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Week 11 · Chapter

Chapter 11 — Rational Expressions & Equations

College Algebra Free sample edition

The module's primary reading · ~40–50 minutes · Objective 7 (analyze polynomial and rational functions (end behavior, zeros and multiplicity, domain, and asymptotes), and simplify and solve rational and radical expressions and equations, including expressions with rational exponents, checking for extraneous solutions), Objective 5 (add, subtract, and multiply polynomials, apply the special products, and factor polynomial expressions by the greatest common factor, trinomial patterns, the difference of squares, and grouping) · Read it first this week; the lecture, the tutorial, the quiz, and the Model/Build Studio all build on it.

In plain language

This week you work with rational expressions (fractions with a polynomial on the top and on the bottom). You will simplify them by cancelling only the pieces that multiply everything, and you will multiply, divide, add, and subtract them. Then you will solve equations that contain them and throw out extraneous solutions (answers that come from correct steps but still fail the original equation). The payoff is real: these equations tell you how long two workers take together and how a shared cost changes as a group grows.

Why you already care about this

Picture a group of friends renting a cabin for a weekend for 600 dollars, split equally. With 4 people, each pays 150 dollars. With 5 people, each pays 120 dollars. With 6 people, each pays 100 dollars. The share is 600 divided by the number of people, and the number of people sits on the bottom of a fraction. That is a rational expression, and it shows up any time a total is shared.

Now the question flips. Say the group knows that two extra friends would cut each share by 25 dollars, and they want to know how many people were in the group at first. That question is an equation with the unknown on the bottom of two fractions. The same kind of equation tells a farmer how long two pipes take to fill a water tank together, and tells a food bank how long two volunteers take to stuff a pile of envelopes as a team. A pharmacy technician working out a dose per kilogram of body weight and a contractor scheduling two crews meet it too.

In this chapter you get the tools for all of it. Each tool is a fraction move you already trust, done with polynomials.

By the end of this chapter you will be able to:

  • find the excluded values of a rational expression before you do anything else;
  • simplify a rational expression by factoring and cancelling common factors, never terms, and carry the restriction;
  • multiply and divide rational expressions by factoring, flipping the divisor, and cancelling;
  • add and subtract rational expressions over a least common denominator built from factored denominators;
  • solve a rational equation, check every candidate, and reject an extraneous solution with its reason;
  • set up and solve a work-rate or cost-sharing problem, adding rates rather than times.

Carry this sentence with you: factor first, then test one input — not the other way round.

Where we are

  • Already covered: Weeks 1 and 2 set the rules for fractions and cleared fractions with the least common denominator. Week 3 found the domain of a rational function, Week 6 taught factoring, Week 7 solved quadratic equations, and Week 10 told vertical asymptotes from holes.
  • This week: simplifying, multiplying, dividing, adding, and subtracting rational expressions, then solving rational equations and work-rate problems.
  • How they connect: every simplification is a Week 6 factoring move followed by a cancel, and every equation becomes a Week 2 or Week 7 equation once the denominators are cleared.
  • Where we're going next: Week 12, Radicals, Rational Exponents & Radical Equations, uses the same check-every-candidate habit when squaring both sides creates answers that do not work.

Each earlier week hands you a tool. Week 2 showed you how to clear fractions: x ⁄ 2 + 1 ⁄ 3 = 5 ⁄ 6 became a whole-number equation once every term was multiplied by 6. Week 3 taught that a rational function's domain leaves out every input that makes its denominator 0. Week 6 taught the factoring patterns you lean on all week, such as x² − 9 = (x + 3)(x − 3). Week 10 showed what happens when a factor cancels from the top and bottom. The graph keeps a hole there: a single missing point on a graph, drawn as an open circle, at an input where a factor in the numerator and denominator cancelled out. This week you put those tools to work on the expressions themselves.

1. Spot a rational expression and its excluded values

You already know that 2 ⁄ 3 is a number and that 2 ⁄ 0 is not. Nobody can split 2 pizzas among 0 people. That one fact drives this whole week: a fraction is undefined when its bottom is 0.

Now let the top and bottom be polynomials. The cabin share 600 ⁄ n is a fraction whose bottom changes with n, and it breaks at n = 0. The expression (x² − 9) ⁄ (x + 3) breaks when x + 3 = 0.

