Chapter 11 of Algebra 1: Concepts and Skills, built for a visual learner. Proportions as scaled bar models, direct versus inverse variation drawn side by side, cancelling factors rather than terms, multiplying and dividing rational expressions, common denominators as shared units, and rational equations with the extraneous-solution check made unmissable.
Subject: Algebra 1 · 63 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
Algebra 1 · Chapter 11
Fractions with variables in them: proportions, variation, and the one check you must never skip
Objectives
A rational expression is a fraction with a variable in it. Everything you know about ordinary fractions still applies — plus one new hazard.
Figure (svg): Two equivalent ratios shown as two bars of different lengths cut into the same proportions, with the cross products marked
Section
Section 11.1
Concept
A proportion says two ratios are equal. Cross multiplying turns it into an ordinary equation with no fractions at all.
Figure (svg): Two equivalent ratios shown as two bars of different lengths cut into the same proportions, with the cross products marked
\[ \frac{a}{b} = \frac{c}{d} \;\Longrightarrow\; ad = bc \]
Worked example
Solve the proportion below.
\[ \frac{x}{12} = \frac{5}{4} \]
Cross multiply
Why: Multiply each numerator by the other denominator. This is really multiplying both sides by both denominators at once.
\[ 4x = 60 \]
Solve the resulting equation
Why: Divide both sides by 4.
\[ x = 15 \]
Figure (svg): Two bars in the ratio five to four, scaled up so the smaller part reads fifteen against twelve
Verify: substitute back and reduce
Why: Fifteen over twelve divides by three to give five over four, which matches the right side exactly.
Socratic
One question, no computation.
\[ \frac{a}{b} = \frac{c}{d} \]
Discussion prompt
What single legal move on this equation produces the cross-multiplied form?
Hint: What would you multiply both sides by to clear both denominators at once?
Answer:
Multiply both sides by b and by d. On the left the b cancels, leaving a times d; on the right the d cancels, leaving c times b.
So cross multiplication is not a special rule — it is clearing denominators, exactly as in Chapter 3. Knowing that matters, because it tells you it only applies when there is a single fraction on each side.
Real world
A recipe for 4 people uses 300 grams of rice. You are cooking for 7.
Discussion prompt
Set up the proportion, solve it, and say what assumption the proportion is making.
Hint: Would twice as many people need twice as long to cook?
Answer:
\[ \frac{300}{4} = \frac{r}{7} \;\Longrightarrow\; r = 525 \]
You need 525 grams of rice.
The assumption is that the relationship is directly proportional — that doubling the people doubles the rice. That is reasonable for rice and unreasonable for, say, cooking time, which is why not every scaling question is a proportion.
Error analysis
Find the error in this setup.
Annotate
On: \( \frac{300\text{ g}}{4\text{ people}} \;\overset{?}{=}\; \frac{7\text{ people}}{r\text{ g}} \)
Label the units before solving. It costs two words and prevents the commonest proportion error there is.
Sorting
Cross multiply mentally and compare. Do not solve anything.
Sort into buckets
Which of these pairs of ratios are actually equal?
Prediction
Commit before computing.
\[ \frac{4}{9} = \frac{x}{45} \]
Predict first
What is x?
Correct: 20 — the denominator was multiplied by 5, so the numerator must be too.
Why: Forty-five is nine times five, so the whole ratio has been scaled by five and the numerator goes from 4 to 20. Cross multiplying gives the same answer: 9x equals 180, so x is 20. Spotting the scale factor is faster when the numbers are friendly, and cross multiplying always works.
Explain it
A classmate can cross multiply but cannot say what a ratio means.
Discussion prompt
Explain in two sentences what a ratio compares, and why 3 to 4 and 6 to 8 count as the same ratio.
Hint: What stayed the same when both numbers doubled?
Answer:
A ratio compares two quantities by how many times bigger one is than the other, rather than by how much bigger.
