Solving a rational equation by cross multiplying when each side is a single fraction, building and using a mixture model, clearing all denominators by multiplying through by the least common denominator, handling equations that become quadratics, and checking every candidate for extraneousness.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 8 — Rational Functions
Solve Rational Equations
Objectives
Five outcomes. Clear the denominators, solve, and then check what you cleared.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.6 Solve Rational Equations §8.6, pp. 589-593 — the lesson these objectives are drawn from
Warm-up
Lesson 8.5 added and subtracted rational expressions, keeping the denominators throughout.
Discussion prompt
Here is an equation rather than an expression: 3 over x equals 6 over 10. What could you do to an equation that you could never do to an expression?
Hint: An equation has two sides you may operate on.
Answer:
You can multiply BOTH SIDES by something and stay balanced. With an expression there is only one side, so the denominators have to stay.
\[ \frac{3}{x} = \frac{6}{10} \;\Longrightarrow\; 30 = 6x \;\Longrightarrow\; x = 5 \]
Every method in this lesson does that: get rid of the denominators, solve what is left, and then check — because multiplying by something that might be zero can create solutions that were never really there.
Concept
Multiply both sides of a rational equation by the least common denominator, so that every denominator disappears and a polynomial equation remains. Then substitute every candidate back, because the multiplier may have been zero.
cross multiplying — Setting the product of one fraction's numerator with the other's denominator equal to the reverse product. It applies when each side of the equation is a single rational expression.
\[ \frac{a}{b} = \frac{c}{d} \;\Longrightarrow\; ad = bc \]
Cross multiplying and multiplying by the least common denominator are the same idea. The first is what the second looks like when there are exactly two fractions.
Figure (svg): Two columns comparing cross multiplying with clearing by the least common denominator
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.6 Solve Rational Equations §8.6, pp. 589-591
Section
Section 1
Concept
When a rational equation is a proportion — a single fraction equal to a single fraction — set the two cross products equal. That clears both denominators in one step.
\[ \frac{3}{x+1} = \frac{9}{4x+5} \;\Longrightarrow\; 3(4x+5) = 9(x+1) \]
The condition matters. If either side has two terms, cross multiplying is not available and the least common denominator method of the third idea is the route.
Figure (svg): A proportion solved by cross multiplying, with the two products shown
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.6 Solve Rational Equations §8.6, pp. 589-589 — Solve a rational equation by cross multiplying
Picture it
Example 1, from proportion to answer.
Figure (svg): A proportion solved by cross multiplying, with the two products shown
After cross multiplying nothing rational remains: what is left is an ordinary linear equation from Lesson 1.3.
Worked example
Example 1.
\[ \text{Solve } \frac{3}{x+1} = \frac{9}{4x+5}. \]
Cross multiply
Why: Each numerator times the other denominator.
\[ 3(4 x + 5) = 9(x + 1) \]
Distribute
Why: Twelve x plus 15 on the left, 9x plus 9 on the right.
\[ 12 x + 15 = 9 x + 9 \]
Collect
Why: Subtract 9x and 15 from both sides.
\[ 3 x = -6 \]
Solve and check
Why: Divide by 3, then substitute back.
\[ x = -2 \]
Figure (svg): A proportion solved by cross multiplying, with the two products shown
\[ x = -2 \]
Verify: substitute into the original
Why: At x equal to negative 2 the left side is 3 over negative 1, which is negative 3, and the right side is 9 over negative 3, also negative 3. Neither denominator is zero there, so the solution is genuine rather than extraneous.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.6 Solve Rational Equations §8.6, pp. 589-589
Sorting
Count the terms on each side.
Sort into buckets
Sort each equation by the method available.
The LCD method works on all five. Cross multiplying is simply faster on the three that qualify, and wrong on the two that do not.
