Adding and subtracting rational expressions with like denominators, building the least common multiple of two polynomials, working with unlike denominators, distributing a subtraction across a whole numerator, simplifying the combined result, and simplifying complex fractions by two methods.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 8 — Rational Functions
Add and Subtract Rational Expressions
Objectives
Five outcomes. Addition needs something multiplication never did: a common denominator.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.5 Add and Subtract Rational Expressions §8.5, pp. 582-587 — the lesson these objectives are drawn from
Warm-up
Lesson 8.4 multiplied and divided rational expressions with no common denominator anywhere in sight.
Discussion prompt
You can compute one half times one third in your head. Now try one half plus one third. What extra work did the second one need?
Hint: Think about what you had to do to the denominators.
Answer:
\[ \tfrac{1}{2} \cdot \tfrac{1}{3} = \tfrac{1}{6}: \text{ multiply straight across} \]
\[ \tfrac{1}{2} + \tfrac{1}{3} = \tfrac{3}{6} + \tfrac{2}{6} = \tfrac{5}{6}: \text{ rebuild both first} \]
Addition required a common denominator and multiplication did not. That is the whole difference between this lesson and the last one, and the polynomials change nothing about it.
Concept
To add or subtract rational expressions, the denominators must match. If they already do, combine the numerators. If they do not, factor both, build the least common denominator, and rewrite each fraction over it first.
complex fraction — A fraction that contains a fraction in its numerator or its denominator. It can be simplified either by combining and dividing, or by multiplying through by the least common denominator of all the small fractions.
\[ \frac{a}{c} + \frac{b}{c} = \frac{a+b}{c}; \qquad \frac{a}{c} + \frac{b}{d} = \frac{ad+bc}{cd} \]
Multiplying the two denominators always produces a common denominator, but not always the least one. Using the least means less simplifying at the end.
Figure (svg): Two columns comparing like and unlike denominators
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.5 Add and Subtract Rational Expressions §8.5, pp. 582-583
Section
Section 1
Concept
When two rational expressions already share a denominator, add or subtract their numerators and place the result over that same denominator. Then check whether the result simplifies.
\[ \frac{a}{c} + \frac{b}{c} = \frac{a+b}{c}, \qquad \frac{a}{c} - \frac{b}{c} = \frac{a-b}{c} \]
The check at the end is not optional. Combining often produces a numerator that shares a factor with the denominator, and occasionally one that cancels it entirely.
Figure (svg): The rule for adding and subtracting rational expressions that already share a denominator
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.5 Add and Subtract Rational Expressions §8.5, pp. 582-582 — Adding or Subtracting with Like Denominators
Picture it
Example 1 and two guided practice items.
Figure (svg): The rule for adding and subtracting rational expressions that already share a denominator
The first simplified by a factor of 2, the second not at all, and the third collapsed to the number 2. All three needed the same look at the end.
Worked example
Example 1, both parts.
\[ \text{Compute } \frac{7}{4x} + \frac{3}{4x} \text{ and } \frac{2x}{x+6} - \frac{5}{x+6}. \]
First: add the numerators
Why: Seven plus 3 is 10, over the shared 4x.
\[ \frac{10}{4 x} \]
First: simplify
Why: Both parts share a factor of 2.
\[ \frac{5}{2 x} \]
Second: subtract the numerators
Why: Two x minus 5, over the shared denominator.
\[ \frac{2 x - 5}{x + 6} \]
Second: check for simplification
Why: The numerator does not factor to include x plus 6.
Figure (svg): The rule for adding and subtracting rational expressions that already share a denominator
\[ \frac{5}{2x}; \qquad \frac{2x-5}{x+6} \]
Verify: test the first at a value
Why: At x equal to 1 the original is seven quarters plus three quarters, which is 10 over 4, or 2.5. The answer gives 5 over 2, also 2.5. And the second at x equal to 0 is 0 minus five sixths, which is negative five sixths, matching negative 5 over 6 from the answer.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.5 Add and Subtract Rational Expressions §8.5, pp. 582-582
Fill the middle
Example 1a.
Fill in the blanks
\frac10___ + \frac______ = \frac___}___
Why: Seven plus 3 is 10, over the unchanged denominator 4x. The result then simplifies to 5 over 2x, which is the step most easily forgotten.