Three to four and six to eight are the same ratio because both parts were multiplied by the same number, so the relative sizes are unchanged even though the absolute amounts doubled.
Ranking
Convert each to a decimal in your head, then order them from smallest to largest.
Put in order
Why: As decimals these are 0.25, about 0.33, 0.4, 0.6 and 0.8. Comparing ratios by turning them into a single number is the reliable method; comparing the pairs of numbers directly is not, since 2 to 5 is smaller than 3 to 5 even though both numerators and denominators look similar.
Section
Section 11.2
Concept
Direct variation means the ratio stays fixed. Inverse variation means the product stays fixed instead.
Figure (svg): A direct variation line through the origin beside an inverse variation curve falling away from both axes
\[ y = kx \qquad \text{versus} \qquad y = \frac{k}{x} \]
Worked example
The quantity y varies inversely with x, and y is 12 when x is 3. Find y when x is 9.
Write the inverse variation model
Why: In inverse variation the product of the two quantities is a constant.
\[ xy = k \]
Substitute the known pair to find the constant
Why: Three times 12 is 36, so the product is always 36.
\[ k = 36 \]
Use the constant with the new input
Why: Nine times y must also be 36.
\[ 9y = 36 \;\Longrightarrow\; y = 4 \]
Figure (svg): An inverse variation curve with two points marked whose coordinates multiply to the same constant
Verify: check the product at both points
Why: Three times 12 is 36 and 9 times 4 is also 36, so both pairs lie on the same inverse relationship.
Discrimination
Ask whether the ratio stays fixed or the product does.
Sort into buckets
Sort each relationship.
Tweak it
Change the constant and see what stays true.
Parameter explorer
Drag k. What happens near the vertical axis, and does the curve ever touch either axis?
\[ y = \frac{{k}}{x} \]
Comparison
Fill the blanks. One word separates these two.
Comparison matrix
| direct variation | inverse variation | |
|---|---|---|
| what stays constant | the ratio y over x | the product x times y |
| double the input and... | the output doubles | the output halves |
| shape of the graph | a line through the origin | a curve approaching both axes |
The reliable test is arithmetic: compute both the ratio and the product from a table, and see which one is constant.
Real world
Six people can paint a fence in 4 hours.
Discussion prompt
How long would 8 people take, what stays constant, and what assumption is the model making?
Hint: What quantity is the same in both scenarios?
Answer:
The constant is the total work: 6 times 4 is 24 person-hours. With 8 people, 8 times t equals 24, so t is 3 hours.
The assumption is that everyone works at the same steady rate and never gets in each other's way. That is why the model would give an absurd answer for a thousand painters on one fence — the mathematics is fine and the assumption has broken.
\[ 6 \cdot 4 = 8 \cdot 3 = 24 \]
Error analysis
Find the flaw in this reasoning.
Annotate
On: \( \text{4 workers take 6 hours, so 8 workers take } 12 \text{ hours} \)
Decide the direction first, then compute. It catches this entire family of errors in one step.
Section
Section 11.3
Concept
A rational expression simplifies by cancelling a whole factor shared by numerator and denominator. A term that is merely present cannot be cancelled.
Figure (svg): A rational expression with a matching factor cancelling from top and bottom
This is the single most consequential rule in the chapter, and the one broken most often.
Worked example
Simplify the expression below.
\[ \frac{x^2 - 9}{x^2 + 7x + 12} \]
Factor the numerator and denominator completely
Why: Nothing can be cancelled until both are products rather than sums.
\[ \frac{(x + 3)(x - 3)}{(x + 3)(x + 4)} \]
Cancel the shared factor
Why: The bracket x plus 3 appears whole in both, so it divides out.
\[ \frac{x - 3}{x + 4} \]
Note the excluded values
Why: The original expression was undefined at x equal to negative 3 and negative 4, and that stays true even though one factor has gone.
