Adding and subtracting rational expressions that share a denominator. Includes the rule for combining numerators over a common denominator, bracketing a subtracted numerator so the negative distributes correctly, simplifying the result by factoring and cancelling, sums that reduce to a constant, and the excluded values such a combination carries.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 11 — Rational Expressions and Equations
Adding and Subtracting with Like 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 11.5 Adding and Subtracting with Like Denominators §11.5, pp. 658-662 — the lesson these objectives are drawn from
Warm-up
Lesson 11.4 multiplied rational expressions, which needed no common denominator. Adding does, and this lesson takes the easy case where they already match.
Discussion prompt
Work out three sevenths plus two sevenths, and then three sevenths minus two sevenths. What happened to the seven each time?
Hint: Count the pieces.
Answer:
\[ \tfrac{3}{7} + \tfrac{2}{7} = \tfrac{5}{7}, \qquad \tfrac{3}{7} - \tfrac{2}{7} = \tfrac{1}{7} \]
The seven stayed exactly as it was in both. It names the size of each piece, and combining three pieces with two changes how many there are rather than how big they are. That is the whole rule, and it carries over to rational expressions unchanged.
Concept
To add or subtract rational expressions with the same denominator, combine their numerators and write the result over that common denominator.
common denominator — A denominator shared by two or more rational expressions, allowing their numerators to be combined directly.
The denominator is never added or subtracted.
Figure (svg): The rule for adding and subtracting with like denominators
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.5 Adding and Subtracting with Like Denominators §11.5, pp. 658-658
Section
Section 1
Concept
If a, b and c are polynomials with c not nought, then a over c plus b over c equals a plus b, all over c. Only the numerators are combined.
\[ \dfrac{a}{c} + \dfrac{b}{c} = \dfrac{a + b}{c} \]
The denominator names the size of the pieces.
Figure (svg): The rule for adding and subtracting with like denominators
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.5 Adding and Subtracting with Like Denominators §11.5, pp. 658-658 — the Adding or Subtracting with Like Denominators rules and Example 1
Picture it
Add or subtract the tops.
Figure (svg): The rule for adding and subtracting with like denominators
The restriction that c is not nought is where this lesson's excluded values come from. It applies from the moment the expressions are written.
Worked example
This is Example 1 from the textbook.
\[ \text{Simplify } \dfrac{5}{2x} + \dfrac{2x - 5}{2x}. \]
Add the numerators
Why: Five plus the bracket.
\[ 5 + (2 x - 5) \]
Combine like terms
Why: The fives cancel.
\[ 2 x \]
Write over the denominator
Why: Unchanged.
\[ \tfrac{2x}{2x} \]
Simplify
Why: A quantity over itself.
\[ 1 \]
Figure (svg): A sum whose numerators cancel to leave a constant
\[ \dfrac{5}{2x} + \dfrac{2x - 5}{2x} = 1 \]
Verify: test at a value
Why: At x equal to one the two fractions are five halves and negative three halves, which add to one. The answer is a constant, so any value of x should give one — and that is worth testing twice to be convinced.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.5 Adding and Subtracting with Like Denominators §11.5, pp. 658-658
Faded example
The denominator is written once.
Fill in the blanks
\dfrac2x1 + \dfrac______ = \dfrac______} = \dfrac______ = ___
Why: The fives are opposites and cancel inside the numerator, leaving the denominator's own expression on top. A quantity divided by itself is one, not nought.
Worked example
Guided Practice, with no dramatic cancellation.
\[ \text{Simplify } \dfrac{x}{x - 2} + \dfrac{3x}{x - 2}. \]
Add the numerators
Why: x plus three x.
\[ 4 x \]
Write over the denominator
Why: Copied down once.
\[ \tfrac{4x}{x - 2} \]
Look for common factors
Why: Four x and x minus two share none.
State the restriction
Why: The denominator forbids two.
\[ x \ne 2 \]
Figure (svg): The rule for adding and subtracting with like denominators
\[ \dfrac{4x}{x - 2}, \quad x \ne 2 \]
Verify: test at a value
Why: At x equal to three the fractions are three and nine, adding to twelve, and the answer gives twelve over one. The x in the numerator cannot cancel with the x inside the denominator, because that x is part of a difference.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.5 Adding and Subtracting with Like Denominators §11.5, pp. 659-659
Trap
\[ \dfrac{3}{x} + \dfrac{2}{x} = \dfrac{5}{2x} \]
Add both the numerators and the denominators
Why: Both parts were combined for symmetry.