Worked example
Guided Practice 1 to 4.
\[ \text{Compute } \frac{7}{12x}-\frac{5}{12x}, \; \frac{2}{3x^2}+\frac{1}{3x^2}, \; \frac{4x}{x-2}-\frac{x}{x-2}, \; \frac{2x^2}{x^2+1}+\frac{2}{x^2+1}. \]
First: combine and simplify
Why: Two over 12x reduces by a factor of 2.
\[ \frac{1}{6 x} \]
Second: combine and simplify
Why: Three over 3x squared reduces by 3.
\[ 1 / x ^{2} \]
Third: combine
Why: Four x minus x is 3x; nothing cancels.
\[ 3 x / (x - 2) \]
Fourth: combine and factor
Why: Two x squared plus 2 is 2 times x squared plus 1.
\[ 2 \]
Figure (svg): The solution to Worked example four more, one that collapses shown as a ladder of expressions, one row per algebraic move
\[ \frac{1}{6x}, \; \frac{1}{x^2}, \; \frac{3x}{x-2}, \; 2 \]
Verify: check the fourth's collapse
Why: At x equal to 3 the original is 18 over 10 plus 2 over 10, which is 20 over 10, or 2. At x equal to 0 it is 0 plus 2 over 1, again 2. The expression really is the constant 2 everywhere, because the combined numerator was exactly twice the denominator.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.5 Add and Subtract Rational Expressions §8.5, pp. 582-582
Trap
\[ \frac{7}{4x} + \frac{3}{4x} \]
Add the tops and the bottoms
Why: Both parts are treated as things to combine.
\[ = \frac{10}{8x} \quad \text{(wrong)} \]
At x equal to 1 the original is 1.75 plus 0.75, which is 2.5, while 10 over 8 is 1.25 — exactly half the right answer.
\[ \frac{7}{4x} + \frac{3}{4x} = \frac{10}{4x} = \frac{5}{2x} \]
Combine numerators only; the denominator names the pieces
Why: Three quarters plus seven quarters is ten quarters, not ten eighths.
\[ \tfrac{7}{4}+\tfrac{3}{4} = \tfrac{10}{4}, \text{ not } \tfrac{10}{8} \]
The denominator says what kind of piece you are counting. Adding it would change the kind of piece halfway through the calculation.
Sorting
Combine first, then look again.
Sort into buckets
Sort each combined result.
Three of five simplified further, which is why the check belongs in the routine rather than being kept for suspicious-looking answers.
Matching
Combine, then simplify.
Match the pairs
Why: The last one collapsed to a constant because the combined numerator was exactly twice the denominator. That kind of complete cancellation only becomes visible after combining, never before.
Prediction
Commit before reasoning.
Predict first
Why do the numerators combine while the denominator stays put?
Correct: Because the denominator names the size of the pieces being counted, and that size does not change.
\[ \tfrac{7}{4x} + \tfrac{3}{4x} = \tfrac{10}{4x}, \text{ not } \tfrac{10}{8x} \]
Why: Three quarters plus seven quarters is ten quarters — you are counting quarters, and adding them does not turn them into eighths. The same holds when the denominator is 4x or x plus 6: it says what kind of piece each numerator counts, so combining counts does not alter it. Adding the denominators would change the piece size in the middle of the sum, which is why it gives an answer half the correct one here.
Section
Section 2
Concept
The least common multiple of two polynomials is built by factoring each one completely — numerical coefficients into primes as well — and then writing each factor to the highest power it reaches in either.
\[ \text{LCM}(4x^2-16, \; 6x^2-24x+24) = 12(x+2)(x-2)^2 \]
Multiplying the two denominators together always gives a common denominator, so the least one is a convenience rather than a necessity. It just leaves less to cancel later.
Figure (svg): Two polynomials factored into primes and binomials, and the least common multiple built from them
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.5 Add and Subtract Rational Expressions §8.5, pp. 583-583 — Find a least common multiple
Picture it
Example 2, both polynomials factored side by side.
Figure (svg): Two polynomials factored into primes and binomials, and the least common multiple built from them
The factor 2 appears squared on the left and singly on the right, so the LCM takes it squared. The x minus 2 appears squared on the right, so the LCM takes that squared too.
