Solving equations that contain rational expressions. Includes cross multiplying when each side is a single fraction, multiplying through by the least common denominator as the general method, factoring denominators before choosing the LCD, checking every candidate against the excluded values, and setting up a work problem in which rates of work add.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 11 — Rational Expressions and Equations
Rational Equations
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.7 Rational Equations §11.7, pp. 670-677 — the lesson these objectives are drawn from
Warm-up
Lesson 11.1 solved proportions and warned that clearing fractions can produce a value the original forbids. This lesson generalises both parts.
Discussion prompt
Solve two over x equals three over four, and say what value of x the equation forbids from the start.
Hint: Cross multiply, then look at the denominator.
Answer:
\[ \tfrac{2}{x} = \tfrac{3}{4} \;\Longrightarrow\; 8 = 3x \;\Longrightarrow\; x = \tfrac{8}{3} \]
The denominator x forbids nought from the beginning, and the solution is not nought, so it stands. This lesson does the same thing with more terms and more denominators, and the forbidden values become considerably easier to trip over.
Concept
A rational equation is an equation containing rational expressions. Both methods for solving one work by clearing the denominators, after which an ordinary linear or quadratic equation remains.
rational equation — An equation that contains one or more rational expressions. Solving it usually begins by clearing the denominators, either by cross multiplying or by multiplying through by the least common denominator.
Every solution must then be checked against the original.
Figure (svg): Two columns comparing the two methods for solving a rational equation
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.7 Rational Equations §11.7, pp. 670-671
Section
Section 1
Concept
If a rational equation has one fraction on each side, it is a proportion and cross multiplying clears both denominators in a single step.
\[ \dfrac{a}{b} = \dfrac{c}{d} \;\Longrightarrow\; ad = bc \]
This is Lesson 11.1's cross product property.
Figure (svg): A rational equation solved by cross multiplying
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.7 Rational Equations §11.7, pp. 670-670 — Example 1, Cross Multiply, and its Study Tip on checking answers
Picture it
Cross, expand, factor.
Figure (svg): A rational equation solved by cross multiplying
The cross multiplying is one line and everything after it is Chapter 10. That is why this method feels short even when the answer is a pair of values.
Worked example
This is Example 1 from the textbook.
\[ \text{Solve } \dfrac{5}{y + 2} = \dfrac{y}{3}. \]
Cross multiply
Why: Five times three, y times the bracket.
\[ 15 = y ^{2} + 2 y \]
Write in standard form
Why: Move everything to one side.
\[ 0 = y ^{2} + 2 y - 15 \]
Factor
Why: Five and negative three.
\[ 0 = (y + 5) (y - 3) \]
Solve and check
Why: Neither makes a denominator nought.
\[ y = -5, \; 3 \]
Figure (svg): A rational equation solved by cross multiplying
\[ y = -5 \text{ and } y = 3 \]
Verify: check both in the original
Why: At y equal to three the left side is five fifths, which is one, and the right is three thirds, also one. At y equal to negative five the left is five over negative three and the right is negative five over three — the same value. Both stand.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.7 Rational Equations §11.7, pp. 670-670
Faded example
Then it is a quadratic.
Fill in the blanks
\dfrac215 = \dfrac______ \;\to\; 15 = y^2 + ___y \;\to\; 0 = y^2 + 2y - ___
Why: The cross product produced a squared term because the variable appeared on both sides. Moving the fifteen across puts it in standard form for factoring.
Worked example
Before any solving.
\[ \text{What value does } \dfrac{5}{y + 2} = \dfrac{y}{3} \text{ forbid?} \]
Look at each denominator
Why: One contains the variable.
\[ y + 2 \text{ and } 3 \]
Set the variable one to nought
Why: Find the forbidden value.
\[ y + 2 = 0 \]
Solve
Why: That value is excluded.
\[ y = -2 \]
Compare with the solutions
Why: Neither is negative two.
Figure (svg): Candidate solutions checked against the excluded values
\[ y \ne -2 \]
Verify: see why the comparison matters
Why: Had one of the solutions been negative two it would have had to be discarded, however correct the algebra. Listing the exclusion first turns that decision into a glance at a list rather than a discovery at the end.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.7 Rational Equations §11.7, pp. 670-670
Trap
\[ \dfrac{2}{x} + \dfrac{1}{3} = \dfrac{4}{x} \;\Longrightarrow\; 2 \cdot 3 = x \cdot ? \]
Cross multiply, since there are fractions on both sides
Why: The method worked before, so it was reached for again.
