Adding and subtracting rational expressions whose denominators differ. Includes finding a least common denominator from prime factorisations, rewriting each expression over that denominator, adding and subtracting once the denominators match, handling binomial denominators whose least common denominator is their product, and modelling a journey's total time as a sum of two rational expressions.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 11 — Rational Expressions and Equations
Adding and Subtracting with Unlike Denominators
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 11.6 Adding and Subtracting with Unlike Denominators §11.6, pp. 663-669 — the lesson these objectives are drawn from
Warm-up
Lesson 11.5 combined expressions that already shared a denominator. This lesson does the work needed when they do not.
Discussion prompt
Add a third and a quarter. What did you have to do first, and how did you choose the number you used?
Hint: The pieces have to be the same size.
Answer:
\[ \tfrac{1}{3} + \tfrac{1}{4} = \tfrac{4}{12} + \tfrac{3}{12} = \tfrac{7}{12} \]
Both fractions were rewritten in twelfths, because twelve is the smallest number both three and four divide into. Once the pieces matched, the numerators could be added — which is exactly Lesson 11.5's rule, reached after one extra step.
Concept
To add or subtract rational expressions with unlike denominators, first rewrite them so that they share a denominator. The one usually chosen is the least common multiple of the originals.
least common denominator — The least common multiple of the denominators of two or more rational expressions, abbreviated LCD. It is the smallest expression that each original denominator divides into.
After that, the rule from Lesson 11.5 finishes the job.
Figure (svg): Finding a least common denominator from prime factorisations
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.6 Adding and Subtracting with Unlike Denominators §11.6, pp. 663-663
Section
Section 1
Concept
Factor each denominator, take the highest power of every factor that appears in either, and multiply those together. The result is the least common denominator.
Every original denominator divides into it exactly.
Figure (svg): Finding a least common denominator from prime factorisations
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.6 Adding and Subtracting with Unlike Denominators §11.6, pp. 663-663 — the definition of the least common denominator and Example 1
Picture it
Three steps, no guessing.
Figure (svg): Finding a least common denominator from prime factorisations
Taking the highest power rather than the product of everything is what makes the denominator least. Multiplying the two denominators together always works but is often needlessly large.
Worked example
This is Example 1 from the textbook.
\[ \text{Find the LCD of } \dfrac{11}{12x} \text{ and } \dfrac{7}{40x^4}. \]
Factor the first denominator
Why: Twelve is four times three.
\[ 2^2 \cdot 3 \cdot x \]
Factor the second
Why: Forty is eight times five.
\[ 2^3 \cdot 5 \cdot x^4 \]
Take the highest power of each
Why: Four factors in all.
\[ 2 ^{3}, \; 3, \; 5, \; x ^{4} \]
Multiply them
Why: Eight times three times five.
\[ 120 x ^{4} \]
Figure (svg): Finding a least common denominator from prime factorisations
\[ \text{LCD} = 120x^4 \]
Verify: check both denominators divide into it
Why: A hundred and twenty x to the fourth divided by twelve x is ten x cubed, and divided by forty x to the fourth is three. Both come out exactly, which is what a common denominator must do.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.6 Adding and Subtracting with Unlike Denominators §11.6, pp. 663-663
Faded example
Each factor, once.
Fill in the blanks
12x = 2^2 \cdot 3 \cdot x, \; 40x^4 = 2^3 \cdot 5 \cdot x^4 \;\Longrightarrow\; \text3 = 2^4} \cdot 3 \cdot 5 \cdot x^___}
Why: For each factor the higher of the two powers is chosen, so that both denominators divide in exactly. Taking the lower power would leave one of them unable to divide.
Worked example
A case where no multiplying is needed on one side.
\[ \text{Find the LCD of } \dfrac{2}{x} \text{ and } \dfrac{1 - 2x}{x^2}. \]
Factor the first
Why: A single x.