Worked example
Example 2.
\[ \text{Find the LCM of } 4x^2-16 \text{ and } 6x^2-24x+24. \]
Factor the first, primes included
Why: Four is 2 squared, and x squared minus 4 is a difference of squares.
\[ 2 ^{2}(x + 2) (x - 2) \]
Factor the second
Why: Six is 2 times 3, and the quadratic is a perfect square.
\[ 2 \cdot 3 \cdot(x - 2) ^{2} \]
Take each factor at its highest power
Why: Two squared, one 3, one x plus 2, and x minus 2 squared.
\[ 2 ^{2} \cdot 3 \cdot(x + 2) (x - 2) ^{2} \]
Write it out
Why: Four times 3 is 12.
\[ 12(x + 2) (x - 2) ^{2} \]
Figure (svg): Two polynomials factored into primes and binomials, and the least common multiple built from them
\[ 12(x+2)(x-2)^2 \]
Verify: check that both divide it
Why: Dividing the LCM by 4x squared minus 16 gives 3 times x minus 2, a polynomial; dividing by 6x squared minus 24x plus 24 gives 2 times x plus 2, also a polynomial. Both go in evenly, and nothing smaller would, since dropping any factor would break one of the two divisions.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.5 Add and Subtract Rational Expressions §8.5, pp. 583-583
Fill the middle
Example 2.
Fill in the blanks
(x-2) \text2 (x-2)^___}
Why: The factor x minus 2 appears to the first power in one polynomial and to the second in the other, so the LCM takes the second. Taking the lower power would give something the second polynomial does not divide.
Worked example
Lesson exercises 9 to 14.
\[ \text{Find the LCM of } 3x \text{ and } 3(x-2); \; 2x^2 \text{ and } 4x+12; \; 9x^2-16 \text{ and } 3x^2-2x-8. \]
First pair
Why: The 3 is shared; x and x minus 2 are not.
\[ 3 x(x - 2) \]
Second pair
Why: Two x squared is 2 x squared; 4x plus 12 is 2 squared times x plus 3.
\[ 4 x ^{2}(x + 3) \]
Third pair: factor both
Why: Nine x squared minus 16 is a difference of squares; the other factors too.
\[ (3 x - 4) (3 x + 4)\text{ and } (3 x + 4) (x - 2) \]
Third pair: combine
Why: Three x plus 4 is shared, so it appears once.
\[ (3 x - 4) (3 x + 4) (x - 2) \]
Figure (svg): The solution to Worked example six more least common multiples shown as a ladder of expressions, one row per algebraic move
\[ 3x(x-2); \; 4x^2(x+3); \; (3x-4)(3x+4)(x-2) \]
Verify: check the second's coefficient
Why: Two x squared contributes one factor of 2, and 4x plus 12 contributes two, so the LCM takes two: 2 squared is 4. Writing 2 rather than 4 would give a multiple of the first polynomial but not of the second, which is exactly the error the prime factorisation prevents.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.5 Add and Subtract Rational Expressions §8.5, pp. 586-586
Error analysis
A student finds the least common multiple of two polynomials.
Annotate
On: \( \text{LCM}(4x^2-16, \; 6x^2-24x+24) = 24(x+2)(x-2)^3 \)
Using the product still works; it just leaves extra factors to cancel at the end. Taking highest powers is what keeps the arithmetic small.
Matching
Factor both, including the numbers.
Match the pairs
Why: The third is the case where one polynomial already divides the other, so the larger one IS the least common multiple. Noticing that saves building it from scratch.
Ranking
Building a least common multiple.
Put in order
Why: Step two is the one most often skipped, and it is where coefficient errors come from: 4 and 6 have least common multiple 12, not 24, and only their prime factorisations make that obvious.
Prediction
Commit before reasoning.
Predict first
You use the product of the two denominators instead of the least common denominator. What happens?
Correct: You get the right answer, but with more simplifying to do at the end.
\[ \text{product} = 24(x+2)(x-2)^3; \quad \text{LCM} = 12(x+2)(x-2)^2 \]
Why: The product is a common denominator, just not the smallest one, so every step is valid and the final answer is correct after cancelling. The cost is larger numbers and a bigger cancellation at the end — the same trade-off as adding numerical fractions over 24 rather than over 12. The book says as much directly: using the least common denominator means you may have less simplifying to do.
Section
Section 3
Concept
Factor both denominators, form the least common denominator, multiply each fraction by whatever it is missing, and only then add the numerators. Each fraction is multiplied by a quantity equal to 1, so its value is unchanged.
\[ \frac{a}{c}+\frac{b}{d} = \frac{ad}{cd}+\frac{bc}{cd} = \frac{ad+bc}{cd} \]
The multiplier for each fraction is the part of the least common denominator it is missing, written over itself so that nothing changes in value.
Figure (svg): Two rational expressions with unlike denominators rewritten over a common denominator and added
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.5 Add and Subtract Rational Expressions §8.5, pp. 583-583 — Add with unlike denominators
Picture it
Example 3, with the least common denominator built explicitly.
Figure (svg): Two rational expressions with unlike denominators rewritten over a common denominator and added
The first fraction was missing x plus 1 and the second was missing 3x. Each was multiplied by exactly that, over itself.