Figure (svg): A rational expression with a matching factor cancelling from top and bottom
Verify: test with x equal to 1
Why: The original gives negative 8 over 20, which is negative two fifths. The simplified form gives negative 2 over 5, which is the same.
Trap
Simplify the expression below.
\[ \frac{x + 3}{x + 5} \]
Cancel the x from top and bottom
Why: An x appears in both places, so it looks cancellable — exactly as it would be in a product.
\[ \frac{x + 3}{x + 5} \;\to\; \frac{3}{5} \]
Test with x equal to 1: the original is 4 over 6, which is two thirds, not three fifths. The cancellation was not legal.
Look at whether the x is a factor or a term.
Check the structure before cancelling anything
Why: In a sum, x is one part being added, not a factor of the whole. Only a factor of the entire numerator and the entire denominator can cancel.
\[ \frac{x + 3}{x + 5} \quad \text{does not simplify} \]
The test that settles it: substitute a number. If the two forms disagree for even one value, the cancellation was illegal.
Sorting
Sort each pair by whether the shared piece is a factor.
Sort into buckets
Which of these cancellations are legal?
Error analysis
Find the illegal step.
Annotate
On: \( \frac{x^2 + 6x}{x^2 + 9} \;\overset{?}{=}\; \frac{6x}{9} \)
Factor first, then look for whole brackets in common. If nothing factors, nothing cancels.
Edge cases
Every rational expression has values it cannot accept.
\[ \frac{x + 1}{x^2 - 4} \]
Discussion prompt
Which values of x must be excluded, and why does that matter even after simplifying?
Hint: What makes a fraction undefined?
Answer:
The denominator factors as x plus 2 times x minus 2, so it is zero at x equal to 2 and x equal to negative 2. Both must be excluded, because division by zero is undefined.
It matters after simplifying because a cancelled factor leaves an invisible hole. Even if x plus 2 cancels away, the original expression was never defined at negative 2, and that restriction survives the simplification.
Fill the middle
Fill each blank.
Fill in the blanks
\frac32 = \frac___})}___})} = \frac______
Why: The numerator factors using the pair 2 and 3, and the denominator is a difference of squares. The shared bracket x plus 2 cancels, leaving x plus 3 over x minus 2. The excluded values are 2 and negative 2, and the negative 2 survives even though its factor cancelled.
Discrimination
Do not simplify. Just name the opening step.
Sort into buckets
Sort each expression by what to do first.
Section
Section 11.4
Concept
Multiplying rational expressions works exactly like multiplying fractions: tops together and bottoms together. Cancelling before multiplying keeps the numbers small.
\[ \frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd} \qquad \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c} \]
To divide, multiply by the reciprocal — flip the second fraction and change the sign to multiplication.
Worked example
Simplify the product below.
\[ \frac{x^2 - 4}{x + 5} \cdot \frac{x + 5}{x - 2} \]
Factor everything first
Why: The first numerator is a difference of squares, and everything else is already factored.
\[ \frac{(x + 2)(x - 2)}{x + 5} \cdot \frac{x + 5}{x - 2} \]
Cancel across the multiplication
Why: In a product, a factor on either top may cancel with a factor on either bottom. Both x plus 5 and x minus 2 disappear.
\[ x + 2 \]
Figure (svg): Two fractions multiplied with matching factors on opposite diagonals crossing out
Verify: test with x equal to 3
Why: The original gives 5 over 8 times 8 over 1, which is 5. The answer gives 3 plus 2, which is also 5.
Prediction
Commit before computing.
\[ \frac{x}{3} \div \frac{x + 1}{6} \]
Predict first
What does this become?
Correct: x over 3, times 6 over x plus 1.
Why: Dividing by a fraction means multiplying by its reciprocal, and it is the SECOND fraction that flips. Flipping the first instead is the classic error and produces a completely different expression. After flipping, the 3 and the 6 reduce, giving 2x over x plus 1.
Fill the middle
Fill each blank.