  • A rational expression is one polynomial divided by another, such as (x² − 9) ⁄ (x + 3); it is defined for every input except the ones that make its denominator 0.
  • 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.

Worked example — the excluded value of (x² − 9) ⁄ (x + 3). You find the one input this expression cannot take.

  1. Look only at the bottom. The denominator is x + 3.
  2. Set it equal to 0. x + 3 = 0.
  3. Solve. You subtract 3 from both sides: x = −3.
  4. Write the exclusion. x ≠ −3.

In words: the expression works for every real number except −3. At x = −3 the bottom is −3 + 3 = 0, and division by 0 is undefined.

Worked example — the excluded value of the week's quotient. The week's parabola, x² − 6x + 5, from Weeks 7, 9, and 10, now sits on top of x − 1. The result is the week's quotient, (x² − 6x + 5) ⁄ (x − 1).

  1. Look at the bottom. x − 1.
  2. Set it equal to 0 and solve. x − 1 = 0, so x = 1.
  3. Write the exclusion first. x ≠ 1. You write this before you factor anything.

In words: the week's quotient is undefined at x = 1, and that fact will matter in Section 3.

Misconception check. Students often look for excluded values by setting the numerator equal to 0. It is tempting because the numerator is the bigger, busier part. The cure: only a zero on the bottom breaks a fraction. A zero on top is fine, since 0 ⁄ 5 = 0. For (x² − 9) ⁄ (x + 3), x = 3 makes the top 0, and the expression equals 0 ⁄ 6 = 0 there, which is a perfectly good value.

2. Simplify by factoring first: cancel factors, never terms

With numbers, you simplify 6 ⁄ 9 by writing it as (2 · 3) ⁄ (3 · 3) and cancelling the 3. Notice what made that legal: the 3 multiplied the whole top and the whole bottom. Rational expressions follow the same rule. You factor, find a piece that multiplies everything on top and everything on the bottom, and cancel it.

  • 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.
  • A restriction is the note written beside a simplified rational expression, such as x ≠ −3, that keeps an excluded value out after the factor that caused it has been cancelled.

Worked example — (x² − 9) ⁄ (x + 3). You simplify it and keep its restriction.

  1. Excluded value first. x ≠ −3, from Section 1.
  2. Factor the top. x² − 9 is a difference of squares from Week 6: x² − 9 = (x + 3)(x − 3).
  3. Rewrite. (x + 3)(x − 3) ⁄ (x + 3).
  4. Cancel the common factor. (x + 3) multiplies the whole top and the whole bottom, so it cancels.
  5. Write the result with its restriction. x − 3, for x ≠ −3.
  6. Check at a legal input. At x = 2 the original is (4 − 9) ⁄ (2 + 3) = −5 ⁄ 5 = −1. The simplified form gives 2 − 3 = −1. They match.

In words: (x² − 9) ⁄ (x + 3) = x − 3 everywhere except x = −3, where the original is undefined.

Worked example — a perfect square on the bottom. Simplify (x² − 4) ⁄ (x² + 4x + 4).

  1. Factor the top. x² − 4 = (x − 2)(x + 2).
  2. Factor the bottom. x² + 4x + 4 = (x + 2)(x + 2), a perfect-square trinomial.
  3. Excluded value. The bottom is 0 when x + 2 = 0, so x ≠ −2.
  4. Cancel one common factor. (x − 2)(x + 2) ⁄ ((x + 2)(x + 2)) = (x − 2) ⁄ (x + 2).
  5. Check at x = 0. The original is −4 ⁄ 4 = −1. The simplified form is (0 − 2) ⁄ (0 + 2) = −2 ⁄ 2 = −1.

In words: (x² − 4) ⁄ (x² + 4x + 4) = (x − 2) ⁄ (x + 2), for x ≠ −2. One (x + 2) is still on the bottom, so x = −2 stays excluded either way.

Worked example — a trinomial on top. Simplify (x² + 5x + 6) ⁄ (x + 2).