Worked example
Example 3.
\[ \text{Add } \frac{7}{9x^2} + \frac{x}{3x^2+3x}. \]
Factor both denominators
Why: Nine x squared is 3 squared x squared; the other is 3x times x plus 1.
\[ 3 ^{2} x ^{2}\text{ and } 3 x(x + 1) \]
Build the least common denominator
Why: Highest power of 3, of x, and the factor x plus 1.
\[ 9 x ^{2}(x + 1) \]
Multiply each fraction up
Why: The first is missing x plus 1; the second is missing 3x.
\[ (7 x + 7)\text{ and } 3 x ^{2} \]
Add the numerators
Why: Three x squared plus 7x plus 7.
\[ \frac{3 x ^{2} + 7 x + 7}{9 x ^{2}(x + 1)} \]
Figure (svg): Two rational expressions with unlike denominators rewritten over a common denominator and added
\[ \frac{3x^2+7x+7}{9x^2(x+1)} \]
Verify: test at a value
Why: At x equal to 1 the original is 7 over 9 plus 1 over 6, which is 14 over 18 plus 3 over 18, or 17 over 18. The answer gives 3 plus 7 plus 7, over 9 times 2, which is 17 over 18. They agree — and the numerator does not factor, so nothing further cancels.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.5 Add and Subtract Rational Expressions §8.5, pp. 583-583
Fill the middle
Example 3.
Fill in the blanks
\fracx+1___ \cdot \frac___}___} \text___ 9x^2(x+1)
Why: The first denominator is missing the factor x plus 1, so the fraction is multiplied by x plus 1 over x plus 1. That is multiplying by 1, so the value is unchanged while the form becomes usable.
Worked example
Lesson exercises 16, 20 and 21.
\[ \text{Compute } \frac{12}{5x}+\frac{7}{6x}, \; \frac{3}{x+4}-\frac{1}{x+6}, \; \frac{9}{x-3}+\frac{2x}{x+1}. \]
First: the LCD is 30x
Why: Multiply the first by 6 over 6 and the second by 5 over 5.
\[ \frac{72 + 35}{30 x} = \frac{107}{30 x} \]
Second: the LCD is the product
Why: Neither denominator factors further.
\[ \frac{3(x + 6) - (x + 4)}{(x + 4) (x + 6)} \]
Second: simplify the numerator
Why: Three x plus 18 minus x minus 4.
\[ 2(x + 7) / ((x + 4) (x + 6)) \]
Third: build and combine
Why: Nine times x plus 1, plus 2x times x minus 3.
\[ \frac{2 x ^{2} - 6 x + 9 x + 9}{(x - 3) (x + 1)} \]
Figure (svg): The solution to Worked example three more additions shown as a ladder of expressions, one row per algebraic move
\[ \frac{107}{30x}, \; \frac{2(x+7)}{(x+4)(x+6)}, \; \frac{2x^2+3x+9}{(x-3)(x+1)} \]
Verify: check the second's subtraction
Why: The whole second numerator, x plus 4, had to be subtracted — both terms, not just the x. Getting 3x plus 18 minus x plus 4 instead would give 2x plus 22, an answer wrong at every value. Distributing the minus sign is the single most common error in this idea.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.5 Add and Subtract Rational Expressions §8.5, pp. 586-586
Trap
\[ \frac{7}{9x^2} + \frac{x}{3x(x+1)}, \; \text{LCD} = 9x^2(x+1) \]
Just write both over the new denominator
Why: The denominators are replaced and the numerators left alone.
\[ = \frac{7 + x}{9x^2(x+1)} \quad \text{(wrong)} \]
Changing a denominator without changing its numerator changes the fraction's value. Seven ninths is not 7 over 18.
\[ \frac{7}{9x^2} \cdot \frac{x+1}{x+1} + \frac{x}{3x(x+1)} \cdot \frac{3x}{3x} \]
Multiply each fraction by a form of 1
Why: Whatever multiplies the denominator must multiply the numerator too.
\[ = \frac{7x+7 + 3x^2}{9x^2(x+1)} \]
Multiplying by x plus 1 over x plus 1 is multiplying by 1, so nothing changes in value while everything changes in form. That is the entire trick.
Ranking
Adding with unlike denominators.
Put in order
Why: Step one cannot be skipped: without factoring, the least common denominator is invisible and the product of the two denominators is the only option. Step five is the one people stop before, and it is often where the answer becomes recognisable.
Comparison
Fill the blanks. Only one needs a common denominator.