Worked example
Guided Practice 1 to 3.
\[ \text{Solve } \frac{3}{5x}=\frac{2}{x-7}, \; \frac{-4}{x+3}=\frac{5}{x-3}, \; \frac{1}{2x+5}=\frac{x}{11x+8}. \]
First: cross multiply and solve
Why: Three x minus 21 equals 10x.
\[ x = -3 \]
Second: cross multiply
Why: Negative 4x plus 12 equals 5x plus 15.
\[ x = -\frac{1}{3} \]
Third: cross multiply
Why: Eleven x plus 8 equals 2x squared plus 5x.
\[ 2 x ^{2} - 6 x - 8 = 0 \]
Third: solve the quadratic
Why: Dividing by 2 gives x squared minus 3x minus 4.
\[ x = 4\text{ or } x = -1 \]
Figure (svg): The solution to Worked example three more proportions shown as a ladder of expressions, one row per algebraic move
\[ x = -3, \; x = -\tfrac{1}{3}, \; x = 4 \text{ or } -1 \]
Verify: check the third's two solutions
Why: At x equal to 4 both sides give 1 over 13. At x equal to negative 1 the left is 1 over 3 and the right is negative 1 over negative 3, also one third. Neither makes a denominator vanish, so both are genuine — a proportion can perfectly well have two solutions when cross multiplying produces a quadratic.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.6 Solve Rational Equations §8.6, pp. 590-590
Trap
\[ \frac{3}{x} + \frac{8}{5} = \frac{-13}{x} \]
Cross multiply the outer fractions
Why: The equation looks close enough to a proportion.
\[ 15 = -13x \quad \text{(wrong)} \]
The left side is a SUM of two fractions, not one fraction. Cross multiplying ignores the middle term entirely.
\[ 5x\left(\tfrac{3}{x}+\tfrac{8}{5}\right) = 5x \cdot \tfrac{-13}{x} \]
Multiply every term by the least common denominator
Why: The general method handles any number of terms.
\[ 15 + 8x = -65 \;\Longrightarrow\; x = -10 \]
Cross multiplying is a shortcut for exactly two fractions. Counting the terms before choosing a method takes a second and prevents this.
Fill the middle
Example 1.
Fill in the blanks
\frac1___ = \frac______ \;\Longrightarrow\; 3(4x+5) = 9(x+___)
Why: Each numerator multiplies the opposite denominator, so the 9 pairs with x plus 1. Writing both products before expanding keeps the pairing straight.
Matching
Cross multiply, then solve.
Match the pairs
Why: The last one produced a quadratic because a variable appeared in a numerator as well as in the denominators. Cross multiplying does not promise a linear equation, only that the denominators will be gone.
Prediction
Commit before reasoning.
Predict first
Why may you replace a over b equals c over d with ad equals bc?
Correct: Because multiplying both sides by bd clears both denominators at once.
\[ bd \cdot \tfrac{a}{b} = ad; \quad bd \cdot \tfrac{c}{d} = bc \]
Why: Multiplying the left by bd cancels the b and leaves ad; multiplying the right cancels the d and leaves bc. So cross multiplying is not a new rule at all — it is the least common denominator method with the least common denominator taken to be the product bd. That also explains its restriction: with a sum on one side, multiplying by bd leaves a term behind that the shortcut never accounts for.
Section
Section 2
Concept
A percentage is a ratio, so a mixture problem written as a percentage equation is a proportion — and cross multiplying solves it. The trick is choosing which ingredient to track.
\[ \frac{7.5}{100} = \frac{0.2(15)}{15+x} \]
Track the ingredient that does not change. Here pure silver is being added, so the copper stays at 3 ounces while the total weight grows.
Figure (svg): Jewelry silver combined with pure silver to reach the copper percentage of sterling silver
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.6 Solve Rational Equations §8.6, pp. 589-589 — Write and use a rational model
Picture it
Example 2: 15 ounces of jewelry silver diluted to sterling.
Figure (svg): Jewelry silver combined with pure silver to reach the copper percentage of sterling silver
Twenty-five ounces of pure silver brings the total to 40, with the same 3 ounces of copper — and 3 over 40 is exactly 7.5 percent.