Factor the second
Why: Two x's.
\[ x ^{2} \]
Take the highest power
Why: x squared covers both.
\[ x ^{2} \]
Note the consequence
Why: The second is already over it.
Figure (svg): A sum with unlike denominators worked through
\[ \text{LCD} = x^2 \]
Verify: confirm the smaller divides the larger
Why: x squared divided by x is x, exactly. Whenever one denominator divides the other, the larger one is the LCD and only the other fraction needs rewriting — which halves the work.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.6 Adding and Subtracting with Unlike Denominators §11.6, pp. 664-664
Trap
\[ \text{LCD of } 12x \text{ and } 40x^4 = 480x^5 \]
Multiply the two denominators
Why: Their product is certainly a common denominator.
It is common but not least: it is four times larger than necessary, because the shared factors of four and x were counted twice. The extra size makes every later step harder and the final simplifying longer.
\[ \text{LCD} = 120x^4 \]
Take the highest power of each factor, counting shared factors once
Why: That is what least means.
The product always works and is worth falling back on, but it is worth trying the factorisation first.
Matching
Factor, then take the highest powers.
Match the pairs
Why: The second is a case where one denominator divides the other, so the larger is already the LCD. The last has no shared factor at all, so the product is the least common denominator.
Elimination
For 6x and 8x squared.
Eliminate the wrong options
What is their LCD?
Survives elimination: A
Why: Six is two times three and eight is two cubed, so the highest powers give two cubed times three, or twenty-four, and x squared. Option D fails the basic test that both denominators must divide in.
Socratic
Both give a common denominator.
Discussion prompt
Explain why the highest-power method gives a smaller denominator than multiplying. Then say when the two methods agree.
Hint: What happens to a shared factor?
Answer:
Multiplying the denominators counts every shared factor twice, once from each. Taking the highest power counts it once, at whichever power is needed to cover both — so any factor common to the two denominators is where the saving comes from.
The two methods agree exactly when the denominators share no factor at all, as with x minus one and x plus six. Then the product is already least, which is why binomial denominators so often have the product as their LCD.
Section
Section 2
Concept
To give a fraction a new denominator, work out what the old denominator must be multiplied by, then multiply the numerator by the same quantity. The value is unchanged.
\[ \dfrac{2}{3y} = \dfrac{2 \cdot 5}{3y \cdot 5} = \dfrac{10}{15y} \]
This is multiplying by a well-chosen form of one.
Figure (svg): Rewriting a fraction with a new denominator
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.6 Adding and Subtracting with Unlike Denominators §11.6, pp. 664-664 — Example 2, Rewrite Rational Expressions
Picture it
The same multiplier for both.
Figure (svg): Rewriting a fraction with a new denominator
Finding the multiplier by dividing the LCD by the old denominator is more reliable than trying to see it. It also works when the expressions are complicated.
Worked example
This is Example 2 from the textbook.
\[ \text{Complete } \dfrac{2}{3y} = \dfrac{?}{15y} \text{ and } \dfrac{3x + 7}{4x^2} = \dfrac{?}{36x^5}. \]
Find the first multiplier
Why: Three y times five is fifteen y.
\[ 5 \]
Multiply the numerator
Why: Two times five.
\[ 10 \]
Find the second multiplier
Why: Four x squared times nine x cubed.
\[ 9 x ^{3} \]
Multiply that numerator
Why: Distribute across the bracket.
\[ 27 x ^{4} + 63 x ^{3} \]
Figure (svg): Working out what each fraction must be multiplied by
\[ \dfrac{10}{15y}, \qquad \dfrac{27x^4 + 63x^3}{36x^5} \]
Verify: check the second by simplifying back
Why: Twenty-seven x to the fourth plus sixty-three x cubed factors as nine x cubed times three x plus seven, and thirty-six x to the fifth is nine x cubed times four x squared. Dividing out the nine x cubed recovers the original, which confirms the rewriting.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.6 Adding and Subtracting with Unlike Denominators §11.6, pp. 664-664
Faded example
Divide the LCD by the old denominator.