  1. Excluded value. x + 2 = 0 gives x ≠ −2.
  2. Factor the top. You need two numbers with product 6 and sum 5: 2 and 3, because 2 · 3 = 6 and 2 + 3 = 5. So x² + 5x + 6 = (x + 2)(x + 3).
  3. Cancel. (x + 2)(x + 3) ⁄ (x + 2) = x + 3.

In words: (x² + 5x + 6) ⁄ (x + 2) = x + 3, for x ≠ −2.

Misconception check — cancelling a term. Students often look at (x + 3) ⁄ x and cross out the two x's to get 3. It is tempting because the same letter appears on top and bottom. The cure: x is a term of the top, added to 3, not a factor that multiplies the whole top. Test one input. At x = 1, (1 + 3) ⁄ 1 = 4, not 3. If a quick test value disagrees, the cancel was illegal. Only cancel a piece that is multiplied, never one that is added.

A memory hook for the whole week: a sum is a single unit; break it into factors before you cancel anything.

3. Read the week's quotient as a table and a graph

The best way to trust a simplification is to watch it work on numbers. Take the week's quotient, (x² − 6x + 5) ⁄ (x − 1), with x ≠ 1 already written down.

Worked example — simplify the week's quotient. You factor the week's parabola on top and cancel.

  1. Factor the top. x² − 6x + 5 needs two numbers with product 5 and sum −6. They are −1 and −5, because (−1)(−5) = 5 and −1 + (−5) = −6. So x² − 6x + 5 = (x − 1)(x − 5).
  2. Cancel the common factor. (x − 1)(x − 5) ⁄ (x − 1) = x − 5. You may cancel (x − 1), because it multiplies everything on top and everything on the bottom.
  3. Attach the restriction. x − 5, for x ≠ 1.

Now check it with a table. For each input, divide the top's value by the bottom's value, then compare with x − 5.

The week's quotient (x² − 6x + 5) ⁄ (x − 1) at six legal inputs: numerator, denominator, quotient, and x − 5
x 0 2 3 4 5 6
numerator x² − 6x + 5 5 −3 −4 −3 0 5
denominator x − 1 −1 1 2 3 4 5
quotient −5 −3 −2 −1 0 1
x − 5 −5 −3 −2 −1 0 1

Two entries worked out in full: at x = 0 the quotient is (0 − 0 + 5) ⁄ (0 − 1) = 5 ⁄ (−1) = −5, and 0 − 5 = −5. At x = 3 it is (9 − 18 + 5) ⁄ (3 − 1) = −4 ⁄ 2 = −2, and 3 − 5 = −2. Every column matches. The six quotients sum to −10.

What happens at x = 1. The top is 1 − 6 + 5 = 0 and the bottom is 1 − 1 = 0, so the quotient is 0 ⁄ 0, which is undefined. The simplified form would give 1 − 5 = −4 there. So the graph of the week's quotient is the line y = x − 5 with a hole at (1, −4), not a zero.

In practice (your tools). Open the free Desmos graphing calculator and type y = (x^2 - 6x + 5)/(x - 1). You see a straight line rising from left to right. Now type y = x - 5 on the next line. The two graphs lie on top of each other. Click the line near x = 1: Desmos shows no point at x = 1 for the first expression, because the quotient is undefined there. Desmos may draw the line without an open circle, so the restriction you wrote by hand is the honest record of the hole. GeoGebra behaves the same way.

Misconception check — a hole read as a zero. Students often see the numerator's factor (x − 1) and say the quotient is 0 at x = 1. It is tempting because x = 1 really is a zero of the week's parabola on top. The cure: the bottom is 0 there too, and 0 ⁄ 0 is undefined. A zero of the quotient needs a 0 on top and a nonzero bottom, which happens at x = 5: 0 ⁄ 4 = 0. At x = 1 there is a hole, not a zero.

4. Multiply and divide: factor, flip, cancel across

Start with numbers. To multiply 2 ⁄ 3 by 9 ⁄ 4, you multiply tops and bottoms: (2 ⁄ 3) · (9 ⁄ 4) = 18 ⁄ 12 = 3 ⁄ 2. You can also cancel first. The 2 on top and the 4 on the bottom share a 2, and the 9 on top and the 3 on the bottom share a 3. That leaves (1 ⁄ 1) · (3 ⁄ 2) = 3 ⁄ 2. Cancelling first keeps the numbers small, and with polynomials it keeps the degrees small.