Worked example
Example 2.
\[ \text{Mix pure silver with } 15 \text{ oz of jewelry silver (20 percent copper) to reach } 7.5 \text{ percent copper.} \]
Decide what stays fixed
Why: Pure silver is added, so the copper never changes.
\[ 0.2(15) = 3\text{ ounces} \]
Write the percentage as a ratio
Why: Copper over total weight equals 7.5 over 100.
\[ \frac{7.5}{100} = \frac{3}{15 + x} \]
Cross multiply
Why: Seven point five times 15 plus x equals 300.
\[ 112.5 + 7.5 x = 300 \]
Solve
Why: Subtract and divide.
\[ x = 25 \]
Figure (svg): Jewelry silver combined with pure silver to reach the copper percentage of sterling silver
\[ x = 25 \text{ ounces} \]
Verify: check the finished alloy
Why: The total weight is 40 ounces, of which 3 are copper — and 3 over 40 is 0.075, or 7.5 percent. The silver is 12 from the original plus 25 added, which is 37, and 37 over 40 is 92.5 percent. Both percentages of sterling silver come out right.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.6 Solve Rational Equations §8.6, pp. 589-589
Fill the middle
Example 2.
Fill in the blanks
\text3 15 \text___ = 0.2(15) = ___ \text___
Why: Twenty percent of 15 is 3 ounces of copper, and adding pure silver never changes it. That fixed quantity becomes the numerator of the proportion.
Worked example
Guided Practice 4.
\[ \text{Repeat with } 10 \text{ ounces of jewelry silver instead of } 15. \]
Find the fixed copper
Why: Twenty percent of 10 ounces.
\[ 2\text{ ounces} \]
Write the proportion
Why: Two over 10 plus x equals 7.5 over 100.
\[ 7.5(10 + x) = 200 \]
Solve
Why: Seventy-five plus 7.5x equals 200.
\[ 7.5 x = 125 \]
Divide and interpret
Why: One hundred twenty-five over 7.5.
\[ \text{about } 16.7\text{ ounces} \]
Figure (svg): The solution to Worked example a smaller batch shown as a ladder of expressions, one row per algebraic move
\[ x = \tfrac{50}{3} \approx 16.7 \text{ ounces} \]
Verify: compare with the larger batch
Why: Fifteen ounces needed 25 added, and 10 ounces needs about 16.7 — a ratio of two thirds in both cases, exactly the ratio of the starting amounts. The model is proportional in the starting weight, which is a useful check and also means the answer for any batch size follows from one calculation.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.6 Solve Rational Equations §8.6, pp. 590-590
Error analysis
A student sets up the alloy problem in terms of silver rather than copper.
Annotate
On: \( \frac{92.5}{100} = \frac{0.8(15)}{15+x} \)
The corrected silver equation gives the same answer, but it needs a variable in both the numerator and the denominator. Tracking what stays fixed keeps the setup simple.
Prediction
Commit before reasoning.
Predict first
Pure silver is being added to an alloy. Which quantity gives the simpler equation?
Correct: The copper, since its amount never changes.
\[ \text{copper: } \frac{3}{15+x}; \qquad \text{silver: } \frac{12+x}{15+x} \]
Why: Tracking copper puts a constant, 3, in the numerator and the only variable in the denominator. Tracking silver puts 12 plus x on top and 15 plus x underneath, which still solves but takes longer and offers two more chances to slip. The general principle: in any mixture problem, write the equation in terms of the component that is not being added.
Comparison
Fill the blanks. Only one column changes.
Comparison matrix
| Quantity | Before mixing | After mixing |
|---|---|---|
| Copper | 3 ounces | still 3 ounces |
| Silver | 12 ounces | 37 ounces |
| Total weight | 15 ounces | 40 ounces |
| Percent copper | 20 percent | 7.5 percent |
The copper row is the one that stays fixed, and it is exactly the row that becomes the numerator of the proportion.
Ranking
Setting up a mixture problem.
Put in order
Why: Steps one and two are the modelling; steps three to five are Lesson 8.6's algebra. Almost every error in mixture problems happens in the first two, where a quantity that changes gets treated as though it were fixed.
Section
Section 3
Concept
When an equation is not a proportion, multiply every term on both sides by the least common denominator of all the rational expressions. Every denominator cancels and a polynomial equation remains.
\[ 5x\left(\frac{3}{x}+\frac{8}{5}\right) = 5x \cdot \frac{-13}{x} \]
Every term means every term, including whole numbers and constants. A term with no visible denominator has an invisible 1, and it gets multiplied like everything else.