Fill in the blanks
\dfrac4x28x = ___ \;\Longrightarrow\; \dfrac______ = \dfrac______ = \dfrac___}___
Why: Dividing the LCD by the existing denominator gives the multiplier directly, with no guessing. The same multiplier is then applied to the numerator.
Worked example
The reliable way to get the multiplier.
\[ \text{What must } \dfrac{7}{6x} \text{ be multiplied by to have denominator } 24x^2? \]
Divide the LCD by the denominator
Why: Twenty-four x squared over six x.
\[ 4 x \]
Multiply the denominator
Why: Six x times four x.
\[ 24x^2 \;\checkmark \]
Multiply the numerator
Why: Seven times four x.
\[ 28 x \]
Write the result
Why: The same value, new form.
\[ \tfrac{28x}{24x^2} \]
Figure (svg): Working out what each fraction must be multiplied by
\[ \dfrac{7}{6x} = \dfrac{28x}{24x^2} \]
Verify: test at a value
Why: At x equal to one the original is seven sixths and the rewritten form is twenty-eight over twenty-four, which reduces to seven sixths. Rewriting changes the form and never the value.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.6 Adding and Subtracting with Unlike Denominators §11.6, pp. 665-665
Error analysis
The student rewrote a fraction with a new denominator.
Annotate
On: \( \begin{aligned} \frac{3x + 7}{4x^2} &= \frac{3x + 7}{36x^5} \cdot 9x^3 \\ \text{or} \quad &= \frac{3x + 63x^3}{36x^5} \end{aligned} \)
The error is the same distribution failure that bracketing prevents when subtracting, appearing here in a different place. Whenever a whole numerator is multiplied by something, putting brackets round it first is the habit that saves the second term.
Faded example
Every term gets multiplied.
Fill in the blanks
9x^3(3x + 7) = 27x^4 + 63x^3
Why: Both terms of the numerator are multiplied by nine x cubed. Multiplying only the second is the error the worked example above diagnoses.
Elimination
The fraction looks completely different afterwards.
Eliminate the wrong options
What justifies multiplying top and bottom by the same quantity?
Survives elimination: A
Why: Four over four, or nine x cubed over nine x cubed, is one, and multiplying by one leaves any value alone. That is the same justification as rationalising a denominator in Lesson 9.3.
Socratic
It is often obvious.
Discussion prompt
Say why finding the multiplier by division is worth doing even when it seems visible. Then say where guessing breaks down.
Hint: Think about complicated denominators.
Answer:
Division gives the multiplier as a computation rather than a recognition, so it works identically whether the denominators are simple or awkward. It also produces a check for free: multiplying back should reproduce the LCD exactly.
Guessing breaks down when the denominators are polynomials with several factors, where it is easy to supply a multiplier that is close but wrong — one that produces a common denominator but not the intended one, so the two fractions still fail to match. Dividing removes that possibility entirely.
Section
Section 3
Concept
Once both expressions are written over the least common denominator, the numerators are added exactly as in Lesson 11.5. The result is then simplified.
Only the first two steps are new.
Figure (svg): A sum with unlike denominators worked through
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.6 Adding and Subtracting with Unlike Denominators §11.6, pp. 664-664 — Example 3, Add with Unlike Denominators, and its Study Tip
Picture it
Rewrite, add, simplify.
Figure (svg): A sum with unlike denominators worked through
Only one fraction needed rewriting here because x divides into x squared. That happens often enough to be worth checking before doing any work.