Division is multiplication by the flipped divisor. (2 ⁄ 3) ÷ (4 ⁄ 9) = (2 ⁄ 3) · (9 ⁄ 4) = 3 ⁄ 2. You flip the divisor. Then you multiply.

The procedure for rational expressions has four steps: factor every top and bottom; flip the divisor if you are dividing; cancel any factor that appears on a top and a bottom anywhere in the product; multiply what is left.

Worked example — a product. Multiply (x + 2) ⁄ (x − 3) · (x² − 9) ⁄ (x² + 4x + 4).

  1. Factor. x² − 9 = (x + 3)(x − 3), and x² + 4x + 4 = (x + 2)(x + 2).
  2. Rewrite. (x + 2) ⁄ (x − 3) · (x + 3)(x − 3) ⁄ ((x + 2)(x + 2)).
  3. Excluded values. The bottoms are 0 at x = 3 and at x = −2, so x ≠ 3 and x ≠ −2.
  4. Cancel across the product. One (x + 2) on top cancels one (x + 2) on the bottom. The (x − 3) on top cancels the (x − 3) on the bottom.
  5. Multiply what is left. (x + 3) ⁄ (x + 2), for x ≠ 3 and x ≠ −2.
  6. Check at x = 0. The original is (2 ⁄ (−3)) · (−9 ⁄ 4) = 18 ⁄ 12 = 3 ⁄ 2. The result is (0 + 3) ⁄ (0 + 2) = 3 ⁄ 2.

In words: the product simplifies to (x + 3) ⁄ (x + 2), and it keeps both restrictions.

Worked example — a quotient. Divide (x² − 4) ⁄ (x + 3) ÷ (x − 2) ⁄ (x² + 5x + 6).

  1. Flip the divisor and multiply. (x² − 4) ⁄ (x + 3) · (x² + 5x + 6) ⁄ (x − 2).
  2. Factor. (x − 2)(x + 2) ⁄ (x + 3) · (x + 2)(x + 3) ⁄ (x − 2).
  3. Excluded values. The original bottoms give x ≠ −3 and x ≠ −2, since x² + 5x + 6 = (x + 2)(x + 3). The divisor itself, (x − 2) ⁄ (x² + 5x + 6), is 0 when x = 2, and you cannot divide by 0, so x ≠ 2 as well.
  4. Cancel. (x − 2) cancels, and (x + 3) cancels.
  5. Multiply what is left. (x + 2)(x + 2) = (x + 2)², for x ≠ −3, x ≠ −2, and x ≠ 2.
  6. Check at x = 0. The original is (−4 ⁄ 3) ÷ (−2 ⁄ 6) = (−4 ⁄ 3) ÷ (−1 ⁄ 3) = (−4 ⁄ 3) · (−3) = 4. The result is (0 + 2)² = 4.

In words: the quotient is (x + 2)², with three excluded values.

Misconception check — flipping the wrong fraction. Students often flip the first fraction, or both, when they divide. It is tempting because "flip and multiply" does not say which one. The cure: only the divisor, the fraction after the ÷ sign, gets flipped. With numbers, (2 ⁄ 3) ÷ (4 ⁄ 9) is 3 ⁄ 2, while flipping the first fraction gives (3 ⁄ 2) · (4 ⁄ 9) = 12 ⁄ 18 = 2 ⁄ 3, a different number. A second caveat: the divisor's own numerator adds an excluded value, which is easy to forget once it has been flipped to the bottom.

5. Add and subtract over a least common denominator built from factors

With numbers, you cannot add 1 ⁄ 6 and 1 ⁄ 4 until they share a bottom. The least common denominator, the smallest number both 6 and 4 divide into, is 12. So 1 ⁄ 6 + 1 ⁄ 4 = 2 ⁄ 12 + 3 ⁄ 12 = 5 ⁄ 12. Adding tops and bottoms gives 2 ⁄ 10. That is wrong. One sixth plus one fourth is more than one fifth.

Your rational expressions work the same way. You build the LCD from the factored denominators: list every factor that appears, each as many times as it appears in any one denominator.