Cross multiplying needs exactly one fraction on each side. With three terms there is nothing to cross with what, so the method has no meaning here — the LCD method is required instead.
Multiply every term by the least common denominator, 3x.
Check the shape of the equation before choosing a method
Why: One fraction each side, or not.
Cross multiplying is a special case of the LCD method, not a rival to it.
Sorting
One fraction on each side, and no more.
Sort into buckets
Sort each equation by whether cross multiplying applies directly.
Half of these are proportions and half are not. Counting the terms on each side before choosing a method takes a second and saves attempting a rule that does not apply.
Elimination
For 5/(y + 2) = y/3.
Eliminate the wrong options
What equation results?
Survives elimination: A
Why: Five times three is fifteen and y times y plus two is y squared plus two y. The variable appearing on both sides is what makes the result quadratic rather than linear.
Socratic
The original equation had no squared term.
Discussion prompt
Explain where the y squared comes from. Then say what that means for how many solutions to expect and to check.
Hint: Which product contains the variable twice?
Answer:
The variable appears in the denominator on the left and in the numerator on the right, so when they are multiplied across, y meets y plus two and produces a squared term. A rational equation with the unknown on both sides in that way is quadratic once cleared.
So up to two solutions should be expected, and both must be checked — not only against the original equation but against its excluded values. Two candidates mean two chances for one of them to be a forbidden value, which is the situation the next sections are about.
Section
Section 2
Concept
Multiplying every term on both sides by the least common denominator clears all the fractions at once. This works for any rational equation, however many terms it has.
Cross multiplying is this method's two-term special case.
Figure (svg): A rational equation solved by multiplying through by the LCD
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.7 Rational Equations §11.7, pp. 671-671 — the Two Methods note and Example 2, Multiply by the LCD
Picture it
Nothing is left out.
Figure (svg): A rational equation solved by multiplying through by the LCD
The one over three had no variable in its denominator, but it still had to be multiplied. Terms without an obvious fraction are the ones most often skipped.
Worked example
This is Example 2 from the textbook.
\[ \text{Solve } \dfrac{2}{x} + \dfrac{1}{3} = \dfrac{4}{x}. \]
Find the LCD
Why: Denominators x and three.
\[ 3 x \]
Multiply every term
Why: All three, on both sides.
\[ 3x \cdot \tfrac{2}{x} + 3x \cdot \tfrac{1}{3} = 3x \cdot \tfrac{4}{x} \]
Simplify each
Why: The fractions clear.
\[ 6 + x = 12 \]
Solve and check
Why: Subtract six; six is not nought.
\[ x = 6 \]
Figure (svg): A rational equation solved by multiplying through by the LCD
\[ x = 6 \]
Verify: substitute back
Why: Two sixths plus a third is a third plus a third, which is two thirds, and four sixths is also two thirds. The excluded value here is nought, and six is not nought, so the solution stands.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.7 Rational Equations §11.7, pp. 671-671
Faded example
Including the one without a variable.
Fill in the blanks
3x \cdot \dfracx12 = 6, \quad 3x \cdot \dfrac______ = ___, \quad 3x \cdot \dfrac______ = ___
Why: Each term loses its denominator when multiplied by the LCD, but each loses a different part of it. The middle term becomes x, not one, which is the term most often mishandled.
Worked example
Guided Practice 6, where one side is a whole expression.
\[ \text{Solve } \dfrac{4}{x - 3} = \dfrac{x}{x - 3} - 1. \]
Find the LCD
Why: Only one distinct denominator.
\[ x - 3 \]
Multiply every term
Why: Including the one.
\[ 4 = x - (x - 3) \]
Simplify
Why: The x terms cancel.
\[ 4 = 3 \]
Interpret
Why: A false statement.
Figure (svg): A rational equation solved by multiplying through by the LCD
\[ 4 = 3 \text{ is false, so there is no solution} \]
Verify: say what a false statement means
Why: The variable vanished and left something untrue, which means no value of x can satisfy the equation. That is a legitimate answer rather than a failure, and it is quite different from an answer of nought.
