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Variable of an equation is in the denominator

  1. Aug 3, 2013 #1
    1. The problem statement, all variables and given/known data

    1/x - 1/a = 1/b

    Solve for x

    2. Relevant equations

    3. The attempt at a solution

    I used to be able to do this in high school, but now I am 43! Normally, I would just cross-multiply and solve for x, to get the x out of the denominator, but you cannot cross-multiply if there is more than one numerator! Does anyone know how to solve this? Thanks!
  2. jcsd
  3. Aug 3, 2013 #2

    Doc Al

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    Staff: Mentor

    Hint: a*(1/a) = 1

    What can you do to both sides to get rid of that 1/x?
  4. Aug 3, 2013 #3


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    Science Advisor

    There are two ways to approach this:
    1: Do the subtraction on the left: since the two denominators are "a" and "x", the "common denominator is ax. Multiplying numerator and denominator of the first fraction by a and the second by x, we have a/ax- x/ax= b so that (a- x)/ax= 1/b. At this point, I would multiply on both sides by abx: b(a- x)= ax so that ba- bx= ax. Can you finish this?

    2. Get the x alone on one side by adding 1/a to both sides: 1/x= 1/a+ 1/b. You can add the two fractions on the right by, as before, getting a common denominator: 1/x= b/ab+ a/ab= (b+ a)/ab. Then get x by taking the reciprocal of both sides.
  5. Aug 3, 2013 #4
    Fabulous! Yes, I can do it:

    ba - bx = ax

    1. Factor out the b:

    b(a - x) = ax

    2. Distribute:

    ab - bx = ax

    3. Get all the x's on one side:

    ab = bx + ax

    4. Factor out the x:

    ab = x(b + a)

    5. Solve for x:

    ab/(b + a) = x

    Thank you! The key was finding the common denominator to make a single fraction out of the two terms (is "term" the correct word here?) or two fractions.

    I just learned a new rule from the above:

    If x/y = a/b, then y/x = b/a, provided that none of the terms here equal zero

    Thank you for that too!
    Last edited: Aug 3, 2013
  6. Aug 3, 2013 #5
    Ok, thanks for the hint. It's been a long time, I wasn't able to solve the above until I learned from HallsofIvy to find the common factor for the two denominators of the fractions on one side.

    So here goes, using your method:

    1/x - 1/a = 1/b

    b(1/x - 1/a) = b(1/b)

    b/x - b/a = b/b

    b/x - b/a = 1

    ba/xa - bx/xa = 1

    (ba - bx)/xa = 1

    1(xa) = 1(ba - bx)

    xa + bx = ba

    x(a + b) = ab

    x = ab/(a + b)

    Wow! Both solutions in both of my posts match. Thank you!
  7. Aug 3, 2013 #6
    Mileena, in your first with a solution your first two steps are unnecessary.

    You said to factor and distribute (virtually undoing the factoring.

    It is not wrong and you came to the right answer, I just wanted to point it out so that you know you can start the problem from step 3.
  8. Aug 8, 2013 #7
    Of course you are right Jufro! Thank you.

    In my preparation the last week for the pre-calculus assessment test I will be taking in a few days, I became more comfortable with that! I don't know what I was thinking when I posted that sequence.

    Also, sorry for my late reply.
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