Worked example — the same denominator. Add 2 ⁄ x + 3 ⁄ x.

  1. Excluded value. x ≠ 0.
  2. The bottoms already match, so add the tops. (2 + 3) ⁄ x = 5 ⁄ x.
  3. Check at x = 1. 2 + 3 = 5, and 5 ⁄ 1 = 5.

In words: 2 ⁄ x + 3 ⁄ x = 5 ⁄ x, for x ≠ 0. The bottom stays x; it does not become 2x.

Worked example — different denominators. Add 1 ⁄ x + 1 ⁄ (x + 1).

  1. Factor each bottom. x and x + 1 share no factor.
  2. Build the LCD. Use each factor once: x(x + 1). Excluded values: x ≠ 0 and x ≠ −1.
  3. Rewrite each fraction over the LCD. 1 ⁄ x = (x + 1) ⁄ (x(x + 1)), and 1 ⁄ (x + 1) = x ⁄ (x(x + 1)).
  4. Add the tops. (x + 1 + x) ⁄ (x(x + 1)) = (2x + 1) ⁄ (x(x + 1)).
  5. Check at x = 1. The original is 1 ⁄ 1 + 1 ⁄ 2 = 3 ⁄ 2. The result is (2 + 1) ⁄ (1 · 2) = 3 ⁄ 2.

In words: 1 ⁄ x + 1 ⁄ (x + 1) = (2x + 1) ⁄ (x(x + 1)).

Misconception check — multiplying denominators instead of factoring. Students often build a common denominator by multiplying the bottoms together. For x ⁄ (x² − 4) − 1 ⁄ (x − 2) that gives (x² − 4)(x − 2), a degree-3 bottom. It is tempting because multiplying always produces some common denominator. The cure: factor first. x² − 4 = (x − 2)(x + 2) already contains (x − 2), so the least common denominator is (x − 2)(x + 2). The product of the bottoms is a limitation, not an error, but it carries an extra factor you would have to cancel later.

6. Distribute the minus sign to every term

Your subtraction step adds one hazard. The minus sign in front of a fraction applies to its entire top, not to the first piece of it.

Worked example — subtracting with a factored LCD. Subtract x ⁄ (x² − 4) − 1 ⁄ (x − 2).

  1. Factor. x² − 4 = (x − 2)(x + 2), so the LCD is (x − 2)(x + 2). Excluded values: x ≠ 2 and x ≠ −2.
  2. Rewrite the second fraction. Multiply its top and bottom by (x + 2): 1 ⁄ (x − 2) = (x + 2) ⁄ ((x − 2)(x + 2)).
  3. Subtract the whole second top. (x − (x + 2)) ⁄ ((x − 2)(x + 2)). The parentheses around x + 2 are the whole point.
  4. Distribute the minus. x − x − 2 = −2, so the result is −2 ⁄ ((x − 2)(x + 2)).
  5. Check at x = 0. The original is 0 ⁄ (−4) − 1 ⁄ (−2) = 0 + 1 ⁄ 2 = 1 ⁄ 2. The result is −2 ⁄ ((−2)(2)) = −2 ⁄ (−4) = 1 ⁄ 2.

In words: x ⁄ (x² − 4) − 1 ⁄ (x − 2) = −2 ⁄ ((x − 2)(x + 2)), for x ≠ 2 and x ≠ −2.

Misconception check — subtracting only the first term. Students often write x − x + 2 in step 3, so the top comes out 2 instead of −2. It is tempting because the minus sign sits right next to the x and seems to belong only to it. The cure: wrap the whole second numerator in parentheses before you subtract, then distribute. The test value catches the slip at once: 2 ⁄ ((x − 2)(x + 2)) at x = 0 is 2 ⁄ (−4) = −1 ⁄ 2, but the original is 1 ⁄ 2.

In practice (your tools). In Desmos, type y = x/(x^2 - 4) - 1/(x - 2) on one line and your answer, y = -2/((x - 2)(x + 2)), on the next. If the two graphs lie exactly on top of each other, your algebra agrees with the original. If they split apart, look first at the minus sign. A graph is a second opinion, never a replacement for the check by substitution.