Figure (svg): A rational equation cleared by multiplying every term by the least common denominator
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.6 Solve Rational Equations §8.6, pp. 590-590 — Using LCDs
Picture it
Example 4, from a rational equation to a quadratic.
Figure (svg): A rational equation cleared by multiplying every term by the least common denominator
The 1 on the left became x times x minus 5. Forgetting to multiply that term is the commonest error in the whole method.
Worked example
Example 3, a multiple-choice item.
\[ \text{Solve } \frac{3}{x} + \frac{8}{5} = \frac{-13}{x}. \]
Find the LCD
Why: The denominators are x and 5, which share nothing.
\[ LCD = 5 x \]
Multiply every term
Why: Five x times each of the three terms.
\[ 15 + 8 x = -65 \]
Collect
Why: Subtract 15 from both sides.
\[ 8 x = -80 \]
Solve and check
Why: Divide by 8, then substitute.
\[ x = -10 \]
Figure (svg): The solution to Worked example clear the denominators shown as a ladder of expressions, one row per algebraic move
\[ x = -10 \]
Verify: substitute back
Why: At x equal to negative 10 the left side is 3 over negative 10 plus 8 over 5, which is negative 0.3 plus 1.6, or 1.3. The right side is negative 13 over negative 10, which is 1.3. And the denominator is not zero there. The book also notes a shortcut: a positive answer would make the left positive and the right negative, ruling out 10 without any arithmetic.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.6 Solve Rational Equations §8.6, pp. 590-590
Fill the middle
Example 4.
Fill in the blanks
x(x-5) \cdot 1 = x^2 - 5x
Why: The 1 becomes the whole LCD, x squared minus 5x. Leaving it as 1 is what turns a quadratic equation into a linear one and loses a solution.
Worked example
Guided Practice 5 to 8.
\[ \text{Solve } \tfrac{7}{2}+\tfrac{3}{x}=3, \; \tfrac{2}{x}+\tfrac{4}{3}=2, \; \tfrac{3}{7}+\tfrac{8}{x}=1, \; \tfrac{3}{2}+\tfrac{4}{x-1}=\tfrac{x+1}{x-1}. \]
First: LCD is 2x
Why: Seven x plus 6 equals 6x.
\[ x = -6 \]
Second: LCD is 3x
Why: Six plus 4x equals 6x.
\[ x = 3 \]
Third: LCD is 7x
Why: Three x plus 56 equals 7x.
\[ x = 14 \]
Fourth: LCD is 2(x - 1)
Why: Three x minus 3 plus 8 equals 2x plus 2.
\[ x = -3 \]
Figure (svg): The solution to Worked example four more with the LCD shown as a ladder of expressions, one row per algebraic move
\[ x = -6, \; 3, \; 14, \; -3 \]
Verify: check the fourth
Why: At x equal to negative 3 the left side is 1.5 plus 4 over negative 4, which is 1.5 minus 1, or 0.5. The right side is negative 2 over negative 4, also 0.5. The denominator x minus 1 is negative 4 there, not zero, so the solution is genuine.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.6 Solve Rational Equations §8.6, pp. 591-591
Trap
\[ 1 - \frac{8}{x-5} = \frac{3}{x}, \; \text{LCD} = x(x-5) \]
Multiply only the fractions by the LCD
Why: The 1 has no denominator, so it is left alone.
\[ 1 - 8x = 3(x-5) \quad \text{(wrong)} \]
That would give a linear equation and one solution. The real equation is quadratic and has two.
\[ x(x-5) \cdot 1 - 8x = 3(x-5) \]
Multiply EVERY term, including the whole number
Why: The 1 is 1 over 1, and it gets the same treatment as everything else.
\[ x^2-16x+15 = 0 \;\Longrightarrow\; x = 1 \text{ or } 15 \]
Writing every term over an explicit denominator of 1 before starting makes the omission impossible.
Ranking
Solving by clearing denominators.
Put in order
Why: Step five is not optional and it must use the original equation, not the cleared one — every candidate satisfies the cleared version by construction, so checking there proves nothing at all.
Matching
Factor the denominators first.
Match the pairs
Why: The last one needed factoring first: x squared minus 9 is x plus 3 times x minus 3, and once that is seen the other two denominators are already among its factors, so nothing extra is needed.