Trap
\[ \dfrac{2}{x} + \dfrac{1}{3} = \dfrac{4}{x} \;\Longrightarrow\; 6 + 1 = 12 \]
Multiply the terms that have variables underneath
Why: The one third looked like it did not need it.
Every term must be multiplied by the LCD, including one third, which becomes x rather than one. This version gives seven equals twelve, a false statement, and would suggest no solution when in fact six works.
\[ 6 + x = 12 \;\Longrightarrow\; x = 6 \]
Multiply every term on both sides, without exception
Why: A term with no variable is still a term.
Writing the multiplication out for each term separately is what stops one being skipped.
Elimination
For 2/x + 1/3 = 4/x.
Eliminate the wrong options
How should it be solved?
Survives elimination: A
Why: The LCD method applies to any rational equation regardless of how many terms it has. Option C is worth noting as a legitimate but longer route, whereas option D actually changes the equation.
Prediction
After clearing the fractions.
Predict first
If you are left with a false statement such as 4 = 3, what does it mean?
Correct: The equation has no solution.
\[ 4 = x - (x - 3) = 3 \quad \text{for every } x \]
Why: The variable cancelling means the equation reduces to a claim about numbers alone, and if that claim is false then no value of the variable can make the original true. Had it reduced to something true, such as three equals three, every permitted value would be a solution instead. Neither outcome is an error; both are genuine answers, and distinguishing them from a mistake is a matter of rechecking the algebra once.
Socratic
Cross multiplying is quicker.
Discussion prompt
Explain why multiplying by the LCD works on equations that cross multiplying cannot handle. Then say how the two methods are related.
Hint: What does cross multiplying assume?
Answer:
Cross multiplying assumes exactly one fraction on each side, because it pairs one numerator with the other denominator. With three terms there is no such pairing to make. Multiplying by the LCD makes no assumption about the number of terms — it simply removes every denominator at once.
Cross multiplying is in fact the LCD method applied to a two-term equation: multiplying a over b equals c over d by b d gives a d equals b c, which is the cross product. So the two are not rival methods but one method and its shortcut, which is why the shortcut can be abandoned without loss whenever it does not fit.
Section
Section 3
Concept
Factor every denominator before deciding what the least common denominator is. Two denominators that look unrelated often share a factor, or one turns out to be a power of the other.
The Study Tip says to look at the equation in factored form.
Figure (svg): A denominator factored before the LCD is chosen
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.7 Rational Equations §11.7, pp. 671-671 — Example 3, Factor First, then Multiply by the LCD, and its Study Tip
Picture it
Once factored, it is obvious.
Figure (svg): A denominator factored before the LCD is chosen
Unfactored, the product of the two denominators would have been chosen — a cube rather than a square, and considerably more work. The factoring saves an entire degree.
Worked example
This is Example 3 from the textbook.
\[ \text{Solve } \dfrac{3}{x + 3} + \dfrac{4}{x^2 + 6x + 9} = 1. \]
Factor the second denominator
Why: A perfect square trinomial.
\[ (x + 3) ^{2} \]
Choose the LCD
Why: The highest power of x plus three.
\[ (x + 3) ^{2} \]
Multiply every term
Why: All three.
\[ 3(x + 3) + 4 = (x + 3) ^{2} \]
Expand and solve
Why: A quadratic in standard form.
\[ 0 = x ^{2} + 3 x - 4 \]
Figure (svg): A denominator factored before the LCD is chosen
\[ x = -4 \text{ and } x = 1 \]
Verify: check both in the original
Why: At x equal to one the terms are three quarters and four sixteenths, which is a quarter, adding to one. At x equal to negative four they are three over negative one, which is negative three, and four over one, which is four, again adding to one. Neither makes x plus three nought, so both stand.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.7 Rational Equations §11.7, pp. 671-671
Faded example
One denominator contains the other.
Fill in the blanks
x^2 + 6x + 9 = (x + 3)^2 \;\Longrightarrow\; \text2 = (x + 3)^___}
Why: The second denominator is the square of the first, so the highest power of x plus three is two. Taking the product instead would give a cube and an unnecessarily hard equation.