7. Solve a rational equation and reject what fails

A rational equation is a fraction problem with an equals sign, and the plan is the one you used in Week 2: clear the fractions. Multiplying both sides by the LCD turns the equation into one with no fractions at all. The new twist is that the cleared equation does not remember which inputs were forbidden. You have to.

  • 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.
  • 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.

The four steps: name the excluded values; multiply every term by the LCD; solve the equation that is left; check every candidate in the original and reject any excluded value.

Worked example — numbers on the bottom. Solve x ⁄ 3 + 1 ⁄ 2 = 5 ⁄ 6.

  1. Excluded values. The bottoms are 3, 2, and 6, never 0, so there are none.
  2. Multiply every term by the LCD 6. 6 · (x ⁄ 3) = 2x, 6 · (1 ⁄ 2) = 3, and 6 · (5 ⁄ 6) = 5, so 2x + 3 = 5.
  3. Solve. 2x = 2, so x = 1.
  4. Check. 1 ⁄ 3 + 1 ⁄ 2 = 2 ⁄ 6 + 3 ⁄ 6 = 5 ⁄ 6.

Worked example — the variable on the bottom. Solve 1 ⁄ x = 4.

  1. Excluded value. x ≠ 0.
  2. Multiply both sides by x. 1 = 4x.
  3. Solve. x = 1 ⁄ 4.
  4. Check. 1 ÷ (1 ⁄ 4) = 4. It works, and 1 ⁄ 4 is not excluded.

Worked example — no solution. Solve x ⁄ (x − 2) = 2 ⁄ (x − 2).

  1. Excluded value. x ≠ 2.
  2. Multiply both sides by x − 2. x = 2.
  3. Check. The only candidate is 2, the excluded value. At x = 2 both bottoms are 2 − 2 = 0, so neither side exists.
  4. Conclude. You write no solution.

Worked example — one kept, one rejected. Solve 1 ⁄ (x − 1) + 1 = 2 ⁄ (x² − 1).

  1. Factor and exclude. x² − 1 = (x − 1)(x + 1), so x ≠ 1 and x ≠ −1. The LCD is (x − 1)(x + 1).
  2. Multiply every term by the LCD. The first term becomes x + 1. The 1 becomes (x − 1)(x + 1) = x² − 1. The right side becomes 2. So (x + 1) + (x² − 1) = 2.
  3. Simplify. x² + x = 2, so x² + x − 2 = 0.
  4. Solve as in Week 7. (x + 2)(x − 1) = 0, so x = −2 or x = 1.
  5. Check x = 1. It is excluded, so it is an extraneous solution and is rejected.
  6. Check x = −2. The left side is 1 ⁄ (−3) + 1 = 2 ⁄ 3. The right side is 2 ⁄ (4 − 1) = 2 ⁄ 3. They match.

In words: the only solution is x = −2. The cleared equation produced x = 1 honestly, but the original equation cannot use it.

Misconception check — trusting the cleared equation. Students often stop at step 4 and report both x = −2 and x = 1. It is tempting because the factoring was correct. The cure: multiplying by an expression that can equal 0 can create a candidate the original never allowed. That is why you write the excluded values first, before the denominators vanish, and why every candidate gets checked against that list.

8. Apply rational equations: rates add, times do not

Here is the setting this section is built on. Pipe A fills a farm's water tank in 3 hours. Pipe B fills the same tank in 6 hours. How long do they take running together? Many people guess 9 hours, or the average, 4.5 hours. Both guesses are too long. A second pipe can only speed the job up, so the answer must be under 3 hours.

The key is to think in rates. Pipe A finishes 1 tank in 3 hours, so each hour it fills 1 ⁄ 3 of a tank.

  • A work rate is the fraction of a job done per unit of time, such as 1 ⁄ 3 of a job per hour for a worker who needs 3 hours; when workers work together, their rates add.

Worked example — two pipes. You find the time for pipes A and B running together.