Worked example
This is Example 3 from the textbook.
\[ \text{Simplify } \dfrac{2}{x} + \dfrac{1 - 2x}{x^2}. \]
Find the LCD
Why: x divides into x squared.
\[ x ^{2} \]
Rewrite the first
Why: Multiply top and bottom by x.
\[ \tfrac{2x}{x^2} \]
Add the numerators
Why: The second is already right.
\[ 2 x + 1 - 2 x \]
Simplify
Why: The two x terms cancel.
\[ \tfrac{1}{x^2} \]
Figure (svg): A sum with unlike denominators worked through
\[ \dfrac{2}{x} + \dfrac{1 - 2x}{x^2} = \dfrac{1}{x^2} \]
Verify: test at a value
Why: At x equal to one the fractions are two and negative one, adding to one, and the answer gives one over one. At x equal to two they are one and negative three quarters, adding to a quarter, and the answer gives a quarter.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.6 Adding and Subtracting with Unlike Denominators §11.6, pp. 664-664
Faded example
The first fraction needs an x.
Fill in the blanks
\dfrac2x1 = \dfrac___}___ \;\Longrightarrow\; \dfrac______ = \dfrac___}___
Why: Multiplying top and bottom by x gives the first fraction the common denominator. The two x terms in the combined numerator are then opposites and vanish.
Worked example
Guided Practice 12, where neither denominator divides the other.
\[ \text{Simplify } \dfrac{5}{3x^2} + \dfrac{1}{9x^3}. \]
Find the LCD
Why: Highest powers of three and x.
\[ 9 x ^{3} \]
Rewrite the first
Why: Multiply by three x.
\[ \tfrac{15x}{9x^3} \]
The second is already right
Why: Nothing to do.
\[ \tfrac{1}{9x^3} \]
Add and simplify
Why: Fifteen x plus one.
\[ \tfrac{15x + 1}{9x^3} \]
Figure (svg): Working out what each fraction must be multiplied by
\[ \dfrac{15x + 1}{9x^3} \]
Verify: check nothing cancels
Why: Fifteen x plus one has no common factor with nine x cubed, since it is not divisible by three or by x. The answer is therefore in simplest form, which is worth confirming rather than assuming.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.6 Adding and Subtracting with Unlike Denominators §11.6, pp. 664-664
Trap
\[ \dfrac{2}{x} + \dfrac{1 - 2x}{x^2} = \dfrac{2 + 1 - 2x}{x^2} \]
Add the numerators and use the larger denominator
Why: One denominator was already x squared, so it was adopted.
The first fraction was not rewritten, so its numerator is still counting pieces of a different size. The result is three minus two x over x squared, which at x equal to one gives one — but so does the correct answer, so a second value is needed: at x equal to two the true sum is a quarter and this gives negative a quarter.
\[ = \dfrac{2x + 1 - 2x}{x^2} = \dfrac{1}{x^2} \]
Rewrite every fraction over the LCD before adding anything
Why: Both numerators must count the same size of piece.
Testing at two values rather than one is worth the extra ten seconds here.
Sorting
Compare each denominator with the LCD.
Sort into buckets
For an LCD of 24x squared, sort each fraction by whether it needs rewriting.
Two of the six are already correct, which is a saving worth spotting before starting. Checking each denominator against the LCD first tells you how much work there actually is.
Hypothesis
The numerators are what get added.
Predict first
What goes wrong if numerators are added over different denominators?
Correct: The numerators count pieces of different sizes, so the total is meaningless.
\[ \tfrac{1}{3} + \tfrac{1}{4} \ne \tfrac{2}{7} \]
Why: A numerator says how many pieces there are and the denominator says how big each one is, so adding two numerators over different denominators is like adding a count of thirds to a count of quarters and calling the result a count of anything. The same reasoning is why a third plus a quarter is not two sevenths. Making the denominators match is what makes the counts comparable.
Socratic
Usually a single substitution suffices.
Discussion prompt
Explain why checking a sum at one value can miss an error. Then say how to choose values that will not.
Hint: Two different expressions can agree somewhere.
Answer:
Two different expressions can happen to agree at a particular value, in the same way that two different lines cross at a point. In the trap above, the correct and incorrect answers both gave one at x equal to one, so that test would have passed a wrong answer.
Choosing two values, ideally not adjacent and neither of them one, makes an accidental agreement very unlikely — two distinct rational expressions can agree at a few points but not at many. Avoiding nought and one is worth doing generally, since those values make so many terms collapse that errors hide in them.