Fill in the blanks
\frac82(x - 1) \div \frac______ = \frac______ \cdot \frac___}___ = ___
Why: Flipping the second fraction gives 8 over x plus 1. The numerator factors as x plus 1 times x minus 1, so the x plus 1 cancels, and 8 over 4 reduces to 2. The result is 2 times x minus 1.
Analogy
Everything here you already do with numbers. Pair each numeric move with its algebraic twin.
Match the pairs
Why: Not one rule in this chapter is new — they are the fraction rules from primary school with expressions in place of numbers. The last pairing is worth dwelling on: nobody would cancel the 3s in four fifths written as 3 plus 1 over 3 plus 2, and the algebraic version is exactly as wrong.
Two truths and a lie
Three claims about multiplying and dividing rational expressions.
Eliminate the wrong options
Which statement is false?
Survives elimination: C
Why: Keep the false statement, which is C. Cancelling across is a property of multiplication only. In a sum the two fractions are separate quantities that have not been combined yet, so there is nothing shared to cancel — you must find a common denominator and add first.
Section
Section 11.5
Concept
When two rational expressions share a denominator, add or subtract the numerators and keep the denominator unchanged.
\[ \frac{a}{c} + \frac{b}{c} = \frac{a + b}{c} \]
Subtraction needs brackets around the second numerator, because the minus reaches all of it.
Worked example
Simplify the expression below.
\[ \frac{3x + 1}{x - 2} - \frac{x + 7}{x - 2} \]
Write the subtraction with brackets around the second numerator
Why: The minus applies to the whole numerator, not only its first term. Brackets are what make that visible.
\[ \frac{(3x + 1) - (x + 7)}{x - 2} \]
Distribute the minus and combine
Why: Three x minus x is 2x, and 1 minus 7 is negative 6.
\[ \frac{2x - 6}{x - 2} \]
Factor and check for cancellation
Why: The numerator factors as 2 times x minus 3, which shares nothing with x minus 2, so this is simplest.
\[ \frac{2(x - 3)}{x - 2} \]
Figure (svg): Two fractions over the same denominator combining into one, with the subtraction bracket highlighted
Verify: test with x equal to 4
Why: The original gives 13 over 2 minus 11 over 2, which is 1. The answer gives 2 times 1 over 2, which is also 1.
Error analysis
Find the error.
Annotate
On: \( \frac{5x}{x+1} - \frac{2x - 3}{x+1} \;\overset{?}{=}\; \frac{5x - 2x - 3}{x+1} \)
Always write the second numerator in brackets before distributing. It is one extra pair of symbols and it removes the error entirely.
Faded example
Fill each blank.
Fill in the blanks
\frac3x + 63 - \frac______ = \frac___}___ = ___
Why: The numerator is 4x plus 5 minus the quantity x minus 1, which is 4x plus 5 minus x plus 1, giving 3x plus 6. That factors as 3 times x plus 3, so the x plus 3 cancels and the whole expression is simply 3. Cancellations like this are common and easy to miss if you stop before factoring.
Prediction
Commit before computing.
\[ \frac{5}{x + 1} + \frac{2}{x + 1} \]
Predict first
What is the sum?
Correct: 7 over (x + 1).
Why: With a shared denominator you add only the numerators and leave the denominator untouched. Adding the denominators as well is the commonest error and would double them; the denominator names the unit you are counting in, and adding two quantities never changes their unit.
Section
Section 11.6
Concept
Two fractions can only be added once they are measured in the same unit, which means the same denominator.
Figure (svg): Two fractions with different denominators being rescaled to a shared one before adding
Find the least common denominator, rescale both fractions, then add the numerators.