Prediction
Commit before reasoning.
Predict first
After multiplying by the least common denominator, what kind of equation is left?
Correct: A polynomial equation, with no denominators at all.
\[ 1 - \tfrac{8}{x-5} = \tfrac{3}{x} \;\Longrightarrow\; x^2-16x+15=0 \]
Why: Every denominator divides the least common denominator, so every one of them cancels and what remains is a polynomial equation — linear, quadratic, or higher. That is the whole point: it converts a problem you cannot solve directly into one you have been solving since Chapter 1. The catch is that the new equation may have solutions the original did not, which is why the check exists.
Section
Section 4
Concept
Clearing denominators often produces a quadratic rather than a linear equation. Write it in standard form, factor or use the quadratic formula, and take both roots forward to the check.
\[ 1 - \frac{8}{x-5} = \frac{3}{x} \;\Longrightarrow\; x^2-16x+15=0 \]
Two roots is not a sign of an error. Nor is it a promise of two solutions: one, both, or neither may survive the check.
Figure (svg): A rational equation cleared by multiplying every term by the least common denominator
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.6 Solve Rational Equations §8.6, pp. 590-590 — Solve a rational equation with two solutions
Picture it
Example 4, where both roots turn out to be genuine.
Figure (svg): A rational equation cleared by multiplying every term by the least common denominator
Both 1 and 15 check out, since neither makes x or x minus 5 vanish. That is the happy case; the next idea covers the other one.
Worked example
Example 4.
\[ \text{Solve } 1 - \frac{8}{x-5} = \frac{3}{x}. \]
Multiply every term by the LCD
Why: X times x minus 5, applied to all three terms.
\[ x(x - 5) - 8 x = 3(x - 5) \]
Expand both sides
Why: X squared minus 5x minus 8x on the left.
\[ x ^{2} - 13 x = 3 x - 15 \]
Write in standard form
Why: Move everything to the left.
\[ x ^{2} - 16 x + 15 = 0 \]
Factor and solve
Why: One and 15 multiply to 15 and add to 16.
\[ x = 1\text{ or } x = 15 \]
Figure (svg): A rational equation cleared by multiplying every term by the least common denominator
\[ x = 1 \text{ or } x = 15 \]
Verify: check both candidates
Why: At x equal to 1 the left is 1 minus 8 over negative 4, which is 1 plus 2, or 3, and the right is 3 over 1, also 3. At x equal to 15 the left is 1 minus 0.8, which is 0.2, and the right is 3 over 15, also 0.2. Neither value makes x or x minus 5 zero, so both are genuine.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.6 Solve Rational Equations §8.6, pp. 590-590
Fill the middle
Example 4.
Fill in the blanks
x^2-13x = 3x-15 \;\Longrightarrow\; x^2-16x+15 = 0
Why: Subtracting 3x gives negative 16x, and adding 15 moves the constant across. Standard form is what makes the factoring visible.
Worked example
Guided Practice 9.
\[ \text{Solve } \frac{3x}{x+1} - \frac{5}{2x} = \frac{3}{2x}. \]
Find the LCD
Why: Two x times x plus 1.
\[ LCD = 2 x(x + 1) \]
Multiply every term
Why: Six x squared, minus 5 times x plus 1, equals 3 times x plus 1.
\[ 6 x ^{2} - 5 x - 5 = 3 x + 3 \]
Write in standard form and divide by 2
Why: Six x squared minus 8x minus 8, halved.
\[ 3 x ^{2} - 4 x - 4 = 0 \]
Factor and solve
Why: Three x plus 2 times x minus 2.
\[ x = -\frac{2}{3}\text{ or } x = 2 \]
Figure (svg): The solution to Worked example another quadratic shown as a ladder of expressions, one row per algebraic move
\[ x = -\tfrac{2}{3} \text{ or } x = 2 \]
Verify: check both
Why: At x equal to 2 the left is 6 over 3 minus 5 over 4, which is 2 minus 1.25, or 0.75, and the right is 3 over 4. At x equal to negative two thirds the left is negative 6 plus 3.75, which is negative 2.25, and the right is negative 2.25. Neither makes 2x or x plus 1 vanish, so both survive.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.6 Solve Rational Equations §8.6, pp. 591-591
Error analysis
A student solves a rational equation that becomes a quadratic.
