Copyright

Solving Rational Equations with Literal Coefficients

Solving Rational Equations with Literal Coefficients
Coming up next: Solving Problems Using Rational Equations

You're on a roll. Keep up the good work!

Take Quiz Watch Next Lesson
 Replay
Your next lesson will play in 10 seconds
  • 0:05 Literal Coefficients
  • 0:55 Rational Equations…
  • 1:27 Step 1: The Common Denominator
  • 3:29 Step 2: Solving the…
  • 4:49 Lesson Summary
Add to Add to Add to

Want to watch this again later?

Log in or sign up to add this lesson to a Custom Course.

Log in or Sign up

Timeline
Autoplay
Autoplay
Speed

Recommended Lessons and Courses for You

Lesson Transcript
Instructor: Yuanxin (Amy) Yang Alcocer

Amy has a master's degree in secondary education and has taught math at a public charter high school.

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)

To unlock this lesson you must be a Study.com Member.
Create your account

Register to view this lesson

Are you a student or a teacher?

Unlock Your Education

See for yourself why 30 million people use Study.com

Become a Study.com member and start learning now.
Become a Member  Back
What teachers are saying about Study.com
Try it risk-free for 30 days

Earning College Credit

Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.

To learn more, visit our Earning Credit Page

Transferring credit to the school of your choice

Not sure what college you want to attend yet? Study.com has thousands of articles about every imaginable degree, area of study and career path that can help you find the school that's right for you.

Create an account to start this course today
Try it risk-free for 30 days!
Create An Account
Support