Worked example
Simplify the sum below.
\[ \frac{2}{x} + \frac{3}{x + 1} \]
Find the least common denominator
Why: The two denominators share no factors, so the smallest common one is their product.
\[ x(x + 1) \]
Rescale each fraction to that denominator
Why: Multiply the first by x plus 1 over x plus 1, and the second by x over x — each of which is one, so no value changes.
\[ \frac{2(x + 1)}{x(x + 1)} + \frac{3x}{x(x + 1)} \]
Add the numerators and simplify
Why: Two x plus 2 plus 3x gives 5x plus 2.
\[ \frac{5x + 2}{x(x + 1)} \]
Figure (svg): Two fractions with different denominators being rescaled to a shared one before adding
Verify: test with x equal to 1
Why: The original gives 2 plus 1.5, which is 3.5. The answer gives 7 over 2, which is also 3.5.
Matching
Find the smallest expression both denominators divide into.
Match the pairs
Why: When the denominators share nothing, the least common denominator is their product. When one divides into the other, the larger one is already enough — multiplying them would work but leaves extra cancelling to do later.
Prediction
Do not add. Just choose.
\[ \frac{1}{x - 3} + \frac{1}{x + 3} \]
Predict first
What is the least common denominator?
Correct: (x - 3)(x + 3), which can also be written as x squared minus 9.
Why: The two denominators share no common factor, so the least common denominator is their product. Writing it in factored form is better here, because it makes any later cancellation visible — expanding to x squared minus 9 hides the factors you may need.
Explain it to yourself
You multiplied a fraction top and bottom by the same thing. Say why that is safe.
Discussion prompt
Why does multiplying a fraction's numerator and denominator by the same expression leave its value unchanged?
Hint: What is x plus 1, over x plus 1, equal to?
Answer:
Because you are multiplying by a fraction that equals one. Anything over itself is one, and multiplying by one changes nothing.
There is one condition worth naming: the expression you multiply by must not be zero, since zero over zero is not one. That is another reason excluded values matter throughout this chapter.
Error analysis
Find the error.
Annotate
On: \( \frac{1}{2} + \frac{1}{3} \;\overset{?}{=}\; \frac{2}{5} \)
This is why the numeric version is worth keeping in mind: it makes an invisible algebraic error visible.
Step zero
Look at the sum below and plan it. Do not compute anything.
\[ \frac{x}{x^2 - 9} + \frac{2}{x + 3} \]
Discussion prompt
What is the least common denominator, and what does each fraction need multiplying by?
Hint: Factor every denominator before deciding anything.
Answer:
Factor the first denominator: x squared minus 9 is x plus 3 times x minus 3. The second denominator, x plus 3, already divides into that.
So the least common denominator is x plus 3 times x minus 3. The first fraction is already over it, and the second needs multiplying top and bottom by x minus 3.
\[ \frac{x}{(x+3)(x-3)} + \frac{2(x-3)}{(x+3)(x-3)} \]
Factoring first is what revealed that the common denominator was already sitting there — multiplying the two denominators together would have worked but left extra cancelling to do.
Section
Section 11.7
Concept
Solve a rational equation by multiplying through by the common denominator — then check every answer, because that move can invent solutions.
Figure (svg): A solving chain producing two candidate answers, with one crossed out because it makes a denominator zero
extraneous solution — A value that satisfies the cleared equation but makes a denominator zero in the original. It is not a solution at all, and only checking exposes it.
Worked example
Solve the equation below.
\[ \frac{3}{x} + \frac{1}{2} = \frac{5}{x} \]
Multiply every term by the common denominator
Why: The denominators are x and 2, so multiply through by 2x. Every term, both sides.
\[ 6 + x = 10 \]
Solve the resulting equation
Why: Subtract 6 from both sides.
\[ x = 4 \]
Check against the original denominators
Why: The only forbidden value is zero, and 4 is not zero, so the answer survives.
Figure (svg): The equation with each term multiplied by two x, and the denominators cancelling away
Verify: substitute 4 into the original equation
Why: Three quarters plus one half is one and a quarter, and five quarters is also one and a quarter. Both sides agree and no denominator is zero.