Annotate
On: \( x^2-16x+15 = 0 \;\Longrightarrow\; (x-1)(x-15)=0 \;\Longrightarrow\; x = 15 \)
A quadratic offers two candidates. Discarding one requires a reason — an extraneous check that fails — not a preference for a single answer.
Sorting
Solve, then check every candidate.
Sort into buckets
Sort each equation by its number of genuine solutions.
All three outcomes are ordinary. The number of roots the cleared equation has says nothing about how many the original one has.
Two truths and a lie
All three are about rational equations that become quadratics.
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 5's quadratic has roots three halves and negative 3, but negative 3 makes two denominators zero and is discarded, leaving one solution. The cleared equation and the original are not equivalent — the cleared one asks less, so it can have more roots.
Prediction
Commit before reasoning.
Predict first
Where does the x squared in Example 4 come from?
Correct: From multiplying the term 1 by the LCD x times x minus 5.
\[ 1 \cdot x(x-5) = x^2-5x \]
Why: The constant term is the only one whose product with the least common denominator keeps both factors, so it contributes x squared minus 5x. The other two terms each cancel one factor and stay linear. That is why the trap of skipping the constant term does not merely lose a term — it changes the degree of the equation and therefore the number of solutions.
Section
Section 5
Concept
Multiplying by the least common denominator is only legitimate where that denominator is not zero. Any candidate that makes it zero is extraneous and must be discarded, however correctly it was derived.
\[ x = -3 \text{ makes } (x+3)(x-3) = 0 \]
The check must use the original equation. Every candidate satisfies the cleared one by construction, so testing there would pass all of them.
Figure (svg): Two candidate solutions of a rational equation tested, one surviving and one failing
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.6 Solve Rational Equations §8.6, pp. 591-591 — Check for extraneous solutions
Picture it
Example 5: three halves and negative 3, tested.
Figure (svg): Two candidate solutions of a rational equation tested, one surviving and one failing
The graph agrees with the algebra: the two curves cross once, at three halves, and not at negative 3, where neither is defined.
Worked example
Example 5.
\[ \text{Solve } \frac{6}{x-3} = \frac{8x^2}{x^2-9} - \frac{4x}{x+3}. \]
Factor and find the LCD
Why: X squared minus 9 is x plus 3 times x minus 3.
\[ LCD = (x + 3) (x - 3) \]
Multiply every term
Why: Six times x plus 3, equals 8x squared minus 4x times x minus 3.
\[ 6 x + 18 = 4 x ^{2} + 12 x \]
Write in standard form and divide by 2
Why: Four x squared plus 6x minus 18, halved.
\[ 2 x ^{2} + 3 x - 9 = 0 \]
Factor and check both
Why: Two x minus 3 times x plus 3; negative 3 kills two denominators.
\[ x = \frac{3}{2}\text{ only} \]
Figure (svg): Two candidate solutions of a rational equation tested, one surviving and one failing
\[ x = \tfrac{3}{2} \]
Verify: test both candidates in the original
Why: At three halves the left side is 6 over negative 1.5, which is negative 4, and the right side is 18 over negative 6.75 minus 6 over 4.5, which is negative 2.667 minus 1.333, or negative 4. At negative 3 the denominators x squared minus 9 and x plus 3 are both zero, so the equation has no meaning there. A graph shows the curves meeting once, at 1.5.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.6 Solve Rational Equations §8.6, pp. 591-591
Fill the middle
Example 5.
Fill in the blanks
\text-3 x-3, \; x^2-9, \; x+3 \text___ x = 3 \text___ x = ___
Why: The forbidden values are 3 and negative 3. Writing them down before solving means the check afterwards takes one glance rather than a full substitution.
Worked example
Guided Practice 10.
\[ \text{Solve } \frac{5x}{x-2} = 7 + \frac{10}{x-2}. \]
Find the LCD
Why: Both fractions share x minus 2.
\[ LCD = x - 2 \]
Multiply every term
Why: Five x equals 7 times x minus 2, plus 10.
\[ 5 x = 7 x - 4 \]
Solve
Why: Subtract 7x and divide.
\[ x = 2 \]
Check
Why: At x equal to 2 the denominator x minus 2 is zero.
