Solving Rational Equations: Examples


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Solving Rational Equations: Examples (page 2 of 3)

Solve the following equation:

The common denominator here will be x(x – 2), and x cannot be zero or 2. I can solve this equation by multiplying through on both sides of the equation by this denominator:

[10/(x(x - 2)] [x(x - 2)/1] + (4/x) [x(x - 2)/1] = [5/(x - 2)] [x(x - 2)/1]

10 + 4(x – 2) = 5(x)
10 + 4x – 8 = 5x
4x + 2 = 5x
2 = x

Or I can solve by converting to the common denominator and then solving the numerators:

10/[x(x - 2)] + (4x - 8)/[x(x - 2)] = 5x/[x(x - 2)]

10 + (4x – 8) = 5x
10 + 4x – 8 = 5x
4x + 2 = 5x
2 = x

Either way, the answer I get is:

However, I need to check this solution with the original equation. Do you see that I'm going to have a problem with x = 2? This value would cause division by zero in the original equation! Since the only possible solution causes division by zero, then this equation really has no solution.

no solution

How did we end up with an invalid solution? We didn't do anything wrong. But notice that, whichever method you use to solve a rational equation, at some point you're going to get rid of the denominators. For rational equations, the difficulties come from those denominators. So whenever you solve a rational, always check the solution against the denominators in the original problem. It is entirely possible (not commonly in homework, but almost always on the test) that a problem will have an invalid ("extraneous") solution.

Solve the following equation:

First I'll need to factor that quadratic, so I can tell what factors I'll have in my common denominator.

x² – 6x + 8 = (x – 4)(x – 2)

That worked out nicely: the factors of the quadratic are duplicates of the other denominators. (This often happens for these problems.) So the common denominator will be (x – 4)(x – 2)., and I'll need to remember (at the end) that x cannot be 2 or 4.

I can convert everything to the common denominator and then solve the numerators:

(x^2 - 4x)/[(x - 2)(x - 4)] + (x - 2)/[(x - 2)(x - 4)] = 2/[(x - 2)(x - 4)]

(x² – 4x) + (x – 2) = 2
x² – 4x + x – 2 = 2
x² – 3x – 4 = 0
(x – 4)(x + 1) = 0
x = 4 or x = –1

...or I can multiply through on both sides by the common denominator and solve the resulting equation:

[x/(x - 2)] [(x - 2)(x - 4)]/1 + [1/(x - 4)] [(x - 2)(x - 4)]/1 = 2/[(x - 2)(x - 4)] [(x - 2)(x - 4)]/1

x(x – 4) + 1(x – 2) = 2
x² – 4x + x – 2 = 2
x² – 3x – 2 = 2
x² – 3x – 4 = 0
(x – 4)(x + 1) = 0
x = 4 or x = –1

Either way, I get the same result: x = 4 and x = –1. Checking these solutions against the denominators of the original equation, I see that "x = 4" would cause division by zero, so I throw that solution out. Then the answer is:

x = –1

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Cite this article as:

Stapel, Elizabeth. "Solving Rational Equations: Examples." Purplemath. Available from
http://www.purplemath.com/modules/solvrtnl2.htm. Accessed

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