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Solving Rational Equations: Introduction (page 1 of 3)

While adding and subtracting rational expressions is a royal pain, solving rational equations is much simpler. (Note that I don't say that it's "simple", just that it's "simpler".) This is because, as soon as you go from a rational expression (with no "equals" sign in it) to a rational equation (with an "equals" sign in the middle), you get a whole different set of tools to work with. In particular, you can multiply through on both sides of the equation to get rid of the denominators.

  • Solve the following equation:
    • 2/3 = x / 3

    This equation is so simple that I can solve it just by looking at it: since I have two-thirds equal to x-thirds, clearly x = 2. The reason this was so easy to solve is that the denominators were the same, so all I had to do was solve the numerators.

      x = 2

  • Solve the following equation:
    • (x - 1) / 15 = 2 / 5




    To solve this, I can convert to a common denominator of 15:

      (x - 1) / 15 = 6 / 15

    Now I can compare the numerators:

      x – 1 = 6
      x = 7

    Note, however, that I could also have solved this by multiplying through on both sides by the common denominator:

      [ (x - 1) / 15] (15/1) = (2/5) (15/1)

      x – 1 = 2(3)
      x – 1 = 6
      x = 7

When you were adding and subtracting rational expressions, you had to find a common denominator. Now that you have equations (with an "equals" sign in the middle), you are allowed to multiply through (because you have two sides to multiply on) and get rid of the denominators entirely. In other words, you still need to find the common denominator, but you don't necessarily need to use it in the same way.

Here are some more complicated examples:

  • Solve the following equation:
    • 3 / (x + 2) - 1 / x = 1 / 5x

    First, I need to check the denominators: they tell me that x cannot equal zero or –2 (since these values would cause division by zero). I'll re-check at the end, to make sure any solutions I find are "valid".

    There are two ways to proceed with solving this equation. I could convert everything to the common denominator of 5x(x + 2) and then compare the numerators:

      15x/[5x(x + 2)] - (5x + 10)/[5x(x + 2)] = (x + 2)/[5x(x + 2)]

    At this point, the denominators are the same. So do they really matter? Not really (other than for saying what values x can't be). At this point, the two sides of the equation will be equal as long as the numerators are equal. That is, all I really need to do now is solve the numerators:

      15x – (5x + 10) = x + 2
      10x – 10 = x + 2
      9x = 12
      x = 12/9 = 4/3

    Since x = 4/3 won't cause any division-by-zero problems in the fractions in the original equation, then this solution is valid. Copyright © Elizabeth Stapel 2003-2011 All Rights Reserved

      x = 4/3

I said there were two ways to solve this problem. The above is one method. Another method is to find the common denominator but, rather than converting everything to that denominator, I'll take advantage of the fact that I have an equation here, and multiply through on both sides by that common denominator. This will get rid of the denominators:

    (3/(x + 2))(5x(x + 2))/1 - (1/x)(5x(x + 2))/1 = (1/5x)(5x(x + 2))/1

    3(5x) – 1(5(x + 2)) = 1(x + 2)
    15x – 5x – 10 = x + 2
    10x – 10 = x + 2
    9x = 12
    x = 12/9 = 4/3

This method gives the same result as the first method. I view this second method as being quicker and easier, but this is only my personal preference. My students have typically been fairly evenly divided in their preferences for the two methods. I will do each of the examples in the following pages both ways. You should pick the method that works best for you.

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Cite this article as:

Stapel, Elizabeth. "Solving Rational Equations: Introduction." Purplemath. Available from Accessed



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