# How to Solve a Rational Equation

A rational equation is one that contains fractions. Yes, we will be finding a common denominator that has ‘x’s. But no worries! Together we will use a process that will help us solve rational equations every time!

Rational Equations

A rational equation is an equation that contains fractions with xs in the numerator, denominator or both. Here is an example of a rational equation: (4 / (x + 1)) – (3 / (x – 1)) = -2 / (x^2 – 1).

Let’s think back for a moment about solving an equation with a fraction. 1/3 x = 8. We think of the 3 in the denominator as being a prisoner, and we want to release it. To set the 3 free, we multiply both sides of the equation by 3. Think of it as 3 letting both sides of the equation know he’s leaving. 3 (1/3 x) = 8 (3).

This process freed our denominator and got rid of the fraction – x = 24. It is also the process we use to solve rational equations with one extra step. In rational equations, sometimes our solution may look good, but they carry a virus; that is, they won’t work in our equation. These are called extraneous solutions. The steps to solve a rational equation are:

Find the common denominator.

Multiply everything by the common denominator.

Simplify.

Check the answer(s) to make sure there isn’t an extraneous solution.

Let’s solve a couple together.

Example #1

Example number one: solve. Remember to check for extraneous solutions. (3 / (x + 3)) + (4 / (x – 2)) = 2 / (x + 3).

Our first step is to figure out the terms that need to be released from the denominators. I look at 3 / (x + 3). I write down (x + 3) as one of my common denominators. I look at 4 / (x – 2). I write down (x – 2) as another part of my common denominator. I look at 2 / (x + 3). Since I already have (x + 3) written in my denominator, I don’t need to duplicate it.

Next, we multiply everything by our common denominator – (x+3)(x-2). This is how that will look: ((3(x + 3)(x – 2)) / (x + 3)) + ((4(x + 3)(x – 2)) / (x – 2)) = (2(x + 3)(x – 2)) / (x + 3))

It isn’t easy for the denominators to be released; there is a battle, and like terms in the numerator and denominator get canceled (or slashed). Slash (or cancel) all of the (x + 3)s and (x – 2)s in the denominator and numerator. Our new equation looks like: 3(x – 2) + 4(x + 3) = 2(x – 2).

In example #1, the first step is finding the common denominator.

Finding the common denominator

Distribute to simplify: (3x – 6) + (4x + 12) = 2x – 4. Collect like terms and solve. 3x + 4x = 7x, -6 + 12 = 6. We end up with 7x + 6 = 2x – 4.

Subtract 2x from both sides: 7x – 2x = 5x. Subtracting from the other side just cancels out the 2x, and we get 5x + 6 = -4. Subtract 6 from both sides: -4 – 6 = -10. Again, subtracting 6 will cancel out the +6, so we end up with 5x = – 10. Divide by 5 on both sides, and we cancel out the 5 and give us x = – 2. It turns out x = – 2.

The reason we check our answers is that sometimes we get a virus, or, in math terms, extraneous solutions. To check, I replace all the xs with -2: (3 / (-2 + 3)) + (4 / (-2 – 2)) = (2 / (-2 + 3)). Let’s simplify: (3 / 1) + (4 / -4) = (2 / 1). Since 3 + -1 = 2 is true, x = – 2 is the solution!

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