The denominator names the size of the pieces, which does not change when pieces are counted together. At x equal to one the true sum is five and this gives two and a half.
\[ \dfrac{3}{x} + \dfrac{2}{x} = \dfrac{5}{x} \]
Combine the numerators and copy the denominator down once
Why: Only the count changes.
Three sevenths plus two sevenths is five sevenths, not five fourteenths, which is the same rule on numbers.
Elimination
Adding two expressions that share one.
Eliminate the wrong options
What should the denominator of the answer be?
Survives elimination: A
Why: The common denominator is copied down unchanged because it describes the size of each piece. Only the numerators, which count the pieces, are combined.
Sorting
Only if the denominators already match.
Sort into buckets
Sort each pair by whether this lesson's rule applies as it stands.
Half of these need the work of Lesson 11.6 before anything can be combined. Checking that the denominators genuinely match is the first thing to do.
Socratic
Both parts of the fraction seem symmetric.
Discussion prompt
Explain why combining fractions changes the numerator but not the denominator. Then say what would have to be true for the denominator to change.
Hint: What does each part of a fraction tell you?
Answer:
The denominator says what size the pieces are and the numerator says how many there are. Putting three pieces together with two gives five pieces of the same size, so the count changes and the size does not. That is why three sevenths plus two sevenths is five sevenths.
The denominator would change only if the pieces themselves were being resized, which is what happens when the fractions are multiplied — a third of a quarter really is a smaller piece, a twelfth. Addition never resizes the pieces, which is exactly why it needs them to be the same size in the first place.
Section
Section 2
Concept
When subtracting, put the second numerator in brackets before combining. The minus sign then distributes to every term inside, rather than only to the first.
\[ \dfrac{a}{c} - \dfrac{b}{c} = \dfrac{a - (b)}{c} \]
This is Lesson 10.1's rule about subtracting polynomials.
Figure (svg): A subtracted numerator written in brackets
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.5 Adding and Subtracting with Like Denominators §11.5, pp. 659-659 — Example 2 and its Study Tip on using parentheses when subtracting
Picture it
Then distribute the minus.
Figure (svg): A subtracted numerator written in brackets
Writing the brackets is a two-second habit that removes the whole class of error. Once they are there, the distribution is mechanical.
Worked example
The method of Example 2 from the textbook.
\[ \text{Simplify } \dfrac{4x}{x - 2} - \dfrac{2x + 4}{x - 2}. \]
Bracket the second numerator
Why: Before combining anything.
\[ 4 x - (2 x + 4) \]
Distribute the minus
Why: Both terms change sign.
\[ 4 x - 2 x - 4 \]
Combine like terms
Why: Two x minus four.
\[ 2 x - 4 \]
Factor and cancel
Why: Two times x minus two.
\[ 2 \]
Figure (svg): The negative distributed across a subtracted numerator
\[ \dfrac{4x}{x - 2} - \dfrac{2x + 4}{x - 2} = 2 \]
Verify: test at a value
Why: At x equal to three the fractions are twelve and ten, whose difference is two. At x equal to four they are eight and six, again differing by two — the answer really is constant.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.5 Adding and Subtracting with Like Denominators §11.5, pp. 659-659
Faded example
Every term inside the brackets.
Fill in the blanks
4x - (2x + 4) = 4x - 2x - 4 = 2x - 4
Why: Both terms inside the brackets change sign, not just the first. The second term is the one that gets left behind, and it is the one worth checking every time.
Worked example
The same problem with the brackets omitted.
\[ \text{What goes wrong in } 4x - 2x + 4 \text{ instead of } 4x - (2x + 4)? \]
Combine without brackets
Why: The four stays positive.
\[ 2 x + 4 \]
Combine with brackets
Why: The four becomes negative.
\[ 2 x - 4 \]
Compare the results
Why: They differ by eight.
\[ \text{wrong by } 8 \]
Check at a value
Why: At x equal to three.
\[ 10 \text{ against } 2 \]
Figure (svg): The negative distributed across a subtracted numerator
\[ 4x - (2x + 4) = 2x - 4 \quad \text{not} \quad 2x + 4 \]
Verify: see how far wrong it goes
Why: Without the brackets the answer would be two x plus four over x minus two, which at x equal to three is ten rather than two. The error is not small, and it survives every later step.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.5 Adding and Subtracting with Like Denominators §11.5, pp. 659-659
Error analysis
The student subtracted two rational expressions with a common denominator.