Section
Section 4
Concept
Subtraction adds one requirement to the addition procedure: once both are over the LCD, the numerator being subtracted must be bracketed so the minus reaches every term.
Both errors of this chapter can appear in one problem.
Figure (svg): A difference with unlike denominators worked through
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.6 Adding and Subtracting with Unlike Denominators §11.6, pp. 665-665 — Example 4, Subtract with Unlike Denominators
Picture it
Each fraction gets its own.
Figure (svg): A difference with unlike denominators worked through
Both fractions needed rewriting here and by different amounts, which is the usual case. The bracketing then happens exactly as in the previous lesson.
Worked example
This is Example 4 from the textbook.
\[ \text{Simplify } \dfrac{7}{6x} - \dfrac{x - 1}{8x^2}. \]
Find the LCD
Why: Six is two times three, eight is two cubed.
\[ 24 x ^{2} \]
Rewrite both
Why: By four x and by three.
\[ \tfrac{28x}{24x^2} - \tfrac{3x - 3}{24x^2} \]
Bracket and subtract
Why: The minus reaches both terms.
\[ 28 x - (3 x - 3) \]
Simplify
Why: Twenty-eight x minus three x plus three.
\[ \tfrac{25x + 3}{24x^2} \]
Figure (svg): A difference with unlike denominators worked through
\[ \dfrac{7}{6x} - \dfrac{x - 1}{8x^2} = \dfrac{25x + 3}{24x^2} \]
Verify: test at a value
Why: At x equal to one the fractions are seven sixths and nought, so the difference is seven sixths — and the answer gives twenty-eight over twenty-four, which is seven sixths. The second fraction vanished at that value, so a second test at x equal to two is worth doing.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.6 Adding and Subtracting with Unlike Denominators §11.6, pp. 665-665
Faded example
Each denominator needs its own.
Fill in the blanks
\dfrac4x3 \cdot \dfrac___}___ = \dfrac______, \qquad \dfrac______ \cdot \dfrac______} = \dfrac______
Why: Each fraction is multiplied by its own form of one, chosen so that both denominators become the LCD. The multipliers differ because the denominators did.
Worked example
Where the sign work actually happens.
\[ \text{Why does } 28x - (3x - 3) \text{ give } 25x + 3? \]
Distribute the minus
Why: Both terms change sign.
\[ 28 x - 3 x + 3 \]
Note the second sign
Why: Minus three becomes plus three.
\[ +3 \]
Combine like terms
Why: Twenty-eight x minus three x.
\[ 25 x \]
Write the numerator
Why: With its positive constant.
\[ 25 x + 3 \]
Figure (svg): A difference with unlike denominators worked through
\[ 28x - (3x - 3) = 25x + 3 \]
Verify: check the constant on its own
Why: Setting x to nought leaves nought minus the quantity nought minus three, which is three — matching the plus three in the answer. Checking just the constant term is a quick partial test that catches most sign errors.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.6 Adding and Subtracting with Unlike Denominators §11.6, pp. 665-665
Trap
\[ \dfrac{28x}{24x^2} - \dfrac{3x - 3}{24x^2} = \dfrac{28x - 3x - 3}{24x^2} \]
Write the second numerator straight after the minus sign
Why: Its terms were copied down in order.
The minus three should become plus three, giving twenty-five x plus three rather than twenty-five x minus three. The difference is six over twenty-four x squared, which is never nought.
\[ = \dfrac{28x - (3x - 3)}{24x^2} = \dfrac{25x + 3}{24x^2} \]
Bracket the second numerator before combining
Why: Then distribute the minus.
This is Lesson 11.5's rule, unchanged; only the rewriting step in front of it is new.
Elimination
Subtracting 3x - 3 from 28x.
Eliminate the wrong options
What is the combined numerator?
