Solving Rational Equations

Solving Rational Equations with Literal Coefficients

Watch this video lesson and be amazed at how you can solve a rational equation with numbers and letters in it. Learn how to solve rational equations with literal coefficients, and you will learn how to solve any rational equation you come across!

Literal Coefficients

When I hear the phrase 'solving rational equations with literal coefficients,' it actually does scare me a bit. It sounds like a BIG problem, but you know what? Once I get into the whole process of it and remember my two basic steps, the big scary problem that used to be an elephant is now a mouse that I can easily handle.

What I need to know first before solving this type of problem and what I want you to understand first is what a literal coefficient is. Simply stated, a literal coefficient is a variable used to represent a number. The number the variable represents can be either known or unknown. It can be our usual x or y, or it can be other letters, such as a, b, or c. Although we can have more than one variable in an equation, we will consider the common case where there is only one literal coefficient.

Rational Equations with Literal Coefficients

We now know what a literal coefficient is, but what about rational equations? An equation with a fraction made up of polynomials is a rational equation. You can identify them easily by looking for fractions with polynomials in the denominator and numerator. If you see one, then you know you have a rational equation. All of these are examples of rational equations:

• t/(t + 2) + 1/2 = 1
• s/2 - 2 = (s + 3)/4
• 1/x + 4/(x + 2) = 5/(x - 1)

Don't get scared, but we are going to work with the largest problem here, the one with the x variable. I'll show you how to go about working with and solving problems like this so they don't become a headache for you.

Step 1: The Common Denominator

• 1/x + 4/(x + 2) = 5/(x - 1)

The first step is to find the common denominator. Our current problem has three denominators we need to look at. We have an x, an x + 2, and an x - 1. Finding common denominators when you have variables involved is slightly different than when you only have numbers involved. I think it's easier when you have variables involved. Why do I say that? Let me show you.

When we have variables in our denominators, to find our common denominator, all we have to do is to write down all the factors we see. For our problem, the first fraction has an x for the denominator, so our common denominator will have an x as well. The second fraction has an x + 2, so we will add that to our common denominator since we don't have that yet. If our common denominator already had it, we won't need to add it. Our common denominator is now x (x + 2). The last fraction has an x - 1. I look at my common denominator and see that I don't have that yet, so I will add that on, too. When adding factors to the common denominator, I multiply them. So, our common denominator is x (x + 2) (x - 1). If our denominators had numbers only, we would go ahead and find the least common multiple of the numbers involved.

• x(x + 2)(x - 1)

We are going to use this common denominator to help us solve our rational equation. What we are going to do is to multiply each term by our common denominator. Watch what happens when we do this. You'll like it. It simplifies the problem into something I know you can manage.

Because we are multiplying each term by the common denominator, we can cancel like terms. The first term has a common x factor in both the top and bottom. The second one has a common x + 2 in both the top and bottom that can be cancelled and the third has a common x - 1 that can be cancelled. Our problem now looks like this:

• (x + 2)(x - 1) + 4x(x - 1) = 5x(x + 2)

I consider this a lot more manageable than what we began with. Don't you agree? No more fractions! Yay! Now we get to use our algebra skills to solve for x.

Step 2: Solving the Rational Equation

We will multiply the factors out making sure to follow the order of operations.

• 5x^2 - 3x - 2 = 5x^2 + 10x

We then move all our variables to the same side and simplify. I'm going to choose the left side. You can choose whichever side you want.

• -13x - 2 = 0

I see that I'm left with a linear equation. This means I can move my number over to the other side and solve for x.

• -13x = 2

We have solved for x, but we are not done. Because a rational equation has fractions, we need to go back and plug in our answer to see if it causes division by zero in any of our fractions. We can do this by plugging our answer into each denominator separately to see if it gives us a zero in the denominator. If all the denominators are okay, meaning none end up being zero, then our answer is valid.

Checking our answer of - 2 / 13, and plugging it into each denominator, we get - 2 / 13 for the first denominator, - 2 / 13 + 2 = 24 / 13 for the second denominator, and - 2 / 13 - 1 = - 15 / 13 for the third denominator. None of these are zero, so my answer is valid.

Also, if we ended up with a quadratic equation after multiplying by the common denominator and simplifying, we would need to use our quadratic equation solving skills such as factoring or the quadratic formula to help us solve the problem.

Lesson Summary

In summary, solving rational equations with literal coefficients is not terribly bad. We recall that a literal coefficient is a variable used to represent a number and that a rational equation is an equation with fractions made up of polynomials. Solving these problems is a two-step process. The first step requires us to find the common denominator and then to multiply the whole equation by it to simplify our equation so we can solve it easier. The second part requires our algebra skills in solving for the literal coefficient or variable.

Learning Outcomes

After you've reviewed this video lesson, you will be able to:

• Define literal coefficient and rational equation
• List the two steps involved in solving rational equations with literal coefficients
• Solve basic problems involving rational equations with literal coefficients