Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Jan 8, 2016 #1
    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?
    $$mass_{\text{mo}} = \frac {mass_{\text{ea}}}{(\text{KLCOM})^2 -2((\text{KLCOM}) (r_{\text{mo}}))}$$
     
  2. jcsd
  3. Jan 8, 2016 #2

    fresh_42

    Staff: Mentor

    Have you tried simply taking reciprocals?
     
  4. Jan 8, 2016 #3
    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?
     
  5. Jan 8, 2016 #4

    fresh_42

    Staff: Mentor

    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.
     
  6. Jan 8, 2016 #5

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    No, not even close.

    You can solve for rmo by recognizing that the denominator on the RHS can be factored. By multiplying both sides by rmo and dividing both sides by massea, you can eventually come up with an expression for rmo in terms of KLCOM and the ratio of the masses.
     
  7. Jan 8, 2016 #6
    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?
     
  8. Jan 8, 2016 #7

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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.
     
  9. Jan 8, 2016 #8
    :H
    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?
     
  10. Jan 8, 2016 #9

    fresh_42

    Staff: Mentor

    Should be correct. My bad eyes just don't see the mass indices very well. I've got "ea / mo".
     
  11. Jan 8, 2016 #10

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    In making my solution of rmo, 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. :smile:
     
  12. Jan 8, 2016 #11
    So I was right after all?
     
  13. Jan 8, 2016 #12

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Yes.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: How do you eliminate one term of several in a denominator?
Loading...