Lecture: Rational Expressions & Equations
Lecture script · about 15–20 minutes spoken · Week 11
Opening
Think about a group of friends splitting a 600-dollar cabin rental. With 6 people, each share is 100 dollars. Add two more friends, and each share drops to 75 dollars. The share is a fraction, 600 over the number of people, and the unknown sits on the bottom. That same kind of fraction is how a contractor figures how long two crews take working together, and how a pharmacy technician works out a dose per kilogram. In this lecture you'll learn to handle fractions like that, and to keep every answer honest.
Here's where we are in this course. In Week 2 you cleared fractions out of an equation by multiplying every term by the least common denominator. In Week 3 you learned that a rational function leaves out any input that makes its denominator zero. In Week 6 you factored, and in Week 7 you solved quadratic equations. In Week 10 you saw that a factor cancelling from the top and bottom leaves a hole in a graph. Now we put all of that to work on the fractions themselves. Next week, in Week 12, Radicals, Rational Exponents & Radical Equations, the same check-every-answer habit comes back.
I'll take eight ideas in order: what a rational expression is, how to simplify one, the week's quotient as a table, multiplying and dividing, adding, subtracting, solving equations, and work-rate problems.
1. Meet a rational expression and its excluded values
Start with something you already believe. You can share 2 pizzas among 3 people. You cannot share 2 pizzas among 0 people. A fraction with zero on the bottom has no meaning at all.
Now let the top and the bottom be polynomials. A rational expression is one polynomial divided by another, such as x squared minus 9, over x plus 3; it is defined for every input except the ones that make its denominator 0. That last part matters so much that it gets its own name. An excluded value is an input that makes a denominator equal 0, so the expression or equation is undefined there; it is found before any simplifying or solving and can never be a solution.
For x squared minus 9 over x plus 3, look only at the bottom. Set x plus 3 equal to zero, and x is negative 3. So you write x is not equal to negative 3, and you write it first, before you touch anything else.
Think of the excluded value as a closed door in a hallway. Whatever you do to the expression later, that door stays shut.
Pause and try this. In the week's quotient, x squared minus 6x plus 5, over x minus 1, which input is excluded?
Look only at the bottom before you read on.
Here's the answer. Setting the bottom, x minus 1, equal to zero gives x equals 1, so 1 is the excluded value. The top doesn't matter for this question.
2. Simplify by factoring first: cancel factors, never terms
With plain numbers, you simplify 6 over 9 by writing the 6 as 2 times 3 and the 9 as 3 times 3, then cancelling a 3. What made that legal? The 3 multiplied the whole top and the whole bottom.
Rational expressions follow the same rule. A common factor is a factor that multiplies the whole numerator and the whole denominator of a rational expression; only a common factor may be cancelled, never a term that is added or subtracted.
So the method is: factor first, then cancel. Take x squared minus 9, over x plus 3. The top is a difference of squares, so it factors as x plus 3, times x minus 3. Now the x plus 3 multiplies the whole top and the whole bottom, so it cancels, and you're left with x minus 3. But the cancelled factor was the one that made negative 3 excluded, so you carry a note along. A restriction is the note written beside a simplified rational expression, such as x is not equal to negative 3, that keeps an excluded value out after the factor that caused it has been cancelled.
Check it at a legal input. At x equals 2, the original is 4 minus 9, over 2 plus 3, which is negative 5 over 5, or negative 1. The simplified form gives 2 minus 3, also negative 1.
Here's the classic slip. Someone looks at x plus 3, over x, and crosses out the two x's to get 3. That x on top isn't a factor. It's added to 3.
Pause and try this. Is x plus 3, over x, equal to 3? Test it at x equals 1.
Try the input before you read on.
Here's the answer. At x equals 1 the expression is 1 plus 3, over 1, which is 4, not 3. One test value is enough to show the cancel was illegal.
3. Read the week's quotient as a table and a graph
Let's watch a simplification work on numbers. The week's parabola, x squared minus 6x plus 5, from Weeks 7, 9, and 10, now sits on top of x minus 1. That's the week's quotient.
Factor the top. You need two numbers whose product is 5 and whose sum is negative 6, and those are negative 1 and negative 5. So the top is x minus 1, times x minus 5. The x minus 1 cancels, and the week's quotient equals x minus 5, for x not equal to 1.
