Rational Functions and Equations

This section covers:


We need to go back and do a little more algebra work in here somewhere; here’s to those who really love playing around with all those variables!  We’ll first start out with more complicated factoring and then go on to adding, subtracting, multiplying and dividing fractions with variables on the bottom (Rationals).  But since factoring is so important in algebra, let’s revisit it first.  Remember that we first learned factoring here in the Solving Quadratics by Factoring and Completing the Square section.  More advanced factoring can be found in the Solving and Factoring section.

Revisiting Factoring

Earlier, we learned how to “unfoil” a trinomial into two binomials:

Factoring Trinomials

Sometimes we have to factor out some stuff before we do the foiling.  We always want to do this first:

Factoring with GCF

And don’t forget “grouping” when we have four terms (but it doesn’t always work – we’ll find other ways to solve in Graphing and Solving Polynomials later).  Again, we are working with GCF’s to do this:

Factoring with 4 Terms

You can factor a difference of squares, but not a sum of squares:

Difference and Sum of Cubes

For cubic binomials (sum or difference of cubes), we can factor:

SOAP

Here are some examples of factoring sums and differences of cubes:

Sum and Difference of Cubes

Introducing Rational Expressions

A rational expression is simply any fraction where either the numerator or denominator or both have a variable in it.  Think of a rational expression as a ratio of two polynomials.

Here are some examples of expressions that are and aren’t rational expressions:

Tables of Rationals

Multiplying and Dividing, and Simplifying Rationals

Frequently, rationals can be simplified by factoring the numerator, denominator, or both, and crossing out factors.  They can be multiplied and divided like regular fractions.

Here are some examples.  Note that these look really difficult, but we’re just using a lot of steps of things we already know.  That’s the fun of math!  Also, note in the last example, we are dividing rationals, so we flip the second and multiply.

Remember that when you cross out factors, you can cross out from the top and bottom of the same fraction, or top and bottom from different factors that you are multiplying.  You can never cross out two things on top, or two things on bottom.

Multiplying Dividing and Simplifying Rationals

Finding the Common Denominator

When we add or subtract two or more rationals, we need to find the least common denominator (LCD), just like when we add or subtract regular fractions.  If the denominators are the same, we can just add the numerators across, leaving the denominators as they are.  We then must be sure we can’t do any further factoring:  Adding Rationals

Just like with regular fractions, we want to use the factors in the denominators in every fraction, but not double-count them if they appear in all denominators.  When nothing is common, just multiply the factors.

Let’s find the least common denominators for the following denominators (ignore the numerators for now).

LCD of Rationals

Now let’s add and subtract the following rational expressions.

Note that one way to look at finding the LCD is to multiply the top by what’s missing in the bottom.  For example, in the first example, the LCD is (x – 3)(x + 4), and we need to multiply the first fraction’s numerator by (x + 4), since that’s missing in the denominator.

Adding and Subtracting Rationals

Restricted Domains of Rational Functions

To find the domain of a rational function, since there are typically variables in the denominator, we must make sure that the denominators won’t end up as “0”.

So the domain of Domain of Rational  This means if we ever get a solution to an equation that contains rational expressions and has variables in the denominator (which they probably will!), we must make sure that none of our answers would make any denominator in that equation “0”.  These “answers” that we can’t use are called extraneous solutions.  We’ll see this in the first example below.

Solving Rational Equations

When we solve rational equations, we can multiply both sides of the equations by the least common denominator (LCD) and not even worry about working with fractions!   The denominators will cancel out and we just solve the equation using the numerators.  Remember that with quadratics, we need to get everything to one side with 0 on the other side and either factor or use the Quadratic Formula.

Again, think of multiplying the top by what’s missing in the bottom from the LCD.

Notice that sometimes you’ll have to solve literal equations, which just means that you have to solve an equation for a variable, but you’ll have other variables in the answer.  The last example shows this.

Solving Rational Equations

Applications of Rationals

There are certain types of word problems that typically use rational expressions.  These tend to deal with rates, since rates are typically fractions (such as distance over time).  We also see problems dealing with plain fractions or percentages in fraction form.

Fraction Problems

The denominator of a fraction is 2 less than twice the numerator.  If 7 is added to both the numerator and denominator, the resulting fraction is Four Fifths  What is the original fraction?

Picture of FractionSolution:

This problem is a little tricky, since we don’t want to set the variable to what the problem is asking – the original fraction.  We want to set n = the numerator, since we’ll have to get both the numerator and the denominator.  Since the denominator is in terms of the numerator, it’s easier to make the variable the numerator.   So, from the first sentence of the problem, the  original fraction is n over 2n - 2.

Now to solve, we just have to add 7 to the numerator and denominator, and set to Equation for Fraction Problem

Now let’s do the math:

Rationals Fraction Word Problem

OK, it was totally coincidental that the answer was the same fraction as in the picture above.  Weird!   😆

Problem: 

Bethany has scored 10 free throws out of 18 tries.  She would really like to bring her free throw average up to at least 68%.  How many consecutive free throws should she score in order to bring up her average to 68%?