Annotate
On: \( \begin{aligned} \frac{4x}{x-2} - \frac{2x + 4}{x-2} &= \frac{4x - 2x + 4}{x - 2} \\ &= \frac{2x + 4}{x - 2} \end{aligned} \)
The student's answer even looks plausible, since it is a tidy rational expression, and nothing about its appearance suggests an error. Substituting x equal to three settles it: the true difference is two and this gives ten.
Elimination
Subtracting 2x + 4 from 4x.
Eliminate the wrong options
What is the combined numerator?
Survives elimination: A
Why: Only the second numerator is subtracted, and all of it is. Option B is the missing-bracket error and option D over-applies the minus to both expressions.
Prediction
In the worked example above.
Predict first
How does the wrong answer compare with the right one?
Correct: It differs by a fixed amount, here eight over the denominator.
\[ \tfrac{2x+4}{x-2} - \tfrac{2x-4}{x-2} = \tfrac{8}{x-2} \]
Why: The error changes plus four into minus four, so the numerator is eight too large and the whole expression is eight over x minus two too large. That gap depends on x and is never nought, so the wrong answer is wrong for every permitted value — there is no range where it happens to be right. At x equal to three the gap is eight, which is why the two answers came out as ten and two.
Socratic
The rule looks symmetric.
Discussion prompt
Explain why subtracting rational expressions causes more errors than adding them. Then say what makes the bracket habit effective.
Hint: How many terms does the sign affect?
Answer:
When adding, every term keeps its sign and the numerators simply run together, so nothing has to be tracked. When subtracting, one sign has to be applied to an unknown number of terms, and the terms after the first are easy to overlook because the minus sign is physically distant from them.
Writing the brackets converts a distributed obligation into a single visible one: instead of remembering to negate each term as you meet it, you negate one bracket in one step. It is the same reason Lesson 10.1 recommended rewriting the whole subtracted polynomial with flipped signs before combining anything.
Section
Section 3
Concept
Nothing can be cancelled while the numerators are separate. After they are combined, factor the new numerator and the denominator and divide out any shared factor.
The combining is what creates the cancellation.
Figure (svg): A difference that simplifies once the numerator is combined
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.5 Adding and Subtracting with Like Denominators §11.5, pp. 659-659 — Example 3, Simplify after Subtracting
Picture it
Combine, factor, cancel.
Figure (svg): A difference that simplifies once the numerator is combined
Neither original fraction could be simplified on its own. The shared factor only appeared once the numerators were combined, which is why simplifying belongs at the end.
Worked example
This is Example 3 from the textbook.
\[ \text{Simplify } \dfrac{4x}{3x^2 + x - 2} - \dfrac{x + 2}{3x^2 + x - 2}. \]
Bracket and subtract
Why: Four x minus the bracket.
\[ 4 x - (x + 2) \]
Combine like terms
Why: Three x minus two.
\[ 3 x - 2 \]
Factor the denominator
Why: A trinomial with a leading coefficient.
\[ (3 x - 2) (x + 1) \]
Cancel
Why: Three x minus two appears above and below.
\[ \tfrac{1}{x + 1} \]
Figure (svg): A difference that simplifies once the numerator is combined
\[ \dfrac{4x}{3x^2 + x - 2} - \dfrac{x + 2}{3x^2 + x - 2} = \dfrac{1}{x + 1} \]
Verify: test at a value
Why: At x equal to one the denominator is two, so the fractions are two and three halves, whose difference is a half — and the answer gives one over two. The two originals were complicated and the answer is very simple, which is typical when a cancellation appears.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.5 Adding and Subtracting with Like Denominators §11.5, pp. 659-659
Faded example
The shared factor appears at the end.
Fill in the blanks
4x - (x + 2) = 3x - 2, \quad 3x^2 + x - 2 = (3x - 2)(x + 1)
Why: The combined numerator turned out to be exactly one factor of the denominator, which is what made the cancellation possible. Neither fraction on its own had that factor on top.