Worked example
Comparing the two possible denominators.
\[ \text{What LCD would you get without factoring } x^2 + 6x + 9? \]
Take the product
Why: The unfactored route.
\[ (x + 3) (x ^{2} + 6 x + 9) \]
Note its degree
Why: One times two.
\[ ^\circ 3 \]
Take the factored LCD
Why: The square.
\[ (x + 3) ^{2} \]
Note its degree
Why: Two.
\[ ^\circ 2 \]
Figure (svg): A denominator factored before the LCD is chosen
\[ (x+3)^3 \text{ against } (x+3)^2 \]
Verify: say what the extra degree costs
Why: The product route gives a cubic equation after clearing, which would then need an extra factor of x plus three divided out before it could be solved. The answers are the same either way; the factored route simply avoids creating work and then undoing it.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.7 Rational Equations §11.7, pp. 671-671
Error analysis
The student solved a rational equation without factoring first.
Annotate
On: \( \begin{aligned} \frac{3}{x+3} + \frac{4}{x^2 + 6x + 9} &= 1 \\ \text{LCD} &= (x+3)(x^2 + 6x + 9) \\ \text{giving a cubic equation} \end{aligned} \)
Nothing here is wrong, only inefficient — and inefficiency in this topic tends to become error, since a cubic invites mistakes that a quadratic does not. Factoring every denominator before choosing an LCD is a two-second habit that prevents the whole detour.
Faded example
Every term by the LCD.
Fill in the blanks
(x+3)^2 \cdot \dfrac4(x+3)^2 = 3(x + 3), \quad (x+3)^2 \cdot \dfrac______ = ___, \quad (x+3)^2 \cdot 1 = ___
Why: Each term loses as much of the LCD as its own denominator contains. The term equal to one has no denominator, so it keeps the whole LCD — which is where the quadratic comes from.
Sorting
Factor before comparing.
Sort into buckets
For each pair of denominators, sort by whether the product is the least common denominator.
Every pair in the left column looked unrelated before factoring. That is exactly why the factoring has to come first rather than being an optional tidy-up.
Socratic
It gives the right answers eventually.
Discussion prompt
Explain why taking the product of the denominators produces correct solutions even when it is not least. Then say what it costs.
Hint: Is it still a common denominator?
Answer:
Any common denominator clears all the fractions, and the product certainly is one — every original denominator divides into it. So the method is valid and the solutions that emerge are the right ones, together with any that the extra factor introduces.
What it costs is degree: multiplying by a cubic rather than a quadratic gives a cubic equation with an extra repeated root that has to be recognised and discarded. That extra root is exactly the excluded value, so the unfactored route makes the checking step harder as well as the solving.
Section
Section 4
Concept
Multiplying by an expression containing the variable can produce values that satisfy the cleared equation but not the original. Every candidate must be tested against the excluded values.
This was already true in Lesson 11.1.
Figure (svg): Candidate solutions checked against the excluded values
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.7 Rational Equations §11.7, pp. 670-671 — the Study Tips in Examples 1 and 2 on checking for zero denominators
Picture it
Exclusions first.
Figure (svg): Candidate solutions checked against the excluded values
The first test is quick and eliminates the candidates that cannot possibly work. Only the survivors need the fuller substitution.
Worked example
An equation whose cleared version has an extra root.
\[ \text{Solve } \dfrac{x}{x - 2} = \dfrac{2}{x - 2} + 3. \]
Note the exclusion
Why: The denominator forbids two.
\[ x \ne 2 \]
Multiply by the LCD
Why: Every term by x minus two.
\[ x = 2 + 3(x - 2) \]
Solve
Why: Expand and collect.
\[ x = 3 x - 4 \]
Finish and check
Why: Two is excluded.
\[ x = 2 \text{, rejected} \]
Figure (svg): Candidate solutions checked against the excluded values
\[ x = 2 \text{ is excluded, so there is no solution} \]
Verify: substitute the candidate
Why: At x equal to two both fractions have a denominator of nought, so the original equation has no value there at all. The cleared equation was perfectly happy with it, which is precisely why the check exists.
Faded example
Before solving anything.
Fill in the blanks
\text2 x - 2 \;\Longrightarrow\; x \ne -3; \quad \text___ (x+3)^2 \;\Longrightarrow\; x \ne ___
Why: Each denominator forbids the value that makes it nought, and a squared denominator forbids the same value as its base. Having the list ready turns the final check into a comparison.
Worked example
Example 3's solutions, tested properly.
\[ \text{Check } x = -4 \text{ and } x = 1 \text{ in } \dfrac{3}{x+3} + \dfrac{4}{(x+3)^2} = 1. \]
List the exclusion
Why: x plus three must not be nought.
\[ x \ne -3 \]
Compare the candidates
Why: Neither is negative three.
