How do you eliminate one term of several in a denominator?

[5P@N]

So I have this fraction:$$\frac {mass_{\text{mo}}}{r_{\text{mo}}^2} = \frac {mass_{\text{ea}}}{(\text{KLCOM})^2 -2((\text{KLCOM}) (r_{\text{mo}})) + (r_{\text{mo}})^2}$$

Don't worry about what these various terms represent - my question has only to do with an algebraic rule I've forgotten. I am trying to isolate one of the $r_{\text{mo}}$ terms. So my question is basically this: if I was to multiply both sides by $r_{\text{mo}}^2$, would it result in the following equation?
$$mass_{\text{mo}} = \frac {mass_{\text{ea}}}{(\text{KLCOM})^2 -2((\text{KLCOM}) (r_{\text{mo}}))}$$

 

[fresh_42]

Have you tried simply taking reciprocals?

 

[5P@N]

No. What sort of reciprocal?

Is my method correct, or do I need to use another method, because I've gotten the algebraic rule wrong?

 

[fresh_42]

Is my method correct, or do I need to use another method, because I've gotten the algebraic rule wrong?

No.
The rules that apply here are $\frac{a}{b} = \frac{c}{d} ⇔ a \cdot d = b \cdot c ⇔ \frac{b}{a} = \frac{d}{c}$ if none of the numbers is zero.
Since your masses are probably not zero you can just exchange numerator and denominator on both sides.

 

[SteamKing]

So I have this fraction:$$\frac {mass_{\text{mo}}}{r_{\text{mo}}^2} = \frac {mass_{\text{ea}}}{(\text{KLCOM})^2 -2((\text{KLCOM}) (r_{\text{mo}})) + (r_{\text{mo}})^2}$$

Don't worry about what these various terms represent - my question has only to do with an algebraic rule I've forgotten. I am trying to isolate one of the $r_{\text{mo}}$ terms. So my question is basically this: if I was to multiply both sides by $r_{\text{mo}}^2$, would it result in the following equation?
$$mass_{\text{mo}} = \frac {mass_{\text{ea}}}{(\text{KLCOM})^2 -2((\text{KLCOM}) (r_{\text{mo}}))}$$

No, not even close.

You can solve for r_mo by recognizing that the denominator on the RHS can be factored. By multiplying both sides by r_mo and dividing both sides by mass_ea, you can eventually come up with an expression for r_mo in terms of KLCOM and the ratio of the masses.

 

[5P@N]

Allright: I factored out the denominator on the RHS, then multiplied both sides by rmo. Then I inverted both fractions, and got the square root of both. Then on the RHS I expressed the fraction as two fractions, simplified, subtracted 1 from both sides, divided both sides by KLCOM, then inverted both fractions once again. Here is my answer:
$$\frac{KLCOM}{\sqrt{\frac{\text{mass}_{\text{ea}}}{\text{mass}_{\text{mo}}}} +1} = r_{\text{mo}}$$

Kosher?

 

[SteamKing]

Allright: I factored out the denominator on the RHS, then multiplied both sides by rmo. Then I inverted both fractions, and got the square root of both. Then on the RHS I expressed the fraction as two fractions, simplified, subtracted 1 from both sides, divided both sides by KLCOM, then inverted both fractions once again. Here is my answer:
$$\frac{KLCOM}{\sqrt{\frac{\text{mass}_{\text{ea}}}{\text{mass}_{\text{mo}}}} +1} = r_{\text{mo}}$$

Kosher?

No, you're close, but you seem to have made some mistakes before you got here.

It would be better if you posted all your work leading up to this.

 

[5P@N]

Very well...

I begin with having the denominator on the RHS factored, to yield:$$\frac{\text{mass}_{\text{mo}}}{r_{\text{mo}}^2} = \frac{\text{mass}_{\text{ea}}}{(\text{KLCOM} - r_{\text{mo}})^2}$$

Then I multiply both sides by $r_{\text{mo}}^2$, and divide both sides by $\text{mass}_{\text{ea}}$ to yield:$$\frac{\text{mass}_{\text{mo}}}{\text{mass}_{\text{ea}}} = \frac{r_{\text{mo}}^2}{(\text{KLCOM} - r_{\text{mo}})^2}$$

Next I switch the numerators and denominators on both sides: $$\frac{\text{mass}_{\text{ea}}}{\text{mass}_{\text{mo}}} = \frac{(\text{KLCOM} - r_{\text{mo}})^2}{r_{\text{mo}}^2}$$

Then I get the square root of both sides:$$\sqrt{\frac{\text{mass}_{\text{ea}}}{\text{mass}_{\text{mo}}}} = \frac{\text{KLCOM} - r_{\text{mo}}}{r_{\text{mo}}}$$

On the RHS, with one common denominator, I then render it into two terms:$$\sqrt{\frac{\text{mass}_{\text{ea}}}{\text{mass}_{\text{mo}}}} = \frac{\text{KLCOM}}{r_{\text{mo}}} -\frac{r_{\text{mo}}}{r_{\text{mo}}}$$

The second term on the RHS simplifies to -1, which I add to both sides viz:$$\sqrt{\frac{\text{mass}_{\text{ea}}}{\text{mass}_{\text{mo}}}} +1 = \frac{\text{KLCOM}}{r_{\text{mo}}}$$

Then I divide both sides by KLCOM:$$\frac{\sqrt{\frac{\text{mass}_{\text{ea}}}{\text{mass}_{\text{mo}}}} +1}{\text{KLCOM}} = \frac{1}{ r_{\text{mo}}}$$

Finally, I invert the numerators and denominators one last time to yield (triumphant trumpet flourish): THE WRONG ANSWER! YAY!:$$\frac{\text{KLCOM}}{\sqrt{\frac{\text{mass}_{\text{ea}}}{\text{mass}_{\text{mo}}}} +1}{} = r_{\text{mo}}$$

All of this more than begs the question: where did I go wrong?

 

[fresh_42]

Should be correct. My bad eyes just don't see the mass indices very well. I've got "ea / mo".

 

[SteamKing]

In making my solution of r_mo, I didn't make as many inversions as you did.

My solution came out:

$$r_{mo}=\frac{KCMOL ⋅ \sqrt{\frac{mass_{mo}}{mass_{ea}}}}{1+\sqrt{\frac{mass_{mo}}{mass_{ea}}}}$$

If you do some more manipulation, I agree with your result, which is more elegant.

 

[5P@N]

So I was right after all?

 

[SteamKing]

So I was right after all?

Yes.