Worked example
Exercise 8's pattern, with the numerators combining to a multiple of the denominator.
\[ \text{Simplify } \dfrac{5x}{x - 4} - \dfrac{20}{x - 4}. \]
Subtract the numerators
Why: Five x minus twenty.
\[ 5 x - 20 \]
Factor it
Why: Common factor five.
\[ 5(x - 4) \]
Compare with the denominator
Why: It is the same bracket.
\[ \tfrac{5(x-4)}{x-4} \]
Cancel
Why: The bracket divides out.
\[ 5 \]
Figure (svg): A sum whose numerators cancel to leave a constant
\[ \dfrac{5x}{x - 4} - \dfrac{20}{x - 4} = 5 \]
Verify: test at two values
Why: At x equal to five the fractions are twenty-five and twenty, differing by five; at x equal to six they are fifteen and ten, again differing by five. The difference really is constant, though the restriction that x is not four still applies.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.5 Adding and Subtracting with Like Denominators §11.5, pp. 660-660
Trap
\[ \dfrac{4x}{3x^2 + x - 2} - \dfrac{x + 2}{3x^2 + x - 2} \;\Longrightarrow\; \text{cancel the } x \text{'s} \]
Look for cancellations in each fraction first
Why: There are x's on top and bottom in both.
The denominators are sums, so nothing in them is a factor until they are factored — and even then, neither numerator shares a factor with them. The cancellation only exists after the numerators are combined.
\[ = \dfrac{3x - 2}{(3x-2)(x+1)} = \dfrac{1}{x + 1} \]
Combine the numerators first, then look for factors
Why: The combining is what creates the shared factor.
This is the reverse of Lesson 11.4, where cancelling first saved work; here it is impossible until later.
Sorting
In a sum or difference of rational expressions.
Sort into buckets
Sort each moment by whether cancelling is possible then.
The last item is the trap: matching letters are not matching factors, and a letter inside a sum is a term. Nothing may be cancelled until both parts of a single fraction are products.
Hypothesis
Neither fraction could be simplified alone.
Predict first
What makes the shared factor appear?
Correct: The combined numerator is a different polynomial from either original.
\[ 4x - (x + 2) = 3x - 2, \quad 3x^2 + x - 2 = (3x-2)(x+1) \]
Why: Four x and x plus two share no factor with three x squared plus x minus two, but their difference, three x minus two, is one of its factors. Subtracting produced a new polynomial that happened to match, which is not a coincidence so much as a design choice by whoever set the problem — but the mechanism is real, and it is why simplifying must wait until after the combining.
Socratic
There, cancelling first was better.
Discussion prompt
Explain why multiplying rewards cancelling first while adding requires cancelling last. Then say what the two cases have in common.
Hint: When is the expression a single fraction?
Answer:
A product of two fractions is already effectively a single fraction — all the numerators multiply together and all the denominators do — so every factor is available to cancel from the start. A sum is not a single fraction until the numerators are combined, so before that step there is nothing above the bar to cancel with.
What they have in common is that cancelling requires one factor above a single fraction bar and one below it. Multiplication reaches that state immediately and addition reaches it only after combining, so the same principle produces opposite advice about ordering.
Section
Section 4
Concept
When the combined numerator turns out to be a constant multiple of the denominator, the whole expression reduces to a number. The restriction from the denominator still applies.
The expression is constant but not defined everywhere.
Figure (svg): A sum whose numerators cancel to leave a constant
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.5 Adding and Subtracting with Like Denominators §11.5, pp. 658-660 — Example 1 and the exercises where the result is a constant
Picture it
The numerator matched the denominator.
Figure (svg): A sum whose numerators cancel to leave a constant
Such an answer looks surprising and is easy to distrust. Testing at two different values is the quickest way to be convinced that it really is constant.
Worked example
The pattern of Exercise 8.
\[ \text{Show that } \dfrac{5x}{x - 4} - \dfrac{20}{x - 4} \text{ is constant.} \]
Combine the numerators
Why: Five x minus twenty.
\[ 5 x - 20 \]
Factor
Why: Five times x minus four.
\[ 5(x - 4) \]
Cancel
Why: The denominator divides out.
\[ 5 \]
Note the restriction
Why: Four is still forbidden.
\[ x \ne 4 \]
Figure (svg): A sum whose numerators cancel to leave a constant
\[ 5, \quad x \ne 4 \]
Verify: test at several values
Why: At x equal to five, six and ten the difference is five each time. At x equal to four both fractions are undefined, so the constant answer has one point missing from its domain.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.5 Adding and Subtracting with Like Denominators §11.5, pp. 660-660
Faded example
The numerator matches the denominator.