Comparison matrix
| Question | Multiplying | Adding |
|---|---|---|
| Common denominator needed? | no | yes |
| First move | factor everything | factor both denominators |
| What factoring is for | finding what cancels | finding the least common denominator |
| Cancelling happens | before multiplying out | after combining the numerators |
Both operations begin by factoring, but for opposite reasons: one to remove factors, the other to discover which ones are missing.
Prediction
Commit before reasoning.
Predict first
Multiplying a fraction by x plus 1 over x plus 1 changes how it looks. Why not what it is worth?
Correct: Because x plus 1 over x plus 1 equals 1, and multiplying by 1 changes nothing.
\[ \frac{x+1}{x+1} = 1 \text{ for } x \neq -1 \]
Why: Any nonzero quantity divided by itself is 1, so the fraction is being multiplied by 1 in disguise. This is the same move as writing one half as 3 over 6: the appearance changes so that the addition becomes possible, and the value does not. It fails only where x plus 1 is zero, and there the original expression was undefined anyway.
Section
Section 4
Concept
A subtraction applies to the entire second numerator, not just its first term. After combining, factor the result and check once more for a cancellation with the denominator.
\[ \frac{x+2}{2x-2} - \frac{-2x-1}{x^2-4x+3} = \frac{x+4}{2(x-3)} \]
The book prints a caution beside this example for the second hazard: once the numerator is simplified, look again for a factor shared with the denominator. Here there is one.
Figure (svg): A subtraction of rational expressions, with the sign of the whole numerator distributed
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.5 Add and Subtract Rational Expressions §8.5, pp. 584-584 — Subtract with unlike denominators
Picture it
Example 4, from factoring to the final cancellation.
Figure (svg): A subtraction of rational expressions, with the sign of the whole numerator distributed
Subtracting negative 4x minus 2 added 4x and 2. Then the numerator factored as x plus 4 times x minus 1, and the x minus 1 cancelled.
Worked example
Example 4.
\[ \text{Subtract } \frac{x+2}{2x-2} - \frac{-2x-1}{x^2-4x+3}. \]
Factor both denominators
Why: Two times x minus 1, and x minus 1 times x minus 3.
\[ 2(x - 1)\text{ and } (x - 1) (x - 3) \]
Build up to the LCD
Why: Two times x minus 1 times x minus 3.
\[ \text{numerators } x ^{2} - x - 6\text{ and } -4 x - 2 \]
Subtract the whole second numerator
Why: Minus negative 4x minus 2 adds 4x and adds 2.
\[ x ^{2} + 3 x - 4 \]
Factor and cancel
Why: X plus 4 times x minus 1, over 2 times x minus 1 times x minus 3.
\[ \frac{x + 4}{2(x - 3)} \]
Figure (svg): A subtraction of rational expressions, with the sign of the whole numerator distributed
\[ \frac{x+4}{2(x-3)} \]
Verify: test at a value
Why: At x equal to 0 the original is 2 over negative 2, minus negative 1 over 3, which is negative 1 plus one third, or negative two thirds. The answer gives 4 over negative 6, also negative two thirds. Both the sign distribution and the final cancellation check out.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.5 Add and Subtract Rational Expressions §8.5, pp. 584-584
Fill the middle
Example 4.
Fill in the blanks
x^2-x-6-(-4x-2) = x^2-x-6+4x+2
Why: Subtracting negative 2 adds 2. Both terms of the second numerator change sign, which turns the expression into x squared plus 3x minus 4.
Worked example
Lesson exercises 18 and 22.
\[ \text{Compute } \frac{x-4}{5x}-\frac{12}{5(x-4)} \text{ and } \frac{x+4}{x^2-4}-\frac{15}{x-2}. \]
First: build the LCD
Why: Five x times x minus 4.
\[ (x - 4) ^{2} - 12 x,\text{ over } 5 x(x - 4) \]
First: expand and combine
Why: X squared minus 8x plus 16, minus 12x.
\[ \frac{x ^{2} - 20 x + 16}{5 x(x - 4)} \]
Second: factor the first denominator
Why: X plus 2 times x minus 2, so the LCD is that product.
\[ LCD = (x + 2) (x - 2) \]
Second: build up and subtract
Why: X plus 4 minus 15 times x plus 2.
\[ (x + 4 - 15 x - 30) \]
Figure (svg): The solution to Worked example two more subtractions shown as a ladder of expressions, one row per algebraic move
\[ \frac{x^2-20x+16}{5x(x-4)}; \quad \frac{-14x-26}{(x+2)(x-2)} \]
Verify: check the second's distribution
Why: Fifteen times x plus 2 is 15x plus 30, and subtracting BOTH terms gives minus 15x minus 30. Subtracting only the 15x would leave plus 30 and an answer wrong at every value. At x equal to 0 the original is 4 over negative 4, minus 15 over negative 2, which is negative 1 plus 7.5, or 6.5; the answer gives negative 26 over negative 4, also 6.5.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.5 Add and Subtract Rational Expressions §8.5, pp. 586-586
Error analysis
A student subtracts two rational expressions over a common denominator.