Substitute one
Why: Three quarters plus a quarter.
\[ 1 \;\checkmark \]
Substitute the other
Why: Negative three plus four.
\[ 1 \;\checkmark \]
Figure (svg): Candidate solutions checked against the excluded values
\[ x = -4 \text{ and } x = 1 \]
Verify: note how quick the first test was
Why: Comparing two candidates with one excluded value took a couple of seconds and would have caught a false solution immediately. The full substitution is worth doing as well, but it is the second line of defence rather than the first.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.7 Rational Equations §11.7, pp. 671-671
Trap
\[ \dfrac{x}{x-2} = \dfrac{2}{x-2} + 3 \;\Longrightarrow\; x = 2 \]
Report the value the algebra produced
Why: Every step was correct, so the answer must be too.
At x equal to two both denominators are nought, so the original equation is undefined there. The algebra was correct and the answer is still not a solution, which is the whole point of this section.
The only candidate is excluded, so the equation has no solution.
Test every candidate against the excluded values before reporting
Why: Correct algebra is not enough here.
This is the one topic in the course where a flawless method can produce a non-answer.
Hypothesis
Every step looks reversible.
Predict first
What makes the cleared equation different from the original?
Correct: Multiplying by an expression that can be zero is not always a valid step.
Such a false candidate is called an extraneous solution in later courses.
Why: Multiplying both sides of an equation by a non-zero quantity preserves its solutions, but the LCD contains the variable and is nought at exactly the excluded values. At those values the multiplication is by nought, which turns any equation into a true statement — so the cleared equation can be satisfied where the original was not even defined. The degree change is a symptom rather than the cause, and factoring introduces nothing.
Elimination
The algebra produced x = 2 and the denominator is x - 2.
Eliminate the wrong options
How should it be reported?
Survives elimination: A
Why: A candidate that makes an original denominator nought is not a solution, and if it was the only one then the equation has none. Saying so explicitly, with the reason, is a complete answer.
Socratic
Linear and quadratic equations never did this.
Discussion prompt
Explain why earlier equations in the course never produced false candidates. Then name one other operation that can.
Hint: What were you allowed to multiply by?
Answer:
In earlier chapters both sides were only ever multiplied or divided by constants, which are never nought, so every step was reversible and every candidate genuine. Rational equations are the first place where the multiplier contains the variable and can therefore be nought for some values.
Squaring both sides does the same thing, which is why Lesson 12.3 on radical equations will need exactly this checking step again. Both operations can turn a false statement into a true one — multiplying by nought, or squaring away a sign difference — and both therefore demand that every candidate be tested against the original.
Section
Section 5
Concept
Someone who takes three hours to do a job does a third of it each hour. When two people work together their hourly shares add, and the shares must total one whole job.
\[ \dfrac{t}{a} + \dfrac{t}{b} = 1 \]
Times do not add; rates do.
Figure (svg): Two workers' hourly shares of a job adding to one whole job
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.7 Rational Equations §11.7, pp. 672-672 — Example 4, Solve a Work Problem, and its verbal model
Picture it
Each hour, a fraction each.
Figure (svg): Two columns separating what adds from what does not in a work problem
The answer must be less than the faster person's time alone, since help can only speed things up. That is the quickest sanity check on any work problem.
Worked example
This is Example 4 from the textbook.
\[ \text{You take } 3 \text{ hours alone and Amy takes } 2. \text{ How long together?} \]
Write each hourly share
Why: One over the time alone.
\[ \tfrac{1}{3} \text{ and } \tfrac{1}{2} \]
Write the model
Why: Shares times time make one job.
\[ \tfrac{t}{3} + \tfrac{t}{2} = 1 \]
Multiply by the LCD
Why: Six times every term.
\[ 2 t + 3 t = 6 \]
Solve
Why: Five t equals six.
\[ t = \tfrac{6}{5} \]
Figure (svg): Two workers' hourly shares of a job adding to one whole job
\[ t = \tfrac{6}{5} \text{ hours} = 1 \text{ h } 12 \text{ min} \]
Verify: check the answer is sensible
Why: Six fifths of an hour is one and a fifth hours, and a fifth of sixty minutes is twelve, so the answer is one hour and twelve minutes. That is less than Amy's two hours alone, which it must be — two people cannot be slower than one.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.7 Rational Equations §11.7, pp. 672-672
Faded example
Each share times the time.