Fill in the blanks
5x - 20 = 5(x - 4) \;\Longrightarrow\; \dfrac5___ = ___
Why: The numerator was five times the denominator, so the quotient is five. Recognising the denominator inside the combined numerator is what makes this visible.
Worked example
A constant function with a hole.
\[ \text{Is } \dfrac{5x}{x - 4} - \dfrac{20}{x - 4} \text{ the same as the constant function } 5? \]
Compare away from four
Why: They agree at every other value.
Compare at four
Why: The original is undefined.
\[ \tfrac{20}{0} - \tfrac{20}{0} \]
Note the difference
Why: The constant is defined there.
\[ 5 \]
State it properly
Why: Equal with a restriction.
\[ 5, \; x \ne 4 \]
Figure (svg): A sum whose numerators cancel to leave a constant
\[ = 5 \text{ for all } x \ne 4 \]
Verify: say what the graph looks like
Why: The graph is a horizontal line at height five with a single point removed at x equal to four. That is the same kind of hole Lesson 11.3 produced by cancelling, and it is why the restriction is part of the answer rather than a footnote.
Trap
\[ \dfrac{5x}{x-4} - \dfrac{20}{x-4} = 5 \]
Report the constant and stop
Why: A number cannot be undefined anywhere, so no restriction seemed necessary.
The constant is not undefined, but the original expression is — at x equal to four both fractions divide by nought. The two are equal everywhere else and not at that one point.
\[ = 5, \quad x \ne 4 \]
Carry the restriction from the original denominator
Why: It does not disappear when the variable does.
A constant answer is exactly the case where the restriction is easiest to forget.
Matching
Combine, then factor.
Match the pairs
Why: Three of these reduce to constants and one does not, because in the fourth the combined numerator shares no factor with the denominator. All four still carry restrictions.
Prediction
An expression that simplifies to five.
Predict first
What is its graph?
Correct: A horizontal line with one point missing.
A missing point like this is called a removable discontinuity in later courses.
Why: The expression equals five wherever it is defined, which is everywhere except the value that makes the denominator nought. So the graph is the line at height five with a single hole punched in it. That hole is invisible in the simplified form and is precisely what the recorded restriction preserves, which is the same situation as a cancelled factor in Lesson 11.3.
Socratic
It looks like something has gone wrong.
Discussion prompt
Say why a constant result feels suspicious and how to become confident in it. Then say what it tells you about the two original expressions.
Hint: Test more than one value.
Answer:
It feels suspicious because both originals plainly depend on x, so it is surprising that their difference does not. The quickest way to be convinced is to evaluate the original difference at two or three unrelated values and see the same number each time — that is far more persuasive than rechecking the algebra.
It tells you the two expressions differ by a constant amount everywhere, so their graphs are the same shape shifted vertically. Five x over x minus four and twenty over x minus four are the same curve five units apart, which is a genuine fact about the pair that the separate expressions do not display.
Section
Section 5
Concept
The common denominator restricts the variable from the start, and that restriction survives every combining and cancelling. Combining two models over a shared denominator is a natural way to compare them.
A shared denominator makes comparison a single subtraction.
Figure (svg): Two quantities with a shared denominator being compared
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.5 Adding and Subtracting with Like Denominators §11.5, pp. 658-662 — the like-denominator rule's condition that c is not zero, and the tennis exercises
Picture it
Compare by subtracting.
Figure (svg): Two quantities with a shared denominator being compared
Because the denominators already agree, the comparison needs no extra machinery. That is exactly the situation this lesson is built for.
Worked example
Reading the exclusions off the shared denominator.
\[ \text{What values are excluded from } \dfrac{4x}{3x^2 + x - 2} - \dfrac{x + 2}{3x^2 + x - 2}? \]
Factor the denominator
Why: It appears in both fractions.
\[ (3 x - 2) (x + 1) \]
Set each factor to nought
Why: Two forbidden values.
\[ 3 x - 2 = 0, \; x + 1 = 0 \]
Solve
Why: One fraction, one integer.
\[ x = \tfrac{2}{3}, \; -1 \]
Note what survives
Why: Both, even after cancelling.
\[ x \ne \tfrac{2}{3}, -1 \]
Figure (svg): Two quantities with a shared denominator being compared
\[ x \ne \tfrac{2}{3} \text{ and } x \ne -1 \]
Verify: check against the simplified answer
Why: The answer is one over x plus one, whose own denominator forbids only negative one. The restriction at two thirds came from the original and would be lost if the exclusions were read off the answer instead.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.5 Adding and Subtracting with Like Denominators §11.5, pp. 659-659
Faded example
Factor the denominator first.