Solution:

Let’s let x = the number of free throws that Bethany should score (in a row) in order to bring up her average.  Again we can use fractions, and this time they will represent the fraction of free throws that she scores.  We’ll start out with her current fraction (rate) of consecutive free throws, and then we’ll add the number she needs to score to both the numerator (number she scores) and denominator (total number of throws):  Free Throw Fraction

Let’s do the math:

Free Throw Rational Problem

Distance/Rate/Time Problems

With rational rate problems, we must always remember:  Distance = Rate x Time.  It seems most of the problems deal with comparing times or adding times.

Problem:

Shalini can run 3 miles per hour faster than her sister Meena can walk.  If Shalini ran 12 miles in the same time it took Meena to walk 8 miles, what is the speed of each sister in this case?

Solution:

This is a “Distance = rate  time” problem, and let’s go ahead and use a table to organize this information like we did in the Algebra Word Problems and Systems of Linear Equations and Word Problems sections.  Let’s let x = Meena’s speed, since Shalini runs faster and it’s easier to add than subtract:

Work Problem with Speed

Now let’s do the fun part – the math:

Work Problem Table

Problem:

Here’s a problem where we use the rates of a boat going upstream (rate in still water rate minus the rate of the water current) and downstream (rate in still water plus the rate of the water current) to add times.  (Think when you’re going downstream, it’s like you’re going down a hill, so it’s faster.)

CanoeThe time it takes for a canoe to go 3 miles upstream and back 3 miles downstream is 4 hours.  The current in the lake is 1 mile per hour.  Find the average speed (rate) of the canoe in still water.

Solution:

We’ll let x = the speed of the boat in still water.  So we know that the rate in the canoe going upstream (again, think that “up” is more difficult so we go slower) is “x – 1”, and the rate going downstream is “x + 1”.   (Think about going “down” a hill; we go faster).

We add the time upstream to the time downstream, and this should equal 4 hours.

Canoe Rational Word Problem


Note:  If we were given the rate of the canoe in still water and had to find the rate of the current, we’d do the problem in a similar way.  We’d have to add and subtract variables (the rate of the current) from numbers (the rates of the canoe in still water) in the denominators.

Work Problems:

Work problems typically have to do with different people or things working together and alone, at different rates.

I find that usually the easiest way to work these problems is to remember: Work Equation (or whatever part of the job or jobs is done; if they do half the job, this equals  \(\frac{1}{2}\);  if they do a job twice, this equals 2).

Note: This is because, for work problems, Rate x Time  =  1 Job.    The individual rates of each worker is One Over Time to do One Job, and when you multiply the rate by the time they work, you get Time Working Together Fraction.  Add all this up and and set it equal to 1 job.

Problem: 

Girl Painting

Erica can paint her room in 5 hours.  If she has her friend Rachel help her, they can paint the room together in 3 hours.  How long would it take for Rachel to paint the room alone, if Erica wanted to go play tennis for the afternoon?

Solution:

We’ll let R = amount of time (in hours) Rachel can paint the room by herself.  We know that the time the girls paint the room together is 3 hours, and the time Erica paints the room by herself is 5 hours.

Work Problem with Painting

Note:  The formula above can also be derived by using the concept that you can figure out how much of the job the girls do per hour (or whatever the time unit is), both together and alone.  Then you can add the individual “rates” to get the “rate” of their painting together.

We are actually adding the Work they complete (together and alone) using formula Work = Rate times Time, where the Time is 1 hour (or whatever the unit is).

In this example, Erica’s rate per hour is One Fifth (she can do One Fifth of the job in 1 hour);  Rachel’s rate per hour is One over R; we can add their rates to get the rate of their painting together: Work Equation.  If we multiply all the terms by 3, we come up with the equation above!  (Sometimes you see this as  One Over Time Alone Equation.)

Problem:

Sandwich

It takes Jill 3 hours to make sandwiches for a charity organization (as many as they need).  Allie starts one hour later and (if she were working alone) it would take her 4 hours to make the sandwiches needed.  How long will it take both of them working together to complete the job?

Solution:

This one’s a little trickier since “Allie starts one hour later”.  Since it takes Jill 3 hours to make sandwiches by herself, after one hour (when Allie starts), she’s already one third of the way through the job (one third of 3 hours is 1 hour).   The amount of job left then is two thirds.

We’ll let T = amount of time (in hours) the girls can make the sandwiches together.  We know it takes Jill 3 hours to make them alone, and Allie 4 hours to make them alone.  So we can use the Job Formula formula, but in this case, we’ll use Two Thirds job instead of 1:

Making Sandwiches Work Problem

Problem:

Pool

Two hoses are used to fill Maddie’s neighborhood swimming pool.  One hose alone can fill the pool in 10 hours; the second hose can fill it in 12 hours.  The pool’s drain can empty the pool in 8 hours.   If the two hoses are working, and the drain is open (by mistake), how long will it take to fill the swimming pool?