Annotate
On: \( \frac{x+4}{(x+2)(x-2)} - \frac{15(x+2)}{(x+2)(x-2)} = \frac{x+4-15x+30}{(x+2)(x-2)} \)
Writing the second numerator in brackets before subtracting makes the distribution unavoidable, and costs one extra pair of brackets.
Two truths and a lie
All three are about subtracting rational expressions.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. Example 4's combined numerator was x squared plus 3x minus 4, which factors as x plus 4 times x minus 1 — and the x minus 1 cancels against the denominator. Stopping at the combined form leaves an unsimplified answer, which is usually marked wrong and always harder to use.
Sorting
After combining the numerators.
Sort into buckets
Sort each combined result by whether it is finished.
The first is the important case: its numerator factors as x plus 4 times x minus 1, and the x minus 1 cancels against the denominator to leave a much shorter answer.
Prediction
Commit before reasoning.
Predict first
What does writing brackets around the subtracted numerator prevent?
Correct: Forgetting to change the sign of every term after the first.
\[ a - (b+c) = a-b-c, \text{ not } a-b+c \]
Why: Without brackets the minus sign visually attaches only to the first term, and the rest sail through unchanged — which is exactly the printed error in lesson exercise 25. With brackets the distribution is unavoidable: minus times each term inside. It costs two characters and removes the most common source of wrong answers in the whole lesson.
Section
Section 5
Concept
A complex fraction contains a fraction in its numerator or denominator. Method one writes each part as a single fraction and then divides. Method two multiplies the whole thing above and below by the least common denominator of every small fraction inside.
\[ f = \frac{1}{\frac{1}{p}+\frac{1}{q}} = \frac{pq}{p+q} \]
Method two is usually shorter when several small fractions appear, because it clears all of them in a single move rather than combining them first.
Figure (svg): The same complex fraction simplified by two different methods
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.5 Add and Subtract Rational Expressions §8.5, pp. 584-585 — Simplify a complex fraction
Picture it
Examples 5 and 6, side by side.
Figure (svg): The same complex fraction simplified by two different methods
The left combines the denominator into one fraction and then flips it. The right multiplies through by x times x plus 4 and every small fraction vanishes at once.
Worked example
Example 5.
\[ \text{Simplify } f = \frac{1}{\frac{1}{p}+\frac{1}{q}}. \]
Combine the denominator
Why: The least common denominator of the two small fractions is pq.
\[ \frac{q}{p q} + \frac{p}{p q} \]
Write it as one fraction
Why: Add the numerators.
\[ \frac{q + p}{p q} \]
Divide by that fraction
Why: Dividing by a fraction is multiplying by its reciprocal.
\[ 1 \cdot p q / (q + p) \]
Write the result
Why: Nothing further cancels.
\[ p q / (p + q) \]
Figure (svg): A thin camera lens with the object distance, image distance and focal length labelled
\[ f = \frac{pq}{p+q} \]
Verify: test with real numbers
Why: Take p equal to 30 centimetres and q equal to 6. Then 1 over 30 plus 1 over 6 is one over 30 plus 5 over 30, which is 6 over 30, or one fifth — so f is 5. And the formula gives 180 over 36, which is 5. The simplified form is what a lens designer computes with, since it needs no reciprocals.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.5 Add and Subtract Rational Expressions §8.5, pp. 585-585
Fill the middle
Example 6.
Fill in the blanks
\textx(x+4) x+4 \text___ x \;\Longrightarrow\; \text___ = ___
Why: The two small denominators share no factor, so their least common multiple is their product. Multiplying above and below by it clears every small fraction in one move.