Figure (svg): A rational equation whose only candidate turns out to be extraneous, leaving no solution
\[ \text{no solution} \]
Verify: see why it had to fail
Why: Rewriting the original as 5x minus 10, all over x minus 2, equals 7 gives 5 times x minus 2, over x minus 2, which is 5 wherever it is defined. So the equation says 5 equals 7, which is never true. The algebra produced a candidate, but the equation was contradictory from the start.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.6 Solve Rational Equations §8.6, pp. 591-591
Trap
\[ 2x^2+3x-9 = 0 \text{ at } x = -3: \; 18-9-9 = 0 \quad \checkmark \]
Conclude that negative 3 is a solution
Why: The candidate satisfies the equation being checked.
\[ x = \tfrac{3}{2} \text{ and } x = -3 \quad \text{(wrong)} \]
Of course it satisfies the cleared equation — it was derived from it. The check has to go back further than that.
\[ \text{substitute } -3 \text{ into } \frac{6}{x-3} = \frac{8x^2}{x^2-9} - \frac{4x}{x+3} \]
Check in the ORIGINAL equation, denominators included
Why: The original is what has restrictions; the cleared version has none.
\[ x^2-9 = 0 \text{ and } x+3 = 0 \;\Longrightarrow\; \text{undefined} \]
A faster version of the same check: list the values that make any original denominator zero, and discard any candidate on that list.
Sorting
Test the original denominators.
Sort into buckets
Sort each candidate.
Every extraneous candidate here is exactly a value that makes the least common denominator zero — which is the only way one can arise from this method.
Comparison
Fill the blanks. Two lessons, one pattern.
Comparison matrix
| Method | The risky move | What it can create |
|---|---|---|
| Rational equations | multiplying by the LCD | roots where the LCD is zero |
| Radical equations | squaring both sides | roots with the wrong sign |
| Logarithmic equations | condensing two logarithms | roots with a negative argument |
| All three | the cure | substitute into the original equation |
In every case a step that was not reversible widened the set of allowed values, and in every case the same check closes the gap.
Prediction
Commit before reasoning.
Predict first
Where exactly does the false root in Example 5 come from?
Correct: From multiplying by (x+3)(x-3), which is zero at x = -3.
\[ \text{at } x=-3: \; (x+3)(x-3) = 0, \text{ so the step was multiplying by } 0 \]
Why: Multiplying both sides of an equation by a nonzero quantity preserves its solutions. Multiplying by something that might be zero does not — at that value the step turns a false statement into the true statement zero equals zero, manufacturing a root. So the extraneous candidates are always exactly the zeros of the least common denominator, which is why they can be predicted before any solving.
Comparison
Fill the blanks. Both clear the denominators.
Comparison matrix
| Question | Cross multiplying | Multiplying by the LCD |
|---|---|---|
| When it applies | one fraction on each side | any rational equation |
| The move | set the cross products equal | multiply every term by the LCD |
| Constants | there are none | get multiplied like everything else |
| Check needed? | yes | yes, always |
The last row is the same in both columns. Whatever route clears the denominators, the candidates it produces have to be tested in the original.
Pattern
One routine, with a shortcut for proportions.
No solution is a legitimate answer. An equation can produce a candidate and still be unsatisfiable.
OpenStax Algebra and Trigonometry 2e, §2.6 Other Types of Equations §2.6
Check
Cross multiplying. One fraction each side.
Check your understanding
Solve 3/(x + 1) = 9/(4x + 5).
Answer: A
Why: Cross multiplying gives 12x + 15 = 9x + 9, so 3x = -6.
Check
The LCD method. Multiply every term.
Check your understanding
Solve 1 - 8/(x - 5) = 3/x.
Answer: A
Why: Clearing gives x^2 - 16x + 15 = 0, and both roots check.
Check
Extraneous roots. Check the original.
Check your understanding
Solve 5x/(x - 2) = 7 + 10/(x - 2).