Fill in the blanks
\dfrac21 + \dfrac______} = ___
Why: Each fraction is the portion of the job that person completes in t hours, and together they must complete exactly one job. The right side is one, not two, however many people are working.
Worked example
Confirming the answer from the rates.
\[ \text{Verify } t = \tfrac{6}{5} \text{ by adding the two hourly rates.} \]
Add the rates
Why: A third plus a half.
\[ \tfrac{5}{6} \text{ per hour} \]
Interpret
Why: Together they do five sixths of the job each hour.
Find the time for one job
Why: One divided by the rate.
\[ \tfrac{6}{5} \]
Compare
Why: The same answer.
\[ \;\checkmark \]
Figure (svg): Two columns separating what adds from what does not in a work problem
\[ \tfrac{1}{3} + \tfrac{1}{2} = \tfrac{5}{6} \;\Longrightarrow\; t = \tfrac{6}{5} \]
Verify: note the reciprocal relationship
Why: The combined time is the reciprocal of the combined rate, which is why the answer is six fifths rather than five sixths. Getting those two the wrong way round is the commonest arithmetic slip at the end of a work problem.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.7 Rational Equations §11.7, pp. 672-672
Trap
\[ t = 3 + 2 = 5 \quad \text{or} \quad t = \tfrac{3 + 2}{2} = 2.5 \]
Combine the two times directly
Why: Two people and two times, so the times were combined.
Both answers are longer than one person working alone, which cannot be right — help makes a job faster, not slower. Times do not add; the amounts of work done per hour do.
\[ \tfrac{1}{3} + \tfrac{1}{2} = \tfrac{5}{6} \;\Longrightarrow\; t = \tfrac{6}{5} \]
Add the hourly rates, then take the reciprocal
Why: Rates are what accumulate.
The answer must be less than the faster person's time, which rules out both wrong versions instantly.
Prediction
Two people working together.
Predict first
How does the combined time compare with the individual times?
Correct: Less than the smaller of the two.
\[ \tfrac{6}{5} = 1.2 < 2 \]
Why: Adding a second worker can only increase the rate at which the job is done, so the time must fall below what the faster person achieves alone. Here Amy alone takes two hours and together they take one hour twelve minutes, comfortably less. This check rejects the two commonest wrong answers — the sum of the times and their average — without any calculation at all.
Sorting
In a work problem.
Sort into buckets
Sort each quantity by whether the two workers' values add.
Rates and amounts add; durations do not. Both workers spend the same t hours, which is why the same t appears in both fractions of the model rather than two different letters.
Socratic
The two people work at different speeds.
Discussion prompt
Explain why the model uses one variable for both workers' times. Then say what would change if one of them started late.
Hint: How long is each of them actually shovelling?
Answer:
They are working together on the same job at the same time, so they both spend the same number of hours at it — the job finishes for both of them at once. Different speeds mean they complete different amounts in that shared time, which the two different denominators record, but the duration itself is common.
If one started late their times would differ and the model would need two variables, with the relationship between them supplied by the problem — something like the second person working an hour less. The equation would then be t over three plus t minus one over two equals one, which is still a rational equation and is solved the same way.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Cross multiplying | Multiplying by the LCD | |
|---|---|---|
| Applies when | one fraction on each side | always |
| Preparation needed | none | find and factor for the LCD |
| Checking required | yes, against excluded values | yes, against excluded values |
The bottom row is identical, because both methods clear denominators and both can therefore invent a solution. The checking step belongs to the topic rather than to either method.
Pattern
To solve any rational equation, these five moves cover it.
Step one is the step that makes step five possible. Without the list written down, an excluded candidate looks exactly like a genuine solution.
OpenStax Elementary Algebra 2e, §8.6 Solve Rational Equations §8.6
Check
One fraction each side.
Check your understanding
Solve 5/(y + 2) = y/3.
Answer: A
Why: Cross multiplying gives fifteen equals y squared plus two y, so y squared plus two y minus fifteen is nought, which factors as y plus five times y minus three.
Check
Every term gets multiplied.
Check your understanding
Solve 2/x + 1/3 = 4/x.
Answer: A
Why: Multiplying every term by three x gives six plus x equals twelve, so x is six — and six does not make any denominator nought.