Fill in the blanks
3x^2 + x - 2 = (3x - 2)(x + 1) \;\Longrightarrow\; x \ne \tfrac2}-1 \text___ x \ne ___
Why: Each factor of the denominator forbids one value, and both restrictions belong to the expression as originally written. Only one of them survives visibly in the simplified answer.
Worked example
The tennis exercises' situation, with a shared denominator.
\[ \text{Two quantities are } \dfrac{7x}{x + 3} \text{ and } \dfrac{21}{x + 3}. \text{ By how much do they differ?} \]
Subtract them
Why: Common denominator already.
\[ \tfrac{7x - 21}{x + 3} \]
Factor the numerator
Why: Seven times x minus three.
\[ 7(x - 3) \]
Look for a cancellation
Why: x minus three and x plus three differ.
State the difference
Why: With its restriction.
\[ \tfrac{7(x - 3)}{x + 3}, \; x \ne -3 \]
Figure (svg): Two quantities with a shared denominator being compared
\[ \dfrac{7(x - 3)}{x + 3}, \quad x \ne -3 \]
Verify: check where the difference is nought
Why: The difference is nought when x is three, since the numerator vanishes there — so the two quantities are equal at that one value and differ everywhere else. A subtraction is often asked precisely to find where two things coincide.
Trap
\[ \dfrac{4x}{3x^2+x-2} - \dfrac{x+2}{3x^2+x-2} = \dfrac{1}{x+1}, \quad x \ne -1 \]
Take the exclusions from the simplified expression
Why: It is the final answer, so its denominator was consulted.
The original denominator also had a factor of three x minus two, which forbids two thirds. Cancelling that factor removed it from view but not from force.
\[ x \ne \tfrac{2}{3} \text{ and } x \ne -1 \]
Read the exclusions from the original denominators, before any cancelling
Why: They are properties of the expression as given.
This is the same rule as in Lessons 11.3 and 11.4, and it will matter most in Lesson 11.7.
Elimination
For a combined rational expression.
Eliminate the wrong options
Which denominator determines the excluded values?
Survives elimination: A
Why: The expression as first written is what defines where it has values, so its denominator is the one to consult. Reading them off the answer loses exactly the restrictions whose factors cancelled.
Sorting
For an expression with denominator (3x - 2)(x + 1).
Sort into buckets
Sort each value by whether it is excluded.
Only two values out of infinitely many are forbidden, which is why it is easy to forget they exist. The last item is worth noting: negative two thirds makes neither factor nought and is permitted.
Socratic
Both could just be evaluated separately.
Discussion prompt
Say what a subtraction of two models tells you that evaluating each one does not. Then say why a shared denominator makes it easy.
Hint: Think about where the difference is nought.
Answer:
The difference is a single expression describing how far apart the two quantities are at every value at once, so it can be factored, simplified and set equal to nought. Evaluating each model separately gives one comparison per value and never reveals the general pattern — whether the gap grows, shrinks or vanishes somewhere.
A shared denominator means the subtraction is a single step with no preparation, so the interesting structure appears immediately in the combined numerator. When the denominators differ, all the work of the next lesson has to happen first before the same question can even be asked.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Adding | Subtracting | |
|---|---|---|
| The numerators | are added directly | the second is bracketed first |
| The signs | all stay as they are | every term of the second flips |
| The denominator | copied down once | copied down once |
The bottom row is identical, which is the point: the denominator is untouched by either operation. Only the treatment of the second numerator differs.
Pattern
To add or subtract rational expressions with the same denominator, these five moves cover it.
Step three is where the brackets go in, and step five is where the restrictions come back out. Both are small habits that prevent the two errors this lesson is most prone to.
OpenStax Elementary Algebra 2e, §8.3 Add and Subtract Rational Expressions with a Common Denominator §8.3
Check
The denominator is copied down.
Check your understanding
Simplify 3/x + 2/x.
Answer: A
Why: The numerators add to five and the denominator is written once, unchanged.
Check
Bracket the second numerator.
Check your understanding
Simplify 4x/(x - 2) - (2x + 4)/(x - 2).