Survives elimination: A
Why: Both terms of the bracket change sign, so minus three becomes plus three. Checking the constant term alone at x equal to nought distinguishes all four options immediately.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Step | Adding | Subtracting |
|---|---|---|
| Find the LCD | yes | yes |
| Rewrite both | yes | yes |
| Bracket the second numerator | not needed | essential |
The first two rows are identical, so subtraction is addition with one extra precaution. That precaution is where nearly all the errors in this lesson occur.
Socratic
There are two independent steps.
Discussion prompt
Name the two distinct errors a subtraction with unlike denominators invites. Then say how to test for each.
Hint: One is about denominators and one about signs.
Answer:
The first is combining numerators before the denominators match, which produces a total of incomparable pieces. The second is failing to bracket the subtracted numerator, which loses the sign on every term after the first.
The first is caught by checking that both denominators are literally identical before any combining, and by testing the answer at two different values. The second is caught by checking the constant term alone: substitute nought for the variable and see whether the constants combine as they should.
Section
Section 5
Concept
When neither denominator can be factored and they share no factor, the least common denominator is their product. Each fraction is then multiplied by the other's denominator.
\[ \text{LCD of } (x-1) \text{ and } (x+6) \text{ is } (x-1)(x+6) \]
The product must contain both factors.
Figure (svg): Two binomial denominators and their least common denominator
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.6 Adding and Subtracting with Unlike Denominators §11.6, pp. 665-665 — Example 5, Add with Unlike Binomial Denominators
Picture it
Nothing shared, so nothing saved.
Figure (svg): Two binomial denominators and their least common denominator
Had the two denominators shared a factor, that factor would appear only once in the LCD. Here they share none, so the product is already least.
Worked example
This is Example 5 from the textbook.
\[ \text{Simplify } \dfrac{x + 2}{x - 1} + \dfrac{12}{x + 6}. \]
Find the LCD
Why: Neither factors; take the product.
\[ (x - 1) (x + 6) \]
Rewrite the first
Why: Multiply by x plus six.
\[ \tfrac{(x+2)(x+6)}{(x-1)(x+6)} \]
Rewrite the second
Why: Multiply by x minus one.
\[ \tfrac{12(x-1)}{(x-1)(x+6)} \]
Add and simplify
Why: Expand and combine.
\[ \tfrac{x^2 + 20x}{(x-1)(x+6)} \]
Figure (svg): Two binomial denominators and their least common denominator
\[ \dfrac{x(x + 20)}{(x - 1)(x + 6)} \]
Verify: check the expansion
Why: x plus two times x plus six is x squared plus eight x plus twelve, and twelve times x minus one is twelve x minus twelve. Adding gives x squared plus twenty x, since the twelves cancel — and factoring out an x gives the answer.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.6 Adding and Subtracting with Unlike Denominators §11.6, pp. 665-665
Faded example
Nothing is shared.
Fill in the blanks
\text1 x - 1 \text6 x + 6 \text___ (x - ___)(x + ___)
Why: The LCD must contain both factors, and since neither divides the other, it contains each exactly once. That product is their least common multiple.
Worked example
The lesson opener's situation, with numbers supplied.
\[ \text{A } 300 \text{ mile trip is driven } 150 \text{ miles at } x \text{ mph and } 150 \text{ at } x + 10. \text{ Find the total time.} \]
Write each time
Why: Distance over speed.
\[ \tfrac{150}{x} + \tfrac{150}{x + 10} \]
Find the LCD
Why: No shared factor.
\[ x(x + 10) \]
Rewrite both
Why: By x plus ten and by x.
\[ \tfrac{150(x+10) + 150x}{x(x+10)} \]
Combine
Why: Three hundred x plus fifteen hundred.
\[ \tfrac{300x + 1500}{x(x + 10)} \]
Figure (svg): A journey split into two stretches at different speeds
\[ \dfrac{300x + 1500}{x(x + 10)} \text{ hours} \]
Verify: test with a realistic speed
Why: At x equal to fifty the two stretches take three hours and two and a half hours, totalling five and a half. The formula gives fifteen thousand plus fifteen hundred, over fifty times sixty — that is sixteen thousand five hundred over three thousand, which is five and a half.