Check
Rates add, not times.
Check your understanding
You take 3 hours and Amy takes 2. How long together?
Answer: A
Why: Their rates are a third and a half of the job per hour, totalling five sixths, so one job takes six fifths of an hour — one hour and twelve minutes.
Real world
This is the snow-shovelling question from the lesson opener. You can clear the driveway in 3 hours and Amy can do it in 2.
Discussion prompt
How long will it take working together? Set up the model, solve it, and explain why the answer must be less than two hours.
Hint: Add the hourly shares, not the times.
Answer:
\[ \dfrac{t}{3} + \dfrac{t}{2} = 1 \;\Longrightarrow\; 2t + 3t = 6 \;\Longrightarrow\; t = \tfrac{6}{5} \]
Six fifths of an hour is one hour and twelve minutes, since a fifth of sixty minutes is twelve.
It must be less than two hours because adding a second person can only make the job go faster, so the combined time cannot exceed what the quicker of the two achieves alone. That check rejects the two most tempting wrong answers immediately: adding the times to get five hours, and averaging them to get two and a half, are both longer than Amy manages by herself. Rates of work add; durations do not.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Solving a rational equation with denominator x - 2 gives the single candidate x = 2. What is the answer?
Correct: There is no solution.
\[ x = 2: \; \tfrac{2}{0} \text{ is undefined} \]
Why: At x equal to two the denominator x minus two is nought, so the original equation has no value there at all and two cannot satisfy it. The algebra really was correct, which is what makes this case worth taking seriously: clearing denominators means multiplying both sides by an expression that is nought at exactly the excluded value, and multiplying by nought turns any equation into a true statement. The cleared equation is therefore a slightly larger problem than the original, and its extra root has to be filtered out by hand. Since this was the only candidate, the equation has no solution — which is a complete and correct answer, not an admission of failure. Any other method would produce the same candidate and the same rejection.
Explain it
They said that two people who each take 3 and 2 hours will take 5 hours together.
Discussion prompt
In no more than four sentences, explain what actually adds. Then give them a check that rejects their answer instantly.
Hint: What does each person do in one hour?
Answer:
A usable answer: what adds is how much of the job each person does per hour, not how long they take. You do a third of it each hour and Amy does a half, so together you do five sixths each hour, and one whole job takes six fifths of an hour.
The check is that two people cannot be slower than one. Amy alone takes two hours, so any answer above two hours must be wrong, and five hours fails that test without any calculation.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: The choice is fixed by counting terms: one fraction each side means cross multiplying, anything else means the LCD. Multiplying through is fixed by writing the multiplication out for every term, including ones without variables underneath. Checking is fixed by listing the excluded values before you start solving. Work problems are fixed by adding rates and remembering the answer must beat the faster worker. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of a page write both methods with the condition for each, and note that cross multiplying is the two-term case of the other. Underneath, solve one equation by cross multiplying and one by the LCD, writing out for the second the multiplication of every single term including any without a variable underneath. In the middle, take an equation whose denominators must be factored first, work out what the LCD would have been without factoring, and compare the degrees of the two resulting equations. Beneath that, solve an equation whose only candidate turns out to be excluded, writing the list of forbidden values at the top before you begin and striking the candidate through at the end with a one-line reason. In the lower half, set up a work problem: write each person's hourly share, form the equation, solve it, convert the answer to hours and minutes, and write beside it why the answer must be less than the faster worker's time. Finally, in the margin, write the reason clearing denominators can invent a solution when earlier chapters' methods never did.
Your excluded-value list should be written before any solving. If you only discover the exclusion at the checking stage, the habit has not formed — and this is the lesson where that costs a whole answer rather than a mark.
Recap
Five things, and the fourth is the one that makes the others safe.
| If the question says | Your first move is |
|---|---|
| One fraction on each side | Cross multiply |
| Three or more terms | Find the LCD and multiply through |
| A denominator is a trinomial | Factor it before choosing the LCD |
| You have a candidate solution | Check it against the excluded values |
| Two people work together | Add their hourly rates, not their times |
That completes Chapter 11. Chapter 12 returns to radicals, and the checking habit built here is needed again immediately: squaring both sides of an equation can invent a solution in exactly the way clearing denominators does.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.7 Rational Equations §11.7, pp. 670-677 — everything on these slides traces back here
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