Solving proportions. Includes the vocabulary of ratios, proportions, extremes and means, the reciprocal property, the cross product property, proportions whose cross product is quadratic, cross multiplying with polynomial expressions and excluding values that make a denominator zero, and modelling a count with a proportion.
Subject: Algebra 1 · 65 slides · symbolic lesson
Open the interactive version of this deck
Title
Algebra 1 · Chapter 11 — Rational Expressions and Equations
Proportions
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.1 Proportions §11.1, pp. 633-638 — the lesson these objectives are drawn from
Warm-up
Chapter 3 solved equations with fractions by multiplying them away. A proportion is that situation in its simplest form.
Discussion prompt
Solve three fifths equals x over twenty by any method you like. What did you multiply by?
Hint: Clear the fraction on the side with the unknown.
Answer:
\[ \tfrac{3}{5} = \tfrac{x}{20} \;\Longrightarrow\; x = 12 \]
Multiplying both sides by twenty clears the right-hand fraction and gives twelve directly. This lesson gives two named shortcuts for that clearing, both of which do the same work in fewer lines.
Concept
A proportion is an equation stating that two ratios are equal. Written as a over b equals c over d, the numbers a and d are the extremes and b and c are the means.
proportion — An equation stating that two ratios are equal, written a over b equals c over d, where b, c and d are not nought. It is read as a is to b as c is to d.
Two properties turn a proportion into an ordinary equation.
Figure (svg): The parts of a proportion named
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.1 Proportions §11.1, pp. 633-633
Section
Section 1
Concept
If two ratios are equal then their reciprocals are equal. Flipping both sides of a proportion is useful when the unknown starts in a denominator.
\[ \text{if } \dfrac{a}{b} = \dfrac{c}{d}, \text{ then } \dfrac{b}{a} = \dfrac{d}{c} \]
Both sides must be flipped, not just one.
Figure (svg): The reciprocal property of proportions
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.1 Proportions §11.1, pp. 633-633 — the Reciprocal Property of Proportions and Example 1
Picture it
Equal numbers have equal reciprocals.
Figure (svg): The reciprocal property of proportions
The property holds because equal non-zero numbers have equal reciprocals, which is a fact about numbers rather than about fractions specifically.
Worked example
This is Example 1 from the textbook.
\[ \text{Solve } \dfrac{5}{2} = \dfrac{60}{x} \text{ using the reciprocal property.} \]
Note where the unknown is
Why: In a denominator.
Flip both sides
Why: The reciprocal property.
\[ \tfrac{2}{5} = \tfrac{x}{60} \]
Multiply both sides by sixty
Why: Clears the remaining fraction.
\[ 24 = x \]
State the solution
Why: The unknown is now alone.
\[ x = 24 \]
Figure (svg): The reciprocal property of proportions
\[ x = 24 \]
Verify: check in the original
Why: Five halves is 2.5, and sixty divided by twenty-four is also 2.5. Substituting into the equation as first written confirms both the flip and the multiplication.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.1 Proportions §11.1, pp. 633-633
Faded example
The property applies to the whole equation.
Fill in the blanks
\dfrac260 = \dfrac______ \;\Longrightarrow\; \dfrac___}___ = \dfrac______}
Why: Both ratios turn over, so the equation stays true. Flipping one side only would change its value and break the equality.
Worked example
The vocabulary applied to a specific proportion.
\[ \text{In } \dfrac{5}{2} = \dfrac{60}{x}, \text{ name the extremes and the means.} \]
Identify the outer terms
Why: The first numerator and last denominator.
\[ 5 \text{ and } x \]
Identify the inner terms
Why: The first denominator and second numerator.
\[ 2 \text{ and } 60 \]
State the restriction
Why: No denominator may be nought.
\[ x \ne 0 \]
Note the cross product
Why: Extremes times extremes.
\[ 5 x = 120 \]
Figure (svg): The parts of a proportion named
\[ \text{extremes } 5, x; \quad \text{means } 2, 60 \]
Verify: check the cross product gives the same answer
Why: Five x equals a hundred and twenty gives x equal to twenty-four, agreeing with the reciprocal route. Two independent methods reaching the same number is the strongest check available here.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.1 Proportions §11.1, pp. 633-633
Trap
\[ \dfrac{5}{2} = \dfrac{60}{x} \;\Longrightarrow\; \dfrac{5}{2} = \dfrac{x}{60} \]
Turn over the side containing the unknown
Why: The aim was to move x upstairs, so that side was flipped.
Flipping one side changes its value, so the two sides are no longer equal. This version gives x equal to a hundred and fifty, which fails the check completely.
\[ \dfrac{2}{5} = \dfrac{x}{60} \]
Flip both sides together
Why: Equal numbers have equal reciprocals, but only if both are flipped.
Substituting the answer into the original equation catches this at once.
Matching
Outer two and inner two.
Match the pairs
Why: The last row is easy to overlook and matters most later in the lesson. A proportion carries restrictions from the moment it is written, not only when a solution is found.
Elimination
Choosing between the two properties.