Worked example
Solve the equation below, and watch what happens.
\[ \frac{x}{x - 2} = \frac{2}{x - 2} + 3 \]
Multiply every term by the common denominator
Why: The common denominator is x minus 2, and the 3 must be multiplied too.
\[ x = 2 + 3(x - 2) \]
Solve the resulting equation
Why: Expanding gives x equals 3x minus 4, so 4 equals 2x and x is 2.
\[ x = 2 \]
Check against the original
Why: Substituting 2 makes the denominator x minus 2 equal to zero, so the original expression is undefined there.
\[ \text{no solution} \]
Figure (svg): A solving chain producing two candidate answers, with one crossed out because it makes a denominator zero
Verify: confirm the equation genuinely has no solution
Why: Every candidate the algebra produces is 2, and 2 is forbidden by the original expression, so there is no value of x that works. The equation is unsolvable rather than mis-solved.
Anomaly
The algebra was correct, and the answer is still wrong.
Predict first
Which step introduced a value that was never a solution of the original equation?
Correct: Multiplying both sides by the denominator.
\[ \text{if } x - 2 = 0, \text{ multiplying by it destroys the equation} \]
Why: Multiplying by an expression containing x is only reversible when that expression is non-zero. When x makes it zero, the multiplication effectively multiplies both sides by zero, which turns a false statement into a true one and manufactures a solution. Every other step is reversible, which is why the check only needs to guard this one.
Sorting
For the equation shown, sort each candidate answer.
\[ \frac{1}{x - 4} = \frac{x}{x - 4} \]
Sort into buckets
Which candidates are genuine solutions?
Pattern
Six steps, and the last one is not optional.
Writing the excluded values down before solving is what makes step five instant rather than an afterthought.
Check
Solve it on paper before you click.
Check your understanding
Simplify (x^2 - 25) divided by (x^2 - 3x - 10).
Answer: A
Why: The numerator factors as x plus 5 times x minus 5, and the denominator as x minus 5 times x plus 2. The x minus 5 cancels, leaving x plus 5 over x plus 2.
Check
Solve it on paper before you click.
Check your understanding
Solve 2/x + 1 = 5/x.
Answer: A
Why: Multiplying every term by x gives 2 plus x equals 5, so x is 3. Checking: two thirds plus 1 is five thirds, and 5 over 3 is also five thirds. No denominator is zero, so the answer stands.
Commit first
Answer, then rate your confidence.
\[ \frac{x + 4}{x + 4} \]
Predict first
What does this expression equal, and are there any restrictions?
Correct: 1, except when x is negative 4.
\[ \frac{x + 4}{x + 4} = 1, \quad x \neq -4 \]
Why: Anything divided by itself is one, so the expression simplifies to 1 for every value it accepts. But at x equal to negative 4 the denominator is zero and the expression is undefined — it is not 1 there, it is nothing at all. Excluded values survive simplification.
Exit ticket
Last commitment of the chapter.
Predict first
Which of these is shakiest right now?
Correct: Whatever you picked is the one to drill first.
Why: The first and last cost the most marks. The first produces confidently wrong simplifications that look tidy, and the last produces answers that are complete fabrications. Both have the same cure: substitute a number, and write down the excluded values before you start.
Connect it up
One page, drawn by you.
Draw it
Write the phrase a fraction with variables in it at the centre. Branch to: proportions, direct and inverse variation, simplifying, multiplying and dividing, adding with like and unlike denominators, rational equations. On every branch, note what the denominator forbids, and circle the two places an answer must be checked.
If excluded values are not on your map at all, add them — they are the one genuinely new idea this chapter contributes.
Recap
You can now work with fractions whose parts are expressions, and you know the one hazard that comes with them.
| if you remember one thing | it should be |
|---|---|
| about cancelling | factors cancel, terms never do |
| about dividing | flip the second fraction, not the first |
| about subtraction | bracket the whole second numerator before distributing |
| about equations | write the excluded values down first, then check every answer against them |
Want this taught 1-on-1? Alexander tutors Algebra 1 — $55/session, free consultation.