around enough to hesitate before answering and ask first whether you mean that or $(x-3)/1=y/(x-2)$? (If the latter, then that's how you should write it. Either that or $\dfrac{x-3}{1}=\dfrac{y}{x-2}$.) –  Michael Hardy Nov 23 '11 at 23:29
@methusaleh: ...and if you do mean $x-\frac 31=\frac yx-2$, first write both sides with a common denominator (i.e. $\frac{x-3}1=\frac{y-2x}x$) and then cross-multiply. (Or just multiply both sides with $x$; multiplying or dividing by 1 makes no difference anyway.) –  Ilmari Karonen Nov 24 '11 at 12:50