Now build a table. At x equals 0, the top is 5 and the bottom is negative 1, so the quotient is negative 5. And 0 minus 5 is negative 5. At x equals 3, the top is negative 4 and the bottom is 2, so the quotient is negative 2. And 3 minus 5 is negative 2. Run the inputs 0, 2, 3, 4, 5, and 6, and the quotients come out negative 5, negative 3, negative 2, negative 1, 0, and 1. Those are exactly the values of x minus 5.
But what about x equals 1? The top is zero and the bottom is zero, and zero over zero is undefined. The simplified form would give 1 minus 5, which is negative 4. So the graph is the line y equals x minus 5, rising one unit for every unit you move right, with a single missing point at 1 comma negative 4. That's a hole, not a zero.
Pause and try this. The week's quotient has a hole at x equals 1. At which input does it actually equal zero?
Check the table of quotients before you read on.
Here's the answer. At x equals 5 the top is zero and the bottom is 4, and zero over 4 is 0. So the quotient's zero is at 5, while 1 is a hole.
4. Multiply and divide: factor, flip, cancel across
With numbers, 2 over 3 times 9 over 4 is 18 over 12, which is 3 over 2. You can also cancel first. The 2 and the 4 share a 2, and the 9 and the 3 share a 3, which leaves 3 over 2 with smaller numbers along the way. Dividing by 4 over 9 is the same as multiplying by 9 over 4, so 2 over 3 divided by 4 over 9 is also 3 over 2.
Rational expressions work the same way. Factor every top and bottom. If you're dividing, flip the divisor. Then cancel any factor that appears on a top and on a bottom anywhere in the product.
Here's a product. Multiply x plus 2, over x minus 3, by x squared minus 9, over x squared plus 4x plus 4. The x squared minus 9 factors as x plus 3, times x minus 3. The x squared plus 4x plus 4 is x plus 2, times x plus 2. One x plus 2 cancels, and the x minus 3 cancels. What's left is x plus 3, over x plus 2, with x not equal to 3 and not equal to negative 2. At x equals 0, the original and the simplified product both come out 3 over 2.
Division adds one thing to watch. When you divide by a fraction, that fraction can't be zero. So the divisor's own top adds an excluded value. In the chapter's division example, the divisor has x minus 2 on top. That top makes 2 an excluded value, even after you flip it to the bottom.
Pause and try this. In 2 over 3 divided by 4 over 9, which fraction do you flip?
Name it before you read on.
Here's the answer. You flip only the divisor, the 4 over 9 after the division sign, which turns the problem into 2 over 3 times 9 over 4, or 3 over 2.
5. Add over a least common denominator built from factors
Here's where people's instincts misfire. One sixth plus one fourth is not 2 over 10. Adding the tops and adding the bottoms gives a number smaller than one fourth, and that can't be right when you added something to one fourth. The fix is a shared bottom. The least common denominator of 6 and 4 is 12, so you get 2 twelfths plus 3 twelfths, which is 5 twelfths.
With polynomials, the shared bottom comes from factoring. You list every factor that appears in the factored denominators, each as many times as it appears in any one of them.
If the bottoms already match, you add the tops: 2 over x plus 3 over x is 5 over x. If they don't, you build the LCD. For 1 over x plus 1 over x plus 1, the bottoms share nothing, so the LCD is x times x plus 1. The first fraction becomes x plus 1 over that LCD, the second becomes x over it, and the tops add to 2x plus 1. At x equals 1, the original is 1 plus one half, which is 3 over 2, and the combined fraction is 3 over 1 times 2, also 3 over 2.
Pause and try this. To subtract 1 over x minus 2 from x over x squared minus 4, what is the least common denominator?
Factor x squared minus 4 before you read on.
Here's the answer. Since x squared minus 4 factors as x minus 2, times x plus 2, it already contains x minus 2. So the LCD is x minus 2, times x plus 2, not the product of the two bottoms.
6. Distribute the minus sign to every term
Subtraction has one trap of its own. A minus sign in front of a fraction belongs to its whole top, not only to the first piece.
Let's finish the problem from the pause. You want x over x squared minus 4, minus 1 over x minus 2. Rewrite the second fraction over the LCD by multiplying its top and bottom by x plus 2. Now its top is x plus 2. Subtract that whole top: x minus the quantity x plus 2. Distribute the minus, and you get x minus x minus 2, which is negative 2. So the difference simplifies to negative 2 over x minus 2, times x plus 2.