Answer: A
Why: The only candidate is 2, which makes x - 2 zero, so it is extraneous.
Real world
One pump fills a tank in 6 hours and a second fills it in 4 hours. Working together, they fill a fraction 1 over t of the tank per hour, where t is the combined time.
Discussion prompt
Write and solve a rational equation for t, and explain why the answer must be less than 4.
Hint: Each pump contributes its own fraction per hour.
Answer:
\[ \frac{1}{6} + \frac{1}{4} = \frac{1}{t} \]
\[ 12t\left(\frac{1}{6}+\frac{1}{4}\right) = 12t \cdot \frac{1}{t} \;\Longrightarrow\; 2t + 3t = 12 \]
\[ 5t = 12 \;\Longrightarrow\; t = 2.4 \text{ hours} \]
Together they fill the tank in 2 hours 24 minutes. It must be less than 4 because the faster pump alone would manage it in 4, and adding a second pump cannot slow it down — a check worth making before trusting any work-rate answer.
Notice the structure: a sum of reciprocals equal to a reciprocal, exactly the lens equation and the parallel-resistor formula from Lesson 8.5. Solving for t gives t equals the product over the sum, so 24 over 10. Work-rate problems, resistors, lenses and springs all reduce to the same rational equation, and clearing the denominators is what makes every one of them a two-line problem.
Commit first
Answer, then rate your confidence honestly.
Predict first
You clear the denominators, solve, and find one candidate. Is it necessarily a solution?
Correct: No — it must be checked in the original, since multiplying by the LCD may have multiplied by zero.
\[ \frac{5x}{x-2} = 7 + \frac{10}{x-2} \;\Longrightarrow\; x = 2, \text{ extraneous} \]
Why: Multiplying both sides by a nonzero quantity preserves solutions; multiplying by something that could be zero does not. At a value where the least common denominator vanishes, the step converts a false statement into zero equals zero and manufactures a root. Guided Practice 10 is exactly this: the algebra is flawless, the candidate is 2, and 2 makes the denominator zero — so the equation has no solution at all. The check is part of the method, not a precaution.
Explain it
They have solved linear equations and are alarmed by denominators containing x.
Discussion prompt
In four sentences or fewer, explain the strategy for solving an equation with fractions in it.
Hint: What would you do to make it look like an equation you already know?
Answer:
Get rid of the fractions. Multiply both sides by something that every denominator divides into, and they all cancel.
What is left is an ordinary equation with no fractions, which you already know how to solve. The one extra step is at the end: put each answer back into the original and make sure it does not make any denominator zero, because if it does, that answer has to be thrown away.
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 method choice, count the terms on each side: two fractions and nothing else means cross multiplying. For constants, write every term over an explicit denominator of 1 before you start. For quadratics, always write standard form before factoring. For extraneous roots, list the forbidden values in the margin before solving, then just compare.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a solving page. Top left: write both methods with the condition on each, and a one-line reason why cross multiplying is a special case of the other. Top right: work Example 1 by cross multiplying and then again by the least common denominator method, showing they agree. Middle left: work Example 4 in full, circling the constant term and writing beside it what happens to the degree if you forget to multiply it. Middle right: work Example 5 in full, and beside it list the forbidden values before the solving starts, so the extraneous root is visible in advance. Bottom: set up the alloy problem and write one sentence explaining why tracking copper is easier than tracking silver. In a margin, write out Guided Practice 10 and the sentence no solution, with the reason.
If any of your checks used the cleared equation rather than the original, redo it: every candidate passes the cleared one by construction, so that check can never fail.
Recap
Five things, and one habit that finishes all of them.
| If you see | Then |
|---|---|
| One fraction equal to one fraction | Cross multiply |
| Three or more terms | Multiply every term by the LCD |
| A constant term | It gets multiplied by the whole LCD |
| A quadratic after clearing | Standard form, factor, take both roots |
| A candidate making a denominator zero | Discard it as extraneous |
| Every candidate discarded | Answer: no solution |
That closes Chapter 8. Chapter 9 turns to conic sections, where the distance formula and the equations of circles, parabolas, ellipses and hyperbolas take over.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.6 Solve Rational Equations §8.6, pp. 589-593 — everything on these slides traces back here
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