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

Either way, the answer I get is:

x = 2 Copyright © Elizabeth Stapel 2003-2011 All Rights Reserved

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.

You can use the Mathway widget below to practice "Rational Expressions and Equations", subtopic "Solving Rational Equations". Try the entered exercise, or type in your own exercise. Then click "Answer" to compare your answer to Mathway's. (Or skip the widget and continue with the lesson.)

To activate, click in the "Enter Problem" box below:

(Clicking on "View Steps" on the widget's answer screen will take you to the Mathway site, where you can register for a free seven-day trial of the software.)

• 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:



(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 – 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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