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[pinkcalculator]

I have to solve this for x.
pq/x-1=p+q/pq

I got this far.
pq/x=p+q/pq +1
pq= x (p+q/pq) + x.

Am I supposed to factor the x out? Because I feel like that's right, but I'm not sure how to combine the terms.

 

[Deleted member 4993]

I have to solve this for x.
pq/x-1=p+q/pq

I got this far.
pq/x=p+q/pq +1
pq= x (p+q/pq) + x.

Am I supposed to factor the x out? Because I feel like that's right, but I'm not sure how to combine the terms.

You need to put some grouping symbol around to make sense of this problem.

As posted - the problem is:

p * q/x - 1 = p + q/p * q

or

\(\displaystyle p\cdot \frac{q}{x} - 1 = p + \frac{q}{p} \cdot q\)

is that correct?

 

[pinkcalculator]

The original is
(pq/x) -1 = (p+q/pq)
Sorry about the confusion.

 

[Deleted member 4993]

The original is
(pq/x) -1 = (p+q/pq)
Sorry about the confusion.

That grouping did not change anything.

DOes your problem look like:

\(\displaystyle \frac{pq}{x} - 1 = \frac{p+q}{pq}\)

 

[pinkcalculator]

Yes, that's correct

 

[Deleted member 4993]

The original is


\(\displaystyle \frac{pq}{x} - 1 = \frac{p+q}{pq}\)

\(\displaystyle \frac{pq}{x} - 1 = \frac{p+q}{pq}\)

\(\displaystyle \frac{pq}{x} = \frac{p+q}{pq} + 1\)

\(\displaystyle \frac{pq}{x} = \frac{p+q + pq}{pq}\)

\(\displaystyle x = \frac{p^2q^2}{p+q + pq}\)

 

[pinkcalculator]

Thank you, thank you, thank you.
My math tutor just transferred to another college. I'm an English/Education major, and I'm in the very last math class I've got to take. It's a bit overwhelming, so I'm glad he referred me to this site.
Math and I do not always get along, so I really appreciate that you'll show how to solve a problem instead of just giving out answers.

 

[mmm4444bot]

I really appreciate that [you] show [all of the steps to solve] a problem instead of just giving out [the answer].

I would bet that you do!