Solving Rational Equations

Now that we know domains and all that jazz, we can solve those pesky rational equations. Not a problem.

Here's the deal.

First, we'll find the LCM of all the denominators. Next, we'll multiply each term of the equation by the LCM. Then we'll solve just like we would with a linear or quadratic equation. No one said this would be excessively easy, but it's doable. Trust us.

One tricky thing to remember: Always be on the lookout for solutions that make the denominator zero. We'll call them extraneous solutions, and they don't count as actual solutions to the equation. They're mathematical Death Stars, destroying everything in their path. Always double check to make sure that this isn't the case.

Sample Problem

Solve

We need to clear the equation of fractions.

The LCM of the denominators is 2(x +8). Multiply each term by the LCM like this:

Cancel identical factors.

6(2) = x + 8

12 = x + 8

Subtract 8 from both sides.

x = 4

Don't forget to check to make sure the solution doesn't blow up the denominator and that it works out properly. Double check using mental math, pen or pencil, or even abacus, but always check.

So does 4 check?

Does ?

Yes, .

An easier way to check would be to see if x = 4 is in the domain. The domain is everywhere but where the denominator is zero. In this example, that would be everywhere except x = -8.

So x = 4 is definitely in the domain, too. Check, double check.

Sample Problem

Solve .

First, multiply both sides by the LCM of the denominators. This time around, that's just (x – 1).

Cancel those (x – 1) terms on the left.

20 = (x – 1)(x –2)

Multiply the right side out.

20 = x² – 3x + 2

Subtract 20 from both sides.

x² – 3x – 18 = 0

Factor the left side.

(x – 6)(x + 3) = 0

Set each factor equal to zero.

x – 6 = 0 and x + 3 = 0

Which give us solutions of: x = 6 and x = -3.

Now, see if they're extraneous solutions by checking if they're living in the domain.

The domain for this equation is everywhere where all denominators are not zero. The only denominator we need to deal with is x – 1.

So, the domain is everywhere except x = 1.

Both x = 6 and x = -3 are legit solutions.

Q.E.D. and bravo. 

Sample Problem

Solve .

First, multiply through by (x – 2).

Now cancel.

2 = 2 + 2(x – 2)

2 = 2 + 2x – 4

2 = 2x – 2

2x = 4

x = 2

Now, check if this is an extraneous root. Does this cause a denominator in the original equation to be zero?

Yes, so x = 2 is an extraneous solution.

The answer is "no solution."

Oh well, can't win 'em all.