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Solving Rational Equations: More Examples (page 3 of 3)


  • Solve the following equation:
    • x + 1 = 72 / x

    There is only one fraction, so the common denominator is just x, and the solution cannot be x = 0. To solve, I can start by multiplying through on both sides by x:

      (x)(x) + (1)(x) = (72/x)(x/1)

      x2 + x = 72
      x2 + x – 72 = 0
      (x + 9)(x – 8) = 0
      x = –9 or x = 8

    ...or I can convert to the common denominator and solve the numerators:

     

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      (x/1)(x/x) + (1/1)(x/x) = 72/x

      x2 + x = 72
      x2 + x – 72 = 0
      (x + 9)(x – 8) = 0
      x = –9 or x = 8

    Either way, the solution is x = –9 or x = 8. Since neither solution causes a division-by-zero problem in the original equation, both solutions are valid.

      x = –9 or x = 8
  • Solve the following equation:
    • 10 / (x + 4) = 15 / [ 4 (x + 1) ]

    First, I note that I cannot have x = –1 or x = –4. Then I notice that this equation is a proportion: the equation is of the form "one fraction equals another fraction". So all I need to do here is cross-multiply.

      cross-multiply

      10(4(x + 1)) = 15(x + 4)
      40x + 40 = 15x + 60
      25x + 40 = 60
      25x = 20
      x = 20/25 = 4/5

    Since this solution won't cause any division-by-zero problems, it is valid:

      x = 4/5


As an aside, you may find it useful to remember that, when solving equations like these, you are (technically) trying to find the intersections of the functions on either side of the "equals" sign. For instance, let's return to this equation: Copyright © Elizabeth Stapel 2003-2011 All Rights Reserved

    x / (x - 2) + 1 / (x - 4) = 2 / (x^2 - 6x + 8)

Let either side of the equation be its own function:

    y = x / (x - 2) + 1 / (x - 4)

    y = 2 / (x^2 - 6x + 8)

Graphing these, you can see that they intersect in one spot:

    graph of rational functions, showing one intersection

This is the solution, x = –1, that we found earlier. But remember how we intially came up with two solutions? That was because we'd gotten to where we were ignoring the denominators. If we do that, we get the following functions:

    y3 = x2 – 3x – 2
    y4 = 2

These graph like this:

    graph of numerators, showing correct and incorrect solutions

As you can see, getting rid of the denominators created an additional (and wrong) solution.

If you have a graphing calculator, by the way, keep in mind that a quick graph can allow you to check your answers (since solutions to the equations are intersections of the graphs) before you hand in your test. Just make sure your solutions match the x-values of the intersection points on the graphs.


Solving rational equations is pretty straightforward if you are careful to write each step completely. But (warning!) as soon as you start skipping steps or doing stuff in your head, you're going to start messing up. So always work neatly and completely. And never forget to check your solutions, because I can just about guarantee that you'll have one of those "no solution" (or "only one solution works") problems on your test.

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Cite this article as:

Stapel, Elizabeth. "Solving Rational Equations: More Examples." Purplemath. Available from
    http://www.purplemath.com/modules/solvrtnl3.htm. Accessed
 

 

 

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