  1. Write each rate. Pipe A: 1 ⁄ 3 of a tank per hour. Pipe B: 1 ⁄ 6 of a tank per hour.
  2. Add the rates. 1 ⁄ 3 + 1 ⁄ 6 = 2 ⁄ 6 + 1 ⁄ 6 = 3 ⁄ 6 = 1 ⁄ 2 of a tank per hour.
  3. Set up the equation. Let t be the hours they run together. Pipe A does t ⁄ 3 of the tank and pipe B does t ⁄ 6, and together they do 1 whole tank: t ⁄ 3 + t ⁄ 6 = 1.
  4. Clear the LCD 6. 2t + t = 6, so 3t = 6.
  5. Solve. t = 2 hours.
  6. Check. In 2 hours pipe A fills 2 ⁄ 3 and pipe B fills 2 ⁄ 6 = 1 ⁄ 3, and 2 ⁄ 3 + 1 ⁄ 3 = 1 full tank.

In words: together the pipes fill the tank in 2 hours, faster than either one alone.

Worked example — the cabin. Friends share a 600-dollar cabin rental equally. If 2 more people join, each person pays 25 dollars less. How many people were in the group at first?

  1. Name the unknown and the shares. Let n be the number of people at first. Each share is 600 ⁄ n dollars, and with 2 more people it is 600 ⁄ (n + 2) dollars.
  2. Write the equation. The old share minus the new share is 25: 600 ⁄ n − 600 ⁄ (n + 2) = 25.
  3. Excluded values. n ≠ 0 and n ≠ −2.
  4. Multiply every term by n(n + 2). 600(n + 2) − 600n = 25n(n + 2).
  5. Simplify. 600n + 1200 − 600n = 25n² + 50n, so 1200 = 25n² + 50n.
  6. Divide by 25 and rearrange. 48 = n² + 2n, so n² + 2n − 48 = 0.
  7. Factor. (n + 8)(n − 6) = 0, so n = −8 or n = 6.
  8. Ask the setting. A number of people cannot be negative. You keep n = 6.
  9. Check. 600 ⁄ 6 = 100 dollars each, 600 ⁄ 8 = 75 dollars each, and 100 − 75 = 25.

In words: 6 friends were in the group at first. With 8, each share drops from 100 dollars to 75 dollars.

In practice (your tools). In Desmos, type y = x/3 + x/6 and y = 1. The line crosses the level y = 1 at x = 2, the 2 hours you found. For the cabin, type y = 600/x - 600/(x + 2) and y = 25, and click the crossing on the right: Desmos labels (6, 25). The graph is a check on the setting as well as the algebra, since only the right-hand crossing describes a real group.

Misconception check — adding times. Students often add the two times, 3 + 6 = 9 hours. It is tempting because adding is what you do when two things combine. The cure: times do not add, rates do. Nine hours is longer than the slower pipe needs alone, so the guess fails a common-sense check before any algebra. Add the rates, 1 ⁄ 3 + 1 ⁄ 6 = 1 ⁄ 2 of a tank per hour, and the time follows.

An honesty note about the model: real pipes can lose pressure when both run at once, so the 2-hour answer assumes each pipe keeps its own steady rate. The algebra is exact; the setting is simplified.

A reflection to choose from. Pick one option, (a) or (b), and write two or three sentences.

  • (a) Think of a job you have shared with someone, such as cooking a meal, cleaning a room, or moving furniture. Describe how the work-rate idea fits it, and say one way the real job did not follow the model.
  • (b) Pick the step of solving a rational equation you are most likely to skip: naming the excluded values, distributing a minus sign, or checking the candidates. Explain what goes wrong when you skip it, using one example from this chapter.

Key terms

  • Rational expression — One polynomial divided by another, such as (x² − 9) ⁄ (x + 3); it is defined for every input except the ones that make its denominator 0. (first used: 1. Spot a rational expression and its excluded values)
  • Excluded value — 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. (first used: 1. Spot a rational expression and its excluded values)
  • Common factor — 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. (first used: 2. Simplify by factoring first: cancel factors, never terms)
  • Restriction — The note written beside a simplified rational expression, such as x ≠ −3, that keeps an excluded value out after the factor that caused it has been cancelled. (first used: 2. Simplify by factoring first: cancel factors, never terms)
  • Rational equation — 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. (first used: 7. Solve a rational equation and reject what fails)
  • Extraneous solution — 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. (first used: 7. Solve a rational equation and reject what fails)
  • Work rate — The fraction of a job done per unit of time, such as 1 ⁄ 3 of a job per hour for a worker who needs 3 hours; when workers work together, their rates add. (first used: 8. Apply rational equations: rates add, times do not)