Answer: A
Why: The numerator is four x minus two x minus four, which is two x minus four, and that factors as two times x minus two, cancelling with the denominator.
Check
Read the restrictions from the original.
Check your understanding
An expression has common denominator (3x - 2)(x + 1) and simplifies to 1/(x + 1). What is excluded?
Answer: A
Why: Both factors of the original denominator forbid a value, and the one whose factor cancelled is still forbidden.
Real world
This is the tennis question from the lesson opener. Quantities describing a ball before and after impact are often modelled over the same denominator, so comparing them is a subtraction.
Discussion prompt
Two quantities are modelled by 7x over x plus 3 and 21 over x plus 3. Find their difference in simplest form, say where it is nought, and state the restriction.
Hint: The denominators already match.
Answer:
\[ \dfrac{7x}{x+3} - \dfrac{21}{x+3} = \dfrac{7x - 21}{x + 3} = \dfrac{7(x - 3)}{x + 3} \]
The numerator factors as seven times x minus three, and nothing cancels because x minus three and x plus three are different factors.
The difference is nought exactly when x is three, so that is the one value at which the two quantities are equal; everywhere else one exceeds the other. The restriction is that x may not be negative three, which comes from the shared denominator and applies to the difference just as it did to each original. Finding where a difference vanishes is usually the point of forming it.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Simplify 4x/(x - 2) minus (2x + 4)/(x - 2).
Correct: 2.
\[ \dfrac{4x - (2x + 4)}{x - 2} = \dfrac{2(x-2)}{x-2} = 2, \quad x \ne 2 \]
Why: The second numerator must be bracketed before the subtraction, so the minus reaches both of its terms: four x minus two x minus four gives two x minus four. That factors as two times x minus two, which cancels with the denominator to leave the constant two. The first option is what happens when the brackets are omitted and the plus four survives unchanged, and it is the most common wrong answer here — it even looks tidy, which is why substituting is worth the ten seconds. At x equal to three the true difference is twelve minus ten, which is two, while that option gives ten. The answer is constant, but the restriction that x may not be two still stands.
Explain it
They wrote that three over x plus two over x is five over two x.
Discussion prompt
In no more than four sentences, explain why the denominator does not change. Then give them a numerical case that makes it obvious.
Hint: What does the denominator describe?
Answer:
A usable answer: the denominator says how big each piece is and the numerator says how many you have. Putting three pieces with two gives five pieces of the same size, so the count changes and the size does not.
Try it on numbers you know: three sevenths plus two sevenths is five sevenths, not five fourteenths. Five fourteenths is smaller than either of the two fractions you started with, which cannot be right for a sum of positive amounts.
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 brackets are fixed by writing them as part of the subtraction step rather than adding them later. The denominator is fixed by remembering it names the size of the pieces. Cancelling afterwards is fixed by refusing to look for factors until the numerators are combined. Exclusions are fixed by recording them from the original denominator before anything is cancelled. 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 rules, and beside them a numerical example with sevenths so you can check the reasoning without algebra. Underneath, work a subtraction twice: once with the second numerator bracketed and once without, and evaluate both answers at one value of the variable to show how far apart they are. In the middle, take a difference whose combined numerator factors into something matching the denominator, writing every line from the bracketing through the cancellation, and note that neither original fraction could be simplified on its own. Beneath that, work an expression that reduces to a constant, test it at three different values to convince yourself, and sketch its graph as a horizontal line with one point missing. In the lower corner, factor a common denominator and list every value it forbids, marking which of them disappears from the simplified answer. Finally, in the margin, write why cancelling comes last here but came first in Lesson 11.4.
Your bracketed and unbracketed versions should differ by a fixed multiple of one over the denominator. If they happen to agree, the second numerator had only one term and the error would not have shown — which is exactly why the habit matters on the ones with two.
Recap
Five things, and the third is the one that costs marks.
| If the question says | Your first move is |
|---|---|
| The denominators match | Combine the numerators |
| Subtract two expressions | Bracket the second numerator |
| Simplify the result | Factor the combined numerator |
| The answer is a constant | State the restriction anyway |
| Give the excluded values | Read them off the original denominator |
Lesson 11.6 removes the assumption that the denominators match. Finding a common denominator brings back the factoring of Chapter 10, and once it is found this lesson's rule finishes the job.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.5 Adding and Subtracting with Like Denominators §11.5, pp. 658-662 — everything on these slides traces back here
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