Check it at x equals 0. The original gives zero over negative 4, minus 1 over negative 2, which is one half. The simplified difference gives negative 2 over negative 4, also one half. They agree.
Now the slip. If you subtract only the x and forget the 2, the top comes out positive 2. At x equals 0 that version gives negative one half, the wrong sign. A tip worth carrying: put parentheses around the second top before you subtract, every single time.
Pause and try this. If the minus reaches only the first term, what top do you get instead of negative 2?
Say it out loud before you read on.
Here's the answer. You get x minus x plus 2, which is positive 2, because the minus never reached the 2. That flipped sign is the slip a test value catches.
7. Solve a rational equation and reject what fails
A rational equation is a fraction problem with an equals sign. Formally, a rational equation is an equation that contains at least one rational expression with the variable in a denominator; it is solved by multiplying every term by the least common denominator and then checking each candidate.
The plan has four steps. Name the excluded values. Multiply every term by the LCD. Solve what's left. Check every candidate in the original.
Start easy. In x over 3 plus 1 over 2 equals 5 over 6, no bottom can be zero. Multiply every term by 6, and you get 2x plus 3 equals 5, so x is 1. In 1 over x equals 4, zero is excluded, and multiplying by x gives x equals one fourth.
Now the twist. In x over x minus 2 equals 2 over x minus 2, the excluded value is 2. Multiply both sides by x minus 2 and you get x equals 2. That's the excluded value. So the equation has no solution. The cleared equation didn't remember the closed door, and you had to.
That's the idea behind the week's most important term. An extraneous solution is a candidate produced by correct solving steps that still fails the original equation, such as an excluded value that appears after the denominators are cleared; it is named and rejected with its reason.
Pause and try this. Clearing 1 over x minus 1, plus 1, equals 2 over x squared minus 1, gives the candidates negative 2 and 1. Which do you keep?
Compare each candidate with the excluded values first.
Here's the answer. Since x squared minus 1 is x minus 1, times x plus 1, both 1 and negative 1 are excluded, so 1 is extraneous. You keep negative 2, which checks: both sides come out to 2 over 3.
8. Use rational equations in settings: rates add, times do not
One pipe fills a farm's water tank in 3 hours. Another fills it in 6 hours. How long do they take together? A lot of people say 9 hours. Stop and picture it. Turning on a second pipe can't make the job slower than the first pipe alone. So the time together has to be under 3 hours.
The trick is to think in rates. A work rate is the fraction of a job done per unit of time, such as one third of a job per hour for a worker who needs 3 hours; when workers work together, their rates add. The first pipe does one third of the tank each hour, and the second does one sixth. Add the rates: one third plus one sixth is 2 sixths plus 1 sixth, which is 3 sixths, or one half of a tank per hour. At half a tank per hour, the full tank takes 2 hours. As an equation, t over 3 plus t over 6 equals 1, and multiplying by 6 gives 3t equals 6, so t is 2.
The cabin from the opening works the same way. If 2 more people would cut each share of 600 dollars by 25 dollars, the equation is 600 over n, minus 600 over n plus 2, equals 25. Clearing the denominators and solving gives n equals 6 or n equals negative 8. A group can't have a negative number of people, so you keep 6. Check it: 600 over 6 is 100, 600 over 8 is 75, and the difference is 25.
Pause and try this. Why is 9 hours impossible for the two pipes together?
Think about the faster pipe working alone.
Here's the answer. The faster pipe alone needs only 3 hours, and a second pipe can only add water. So together they must finish in under 3 hours, and adding the rates gives 2.
Closing
Let's pull it together, in the order we took it. A rational expression is a polynomial over a polynomial, and its excluded values come first. You simplify by factoring and cancelling common factors, never terms, and you carry the restriction. The week's quotient matches x minus 5 everywhere except its hole at 1. To multiply or divide, you factor, flip only the divisor, and cancel. To add or subtract, you build the LCD from factored bottoms, and a minus sign reaches the whole second top. To solve, you name the excluded values, clear the LCD, and check every candidate, rejecting the extraneous ones. And in a work problem, rates add while times don't.
Next in this module, you'll find the slide deck, then the Readings & Resources page, the lecture tutorial, the practice set, and the quiz. In Week 12, Radicals, Rational Exponents & Radical Equations, squaring both sides will create candidates that fail, and the same checking habit will catch them.
Built with The Course Maker · samples.thecoursemaker.com