Solution:

This is the same type of “work” problem where we have “alone” times, but want to find the “together” time.  The difference with this problem is that we have 3 different things working to fill the pool, one of which is working against filling the pool (the drain), which we need to make negative.

Let’s let x = the time it takes for the 2 hoses (positive work) and the drain (negative work) all together to fill the pool:

Work Problem with Positive and Negative

Problem

8 women and 12 girls can paint a large mural in 10 hours.  6 women and 8 girls can paint it in 14 hours.  Find the time by 1 woman alone, and 1 girl alone to paint the mural.

Solution:

This is actually a systems problem and also a work problem at the same time.  We did this problem without using rationals here in the Systems of Linear Equations and Word Problems section (and be careful, since the variables we assigned were different).

Let’s use the approach that you can figure out what per hour rates are (together and alone), and add the individual “rates” to get the “rate” of their painting together.  (We’ll not use theJob Formulathis time).

Let’s let w = the time (in hours) it takes for one woman to paint the mural alone, and g = the time it takes for one of the girls to paint the mural alone.  Set up the systems, and solve:

Work Problem with Systems

Cost Problems

A third type of problem you might get while studying rational has to do with costs per person (or unit cost), which is again a rate.

Problem:

Girls at Beach

A group of girls decided to go on a trip to the beach and the organizer said that the bus would cost $360 to rent.  The organizer also told them that if they got 3 more girls to go on the trip, each girl could pay $6 less (which they ended up doing).  How many girls ended up going on the trip?

Solution:

Let n be the original number of girls going on the trip.  The cost per girl for the original trip (like a rate) is 360 over n (use real numbers to see this – if 10 girls were going, it would be $36 per girl).

If they added 3 more girls, each girl would have to pay 360 over n + 3 since the total cost is still $360, but the number of girls isn + 3

So now with n + 3 again girls, the new cost is 6 dollars less, or the original cost minus 6, so we can set up an equation like this: New Formula

Let’s do the math:

Cost Problem with Rationals


Understand these problems, and practice, practice, practice!


Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.

If you click on “Tap to view steps”, you will go to the Mathway site, where you can register for the full version (steps included) of the software.  You can even get math worksheets.

You can also go to the Mathway site here, where you can register, or just use the software for free without the detailed solutions.  There is even a Mathway App for your mobile device.  Enjoy!

On to Graphing Rational Functions, including Asymptotes  – you’re ready! 

26 thoughts on “Rational Functions and Equations

  1. Hi, I absolutely love this site. However, I’d love it if you could work a bit faster! I’m starting Calculus in a couple of weeks, and I would love more resources. Thanks!

    • Amy,
      Thanks so much for your comment. You are absolutely right; I need to write faster! I’ve been out of town a lot lately (my 3 kids moved out of state and I visited all of them 😉 and am busy now with my tutoring, but want to finish this year!

      Your comment makes me want to work harder – thanks again 😉
      Lisa

      • THANK YOU SO MUCH!!!!!!!!!
        YOU’RE POST IS REALLY HELPFUL FOR MY HOMEWORK, NOW I CAN FULLY UNDERSTAND THESE PROBLEMS!! THANK YOU SO MUCH!

  2. I just recently found your website and I love it, thank you! I’m in college algebra and I really need help and I came across your site last week and this has been a miracle for me!!! I do great in class but when I get home its a different story! Again, thank you from the bottom of my heart!
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    • Kristie,
      Thank you so so much for taking the time to write me! I’m so glad you found the site, and your message wants me to keep writing!! Please let me know if you have any questions on anything, and spread the word about She Loves Math 😉
      Lisa

    • So sorry the section confused you. If you have a specific question or problem, could you let me know and I’ll try to clarify it? Thanks, Lisa

  3. i love math because of this ..but can you help me understand my lesson because my assignment is really hard for me to answer those worded problem . may i get more example of word problem about rational function

  4. This Math is a torture but when I started reading your lectures I made my bad math a big change. Thanks a bunch! XD
    (PS: My teacher is a 3 degree one and she teaches us as if we were already collage students)

  5. This is great !
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    i don’t know how to graph that, since my teacher is asking for a graphing image:( could this site help ?

  6. You are the head of the general services division of a famous resort in your locality with the widest swimming pool. Five pipes we’re installed in the pool. Three of which one used for filling and the other two for draining. The filling pipes A,B, and C can fill the pool in 9 hours, 7 hours and 10 hours respectively. The pool can be drained in 12 hours using pipe D and 15 hours using pipe E.
    The owner want you to change the water of the pool immediately because of some impurities. But it takes a longer time for you to make it filled up than the normal time needed. Then you found out the pipes D and E were open while pipes A, B, and C were filling the pool. Your report should contain the computation for the total number of hours it took to fill the swimming pool and it’s difference from the normal of hours required to fill the pool. ??

  7. You’re a gift of God. Thankyou for this. It’s our midterm tomorrow and it really helps me a lot! 🙂 Continue helping other people. Godbless! 🙂

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