Trap
\[ \text{total time} = \dfrac{300}{x + (x + 10)} \]
Divide the whole distance by the combined speed
Why: Distance over speed gives time, so the total distance was divided by the total speed.
Speeds do not add like that: driving at fifty and then at sixty is not the same as driving at a hundred and ten. Each stretch has its own time, and it is the times that add.
\[ \dfrac{150}{x} + \dfrac{150}{x + 10} \]
Compute each stretch's time separately, then add
Why: Time is what accumulates.
The check with fifty miles an hour exposes the wrong version instantly: it would give less than three hours for a three-hundred-mile trip.
Sorting
Only when nothing is shared.
Sort into buckets
Sort each pair of denominators by whether their product is the LCD.
Every pair of distinct binomials here has the product as its LCD, and every pair of monomials shares something. That is a useful rule of thumb, though it is worth checking rather than assuming.
Prediction
Such as x - 1 and x squared - 1.
Predict first
What would the LCD be?
Correct: x squared - 1, since x - 1 divides into it.
\[ x^2 - 1 = (x-1)(x+1) \]
Why: Factoring gives x minus one and the pair x minus one times x plus one, so the second denominator already contains the first. The LCD is therefore the larger one, and only the first fraction needs rewriting. Taking the product would give a denominator with x minus one squared in it, which is a common denominator but four times more work than necessary — and it would need simplifying at the end.
Socratic
It is one trip.
Discussion prompt
Explain why the total time for a two-speed journey is a sum of rational expressions. Then say what makes their denominators differ.
Hint: How is time computed from distance and speed?
Answer:
Time is distance divided by speed, so each stretch contributes a fraction with its own speed underneath. The total time is the sum of those two fractions, because time is what accumulates across the journey — unlike speed, which does not.
The denominators differ precisely because the speeds do, so a journey at two speeds is a natural source of unlike denominators. Combining them into one expression is what lets you ask questions about the trip as a whole, such as what speed would be needed to finish within a given time.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Like (11.5) | Unlike (11.6) | |
|---|---|---|
| First step | combine the numerators | find the least common denominator |
| Rewriting needed | none | each fraction over the LCD |
| After that | simplify | exactly the same as Lesson 11.5 |
The right-hand column is the left-hand one with two steps in front of it. Once the denominators match, this lesson becomes the previous one.
Pattern
To add or subtract rational expressions with unlike denominators, these five moves cover it.
Steps one to three are all that distinguish this lesson from the last one. Step four is Lesson 11.5's rule applied unchanged.
OpenStax Elementary Algebra 2e, §8.4 Add and Subtract Rational Expressions with Unlike Denominators §8.4
Check
Highest power of each factor.
Check your understanding
What is the LCD of 11/(12x) and 7/(40x to the fourth)?
Answer: A
Why: Twelve is two squared times three and forty is two cubed times five, so the highest powers give eight times three times five, or a hundred and twenty, with x to the fourth.
Check
Rewrite before combining.
Check your understanding
Simplify 2/x + (1 - 2x)/x squared.
Answer: A
Why: The first fraction becomes two x over x squared, and two x plus one minus two x is one.
Check
Bracket the second numerator.
Check your understanding
Simplify 7/(6x) - (x - 1)/(8x squared).
Answer: A
Why: The LCD is twenty-four x squared, giving twenty-eight x minus the quantity three x minus three, which is twenty-five x plus three.
Real world
This is the road trip question from the lesson opener. A 300 mile journey is driven in two equal stretches at different speeds, and the total time is what you want to know.
Discussion prompt
The first 150 miles are driven at x miles per hour and the second 150 at ten miles per hour faster. Write the total time as a single rational expression and test it at a realistic speed.
Hint: Time is distance over speed, and times add.