Worked example
Example 6 and Guided Practice 12 and 13.
\[ \text{Simplify } \frac{\frac{5}{x+4}}{\frac{1}{x+4}+\frac{2}{x}}, \; \frac{\frac{2}{x}-4}{\frac{2}{x}+3}, \; \frac{\frac{3}{x+5}}{\frac{2}{x-3}+\frac{1}{x+5}}. \]
First: find the LCD of everything inside
Why: X times x plus 4 covers both small denominators.
\[ LCD = x(x + 4) \]
First: multiply above and below by it
Why: The numerator becomes 5x; the denominator x plus 2 times x plus 4.
\[ 5 x / (3 x + 8) \]
Second: the LCD is just x
Why: Two minus 4x over 2 plus 3x.
\[ \frac{2 - 4 x}{2 + 3 x} \]
Third: the LCD is x plus 5 times x minus 3
Why: Three times x minus 3, over 2 times x plus 5 plus x minus 3.
\[ 3(x - 3) / (3 x + 7) \]
Figure (svg): The solution to Worked example method two, three times shown as a ladder of expressions, one row per algebraic move
\[ \frac{5x}{3x+8}, \; \frac{2-4x}{2+3x}, \; \frac{3(x-3)}{3x+7} \]
Verify: test the first at a value
Why: At x equal to 1 the original is five fifths on top, which is 1, over one fifth plus 2, which is 2.2 — giving about 0.4545. The answer gives 5 over 11, which is 0.4545. Multiplying above and below by the same quantity is multiplying by 1, so the value could not have changed.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.5 Add and Subtract Rational Expressions §8.5, pp. 585-585
Trap
\[ \frac{\frac{5}{x+4}}{\frac{1}{x+4}+\frac{2}{x}} \]
Multiply the numerator by x times x plus 4
Why: The clearing is applied where the fractions look worst.
\[ = \frac{5x}{\frac{1}{x+4}+\frac{2}{x}} \quad \text{(wrong)} \]
Multiplying only one part multiplies the whole expression by x times x plus 4, changing its value entirely.
\[ \cdot \frac{x(x+4)}{x(x+4)} \]
Multiply the numerator AND the denominator
Why: The multiplier must be written over itself, so that it equals 1.
\[ = \frac{5x}{x+2(x+4)} = \frac{5x}{3x+8} \]
This is the same principle as building fractions up to a common denominator: any change of form has to be a multiplication by 1.
Comparison
Fill the blanks. Same answer, different route.
Comparison matrix
| Question | Method 1 | Method 2 |
|---|---|---|
| First move | combine top and bottom separately | find the LCD of every small fraction |
| Second move | divide, by multiplying by the reciprocal | multiply top and bottom by that LCD |
| Best when | each part is already one fraction | several small fractions appear |
| Risk | forgetting to flip | multiplying only one part |
Neither is more correct. Method one is natural when the bar is clearly a division; method two is faster when there are four small fractions to clear.
Matching
Clear the small fractions.
Match the pairs
Why: The last one has four small fractions and a numerical least common denominator of 30, which is where method two saves the most work: one multiplication clears all four at once.
Prediction
Commit before reasoning.
Predict first
Method two multiplies a complex fraction above and below by the same quantity. Why is that allowed?
Correct: Because the quantity over itself equals 1, and multiplying by 1 preserves value.
\[ \frac{A}{B} = \frac{A}{B} \cdot \frac{L}{L} = \frac{AL}{BL}, \; L \neq 0 \]
Why: The move is exactly the same as building fractions up to a common denominator, and it rests on the same fact: any nonzero quantity divided by itself is 1. The one condition is that both parts get multiplied — multiplying only the numerator scales the whole expression instead. And the multiplier must not be zero, which is why the excluded values of the original expression remain excluded.
Comparison
Fill the blanks. The denominators decide the work.
Comparison matrix
| Situation | What to do | Then |
|---|---|---|
| Like denominators | combine the numerators | simplify the result |
| Unlike denominators | factor, build the LCD, multiply each up | combine and simplify |
| A subtraction | bracket the second numerator | distribute the minus to every term |
| A complex fraction | combine and divide, or clear with the LCD | simplify the result |
Every row ends the same way, and that last step is the one most often skipped — Example 4's answer halves in length because of it.
Pattern
One routine, with a shortcut when the denominators already match.
For a complex fraction, either combine each part into one fraction and divide, or multiply above and below by the least common denominator of every small fraction inside.
OpenStax Algebra and Trigonometry 2e, §1.6 Rational Expressions §1.6
Check
Like denominators. Combine, then look again.
Check your understanding
Simplify 2x^2/(x^2 + 1) + 2/(x^2 + 1).
Answer: A
Why: The combined numerator is 2 times x squared plus 1, which cancels the denominator entirely.
Check
Least common multiples. Factor the numbers too.
Check your understanding
What is the least common multiple of 4x^2 - 16 and 6x^2 - 24x + 24?