Summary

  • A rational expression is a polynomial over a polynomial; find its excluded values first.
  • Factor the top and the bottom, then cancel common factors only, never terms.
  • Write the restriction beside every simplified form.
  • A cancelled factor leaves a hole in the graph, not a zero.
  • To divide, flip only the divisor, then multiply and cancel.
  • Build the LCD from factored denominators, not from the product of the bottoms.
  • A minus sign in front of a fraction applies to its whole numerator.
  • Clear a rational equation with the LCD, then check every candidate against the excluded values.
  • In work problems, rates add and times do not.

Check your understanding

Work these before looking at the answers — they are ungraded, and the answers (with reasoning) are at the bottom of the page.

  1. Simplify (x² − 25) ⁄ (x² + 2x − 15), state the restrictions, and check your answer at x = 0.
  2. Subtract 4 ⁄ (x² − 1) − 2 ⁄ (x − 1) and simplify fully.
  3. Solve x ⁄ (x − 4) = 4 ⁄ (x − 4) + 3, and explain your answer.
  4. Two volunteers at a food bank stuff a stack of envelopes. One alone takes 6 hours, and the other alone takes 30 hours. How long do they take working together?

Answers (worked reasoning — read after attempting)

  1. Factor the top: x² − 25 = (x − 5)(x + 5). Factor the bottom: you need a product of −15 and a sum of 2, which is 5 and −3, so x² + 2x − 15 = (x + 5)(x − 3). The bottom is 0 at x = −5 and x = 3, so x ≠ −5 and x ≠ 3. Cancel (x + 5): the result is (x − 5) ⁄ (x − 3). Check at x = 0: the original is −25 ⁄ (−15) = 5 ⁄ 3, and the result is −5 ⁄ (−3) = 5 ⁄ 3. Crossing out the x² terms would be cancelling terms, the slip from Section 2.
  2. Factor: x² − 1 = (x − 1)(x + 1), so the LCD is (x − 1)(x + 1), with x ≠ 1 and x ≠ −1. Rewrite the second fraction: 2 ⁄ (x − 1) = 2(x + 1) ⁄ ((x − 1)(x + 1)). Subtract the whole second top: (4 − (2x + 2)) ⁄ ((x − 1)(x + 1)) = (4 − 2x − 2) ⁄ ((x − 1)(x + 1)) = (2 − 2x) ⁄ ((x − 1)(x + 1)). Factor the top: 2 − 2x = −2(x − 1). Cancel (x − 1): the result is −2 ⁄ (x + 1). Check at x = 0: the original is 4 ⁄ (−1) − 2 ⁄ (−1) = −4 + 2 = −2, and −2 ⁄ 1 = −2.
  3. The excluded value is x = 4. Multiply every term by x − 4: x = 4 + 3(x − 4), so x = 4 + 3x − 12, so x = 3x − 8. Subtract 3x: −2x = −8, so x = 4. The only candidate is the excluded value, so it is extraneous and the equation has no solution. At x = 4 every bottom is 0, so neither side exists.
  4. The rates are 1 ⁄ 6 and 1 ⁄ 30 of the stack per hour. They add: 1 ⁄ 6 + 1 ⁄ 30 = 5 ⁄ 30 + 1 ⁄ 30 = 6 ⁄ 30 = 1 ⁄ 5 of the stack per hour. With t hours together, t ⁄ 6 + t ⁄ 30 = 1. Multiply by 30: 5t + t = 30, so 6t = 30, and t = 5 hours. Check: in 5 hours they do 5 ⁄ 6 + 5 ⁄ 30 = 5 ⁄ 6 + 1 ⁄ 6 = 1 full stack. Adding the times, 6 + 30 = 36 hours, is the trap from Section 8.

Next up in this module: the Readings & Resources page, the lecture tutorial, the practice set, the quiz, the discussion, and the assignment — and next week the same check-every-candidate rule returns when squaring both sides of a radical equation creates answers that fail.

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