Answer:
\[ \dfrac{150}{x} + \dfrac{150}{x + 10} = \dfrac{150(x + 10) + 150x}{x(x + 10)} = \dfrac{300x + 1500}{x(x + 10)} \]
At fifty miles an hour the two stretches take three hours and two and a half hours, totalling five and a half — and the formula gives sixteen thousand five hundred over three thousand, which is five and a half.
Note that the speeds themselves cannot be added: driving at fifty and then at sixty is not a journey at a hundred and ten. It is the times that accumulate, which is exactly why the total is a sum of two fractions with different denominators rather than a single division. Having one expression for the whole trip then lets you ask what x would be needed to finish in, say, five hours.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
What is the least common denominator of 6x and 8x squared?
Correct: 24x squared.
\[ 6x = 2 \cdot 3 \cdot x, \quad 8x^2 = 2^3 \cdot x^2 \;\Longrightarrow\; 2^3 \cdot 3 \cdot x^2 = 24x^2 \]
Why: Six is two times three and eight is two cubed, so the highest power of two is two cubed and the three appears once, giving twenty-four; the highest power of x is x squared. The product, forty-eight x cubed, is a common denominator but not the least, because it counts the shared factor of two x twice — using it works but leaves an answer that needs simplifying at the end. Adding the numbers to get fourteen has no justification at all. The last option fails the basic requirement: eight x squared does not divide into twenty-four x, so the two fractions could not both be rewritten over it. The test for any candidate is whether both denominators divide into it exactly.
Explain it
They added the numerators while the denominators were still different.
Discussion prompt
In no more than four sentences, explain why that cannot work. Then tell them the two steps that come first.
Hint: Use a numerical example.
Answer:
A usable answer: a numerator counts pieces and the denominator says how big they are, so adding counts of different-sized pieces gives a number of nothing in particular. A third plus a quarter is not two sevenths, and the same goes for expressions.
First find the least common denominator by factoring both denominators and taking the highest power of each factor. Then rewrite each fraction over it by multiplying top and bottom by whatever its denominator needs, and only then combine the numerators.
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 LCD is fixed by factoring both denominators and taking the highest power of each factor. The multiplier is fixed by dividing the LCD by the existing denominator rather than guessing. Distributing is fixed by writing brackets round the numerator before multiplying. Bracketing when subtracting is fixed by making the brackets part of the subtraction step itself. 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 find a least common denominator by writing both denominators as products of primes and powers, ringing the highest power of each factor, and multiplying them — then check that both originals divide into your answer exactly. Underneath, rewrite two fractions over that denominator, showing for each one the division that produced its multiplier and the distribution of that multiplier across the whole numerator. In the middle, work an addition in full, and beside it work a subtraction in full with the second numerator bracketed, circling every sign that changed. Beneath that, take two binomial denominators sharing no factor, form their product as the LCD, add the two fractions, and expand and combine the numerator carefully. In the lower corner, write the two-speed journey model, test it at a realistic speed against a direct calculation, and note why the speeds themselves cannot be added. Finally, in the margin, write the two independent errors this lesson invites and the test for each.
Test every answer at two different values, neither of them nought or one. A single test can pass a wrong answer, which is exactly what happens in the trap where an unrewritten fraction happens to agree at x equal to one.
Recap
Five things, and the first two are the only genuinely new ones.
| If the question says | Your first move is |
|---|---|
| The denominators differ | Factor them and find the LCD |
| Rewrite over a new denominator | Divide to find the multiplier |
| Multiply a numerator by something | Bracket it, then distribute |
| Subtract two expressions | Rewrite, then bracket the second numerator |
| Two binomials share no factor | Their product is the LCD |
Lesson 11.7 solves equations containing rational expressions. Everything from this chapter is needed there — and so is Lesson 11.1's warning, because clearing the denominators can produce a solution the original equation forbids.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.6 Adding and Subtracting with Unlike Denominators §11.6, pp. 663-669 — everything on these slides traces back here
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