- #1

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How is it done?

Having to solve for X when X is in the denominator really confuses me. Please help!

Having to solve for X when X is in the denominator really confuses me. Please help!

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- Thread starter 939
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- #1

- 111

- 2

How is it done?

Having to solve for X when X is in the denominator really confuses me. Please help!

Having to solve for X when X is in the denominator really confuses me. Please help!

- #2

jedishrfu

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How is it done?

Having to solve for X when X is in the denominator really confuses me. Please help!

Just multiply both sides by 120*x and remember that this works as long as 120*x =/= 0 which means

that x=/=0 in your final solution because dividing by zero is not defined.

- #3

Mentallic

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[tex]a\cdot\frac{b}{c} = \frac{a}{c}\cdot b =\frac{ab}{c}[/tex]

(The dot means multiply) so if we multiply

[tex]\frac{23}{120x}=\frac{1}{5}[/tex]

by the denominator in the left fraction (which is 120x) then we get

[tex]120x\cdot\frac{23}{120x}=120x\cdot\frac{1}{5}[/tex]

And now we can cancel 120x from the left side because

[tex]120x\cdot\frac{23}{120x} = \frac{120x}{120x}\cdot 23 = 1\cdot 23=23[/tex]

So we now have

[tex]23=\frac{120x}{5}[/tex]

Now you can solve the rest in the usual way you've been solving equations.

Also, keep in mind that you can have just multiplied through by x rather than 120x.

- #4

Student100

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- #5

Mark44

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IOW, if a/b = c/d, then b/a = d/c, barring of course the possibility that any of the numbers are zero.

In the context of this problem, you can rewrite 23/(120x) = 1/5 as 120x/23 = 5.

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