Answer: A
Why: Take 2 squared, one 3, one x + 2 and x - 2 squared.
Check
Subtracting. Distribute, then simplify.
Check your understanding
Simplify (x + 2)/(2x - 2) - (-2x - 1)/(x^2 - 4x + 3).
Answer: A
Why: The combined numerator factors as (x + 4)(x - 1), and the x - 1 cancels.
Real world
Two resistors of resistance R1 and R2 wired in parallel behave as a single resistor R, where 1 over R equals 1 over R1 plus 1 over R2.
Discussion prompt
Solve for R as a single fraction, then find the combined resistance of a 6 ohm and a 3 ohm resistor and explain why it is smaller than either.
Hint: This is the lens equation with different letters.
Answer:
\[ R = \frac{1}{\frac{1}{R_1}+\frac{1}{R_2}} = \frac{R_1R_2}{R_1+R_2} \]
\[ R = \frac{6 \cdot 3}{6+3} = \frac{18}{9} = 2 \text{ ohms} \]
The pair behaves as a 2 ohm resistor — smaller than either one alone, which sounds wrong until you see what the algebra is describing: adding a second path for the current makes it easier to flow, not harder.
The formula is identical to Example 5's lens equation, letter for letter, and simplifying it once serves both. That happens constantly: the reciprocal-sum structure turns up wherever quantities combine by sharing rather than stacking — parallel resistors, lenses, springs in series, and workers finishing a job together. Recognising the shape means you already know the simplified form.
Commit first
Answer, then rate your confidence honestly.
Predict first
Does 7 over 4x plus 3 over 4x equal 10 over 8x?
Correct: No — the denominator names the piece size and stays put, so the answer is 10 over 4x.
\[ \tfrac{1}{2}+\tfrac{1}{2} = 1, \text{ not } \tfrac{2}{4} \]
Why: Seven quarters plus three quarters is ten quarters, not ten eighths: you are counting quarters, and adding counts does not change what a quarter is. Testing at x equal to 1 gives 2.5 for the correct answer and 1.25 for the wrong one, exactly half. The same instinct that adds denominators here would say one half plus one half is 2 over 4, which is one half — visibly wrong. The correct answer simplifies further, to 5 over 2x.
Explain it
They can multiply rational expressions but freeze when they see a plus sign between two of them.
Discussion prompt
In four sentences or fewer, explain why addition needs a common denominator when multiplication does not.
Hint: Think about what a denominator tells you.
Answer:
A denominator says what size the pieces are, and you can only add counts of pieces that are the same size.
Multiplying does not care: a third of a half is a sixth, and the sizes combine on their own. But adding a third and a half means first rewriting both as sixths, and then the counts can finally be added. With polynomials it is the same idea, except that finding the common size means factoring both denominators first.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the LCD, list every distinct factor in a column and write its highest power beside it. For building up, ask what each denominator is missing and multiply by that over itself. For the minus sign, bracket the second numerator before you subtract anything. For complex fractions, pick method two and write the LCD above the bar and below it before doing anything else.
Connect it up
Paper. Fifteen minutes.
Draw it
Build an addition page. Top left: write the like-denominator rule with one numerical example and one algebraic one, and beside it the wrong version that adds denominators, with a test value showing it fail. Top right: factor 4x squared minus 16 and 6x squared minus 24x plus 24 in a column each, then build the least common multiple, circling the factor whose power you had to raise. Middle: work Example 3 in full, labelling each fraction's missing multiplier. Bottom left: work Example 4 in full, bracketing the subtracted numerator and boxing the final cancellation. Bottom right: simplify the lens equation by method one and Example 6 by method two, and write one sentence on when you would choose each.
If your Example 4 answer still has a quadratic numerator, factor it: x squared plus 3x minus 4 is x plus 4 times x minus 1, and that second factor cancels.
Recap
Five things, all governed by the denominators.
| If you see | Then |
|---|---|
| Matching denominators | Combine numerators; keep the denominator |
| Different denominators | Factor both, build the LCD, multiply each up |
| A minus between two fractions | Bracket the second numerator before subtracting |
| A combined numerator | Factor it and check for a cancellation |
| A fraction inside a fraction | Clear it with the LCD of every small fraction |
| A reciprocal sum like 1/p + 1/q | Its reciprocal is pq over p plus q |
Lesson 8.6 solves rational equations, where the whole point is to clear the denominators entirely — and where the solutions have to be checked, because clearing can introduce ones that do not work.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.5 Add and Subtract Rational Expressions §8.5, pp. 582-587 — everything on these slides traces back here
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