Eliminate the wrong options
For which proportion does the reciprocal property save the most work?
Survives elimination: A
Why: Flipping helps when the unknown is alone in a denominator, since it moves it to the top in one step. In every other arrangement the cross product is at least as quick.
Socratic
It looks like a large change.
Discussion prompt
Explain why two equal ratios have equal reciprocals. Then say why the property needs both numbers to be non-zero.
Hint: Think about the numbers rather than the fractions.
Answer:
Two equal ratios are just two names for the same number, and a number has only one reciprocal. So dividing one into one gives the same result whichever name is used, and the two reciprocals must be equal.
The reciprocal of nought does not exist, since nothing multiplied by nought gives one, so the property has nothing to say if either ratio is nought. That is why the definition of a proportion excludes a zero in the position that would end up in a denominator after flipping.
Section
Section 2
Concept
In any proportion the product of the extremes equals the product of the means. Multiplying across clears both fractions in a single step.
\[ \text{if } \dfrac{a}{b} = \dfrac{c}{d}, \text{ then } ad = bc \]
It follows from writing both fractions over the denominator b d.
Figure (svg): The cross product property
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.1 Proportions §11.1, pp. 634-634 — the Cross Product Property of Proportions and Example 2
Picture it
Extremes one way, means the other.
Figure (svg): The cross product property
The two arcs are the two products being equated. Drawing them once makes the rule hard to misremember, since the pattern is visually symmetric.
Worked example
This is Example 2 from the textbook.
\[ \text{Solve } \dfrac{3}{y} = \dfrac{5}{8}. \]
Multiply the extremes
Why: Three and eight.
\[ 24 \]
Multiply the means
Why: y and five.
\[ 5 y \]
Set them equal
Why: The property.
\[ 24 = 5 y \]
Divide
Why: By five.
\[ y = \tfrac{24}{5} \]
Figure (svg): The cross product property
\[ y = \dfrac{24}{5} \]
Verify: substitute back
Why: Three divided by twenty-four fifths is three times five over twenty-four, which is fifteen twenty-fourths, and that reduces to five eighths. The fractional answer checks exactly, which a decimal approximation could not.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.1 Proportions §11.1, pp. 634-634
Faded example
Extremes and means.
Fill in the blanks
\dfrac85 = \dfrac______ \;\Longrightarrow\; 3 \cdot ___ = y \cdot ___
Why: Each product pairs a numerator with the denominator on the other side. Pairing the two numerators instead is the classic misuse and gives the reciprocal of the right answer.
Worked example
Deriving it rather than accepting it.
\[ \text{Show that } \dfrac{a}{b} = \dfrac{c}{d} \text{ leads to } ad = bc. \]
Write both over a common denominator
Why: Multiply each by what it lacks.
\[ \tfrac{ad}{bd} = \tfrac{bc}{bd} \]
Note the denominators match
Why: Both are b d.
Compare the numerators
Why: Equal fractions with equal denominators.
\[ a d = b c \]
State the conclusion
Why: The property.
\[ \text{extremes } =\text{ means} \]
Figure (svg): The cross product property
\[ \dfrac{ad}{bd} = \dfrac{bc}{bd} \;\Longrightarrow\; ad = bc \]
Verify: test the derivation on numbers
Why: For two thirds equals four sixths, the common denominator is eighteen and the numerators become twelve and twelve. The rule is a shortcut for a step you could always do longhand, which is why it is safe to trust.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.1 Proportions §11.1, pp. 634-634
Error analysis
The student used cross products on a proportion.
Annotate
On: \( \begin{aligned} \frac{3}{y} &= \frac{5}{8} \\ 3 \cdot 5 &= y \cdot 8 \\ 15 &= 8y \\ y &= \frac{15}{8} \end{aligned} \)
This error produces the reciprocal of the correct answer in many cases, which is why the check is so effective at catching it. Drawing the two diagonal arcs before multiplying makes the correct pairing visible rather than remembered.
Translation
Extremes times extremes.
Match the pairs
Why: The third gives a true statement with no unknown, confirming the proportion rather than solving it. The fourth gives a quadratic, which is the subject of the next section.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Reciprocal | Cross product | |
|---|---|---|
| What it does | flips both ratios | multiplies across the equals sign |
| Best when | the unknown is alone in a denominator | always applicable |
| Result | usually one more step | both fractions cleared at once |
The cross product is the general tool and the reciprocal property is a convenience in one shape. Knowing both means never having to fight a proportion into a particular form.
Socratic
It is given its own name and box.
Discussion prompt
Explain what the cross product property is a shortcut for. Then say why that matters for trusting it.
Hint: What would you do without it?
Answer:
Without it you would write both fractions over the common denominator b d and then compare numerators, since equal fractions with equal denominators must have equal numerators. The property packages those two steps into one multiplication.
Knowing the derivation means the rule can be rebuilt if it is misremembered, which matters because the wrong pairing — numerators with numerators — looks just as symmetric. A rule you can derive is one you can check, and this one takes about fifteen seconds to rebuild.
Section
Section 3
Concept
If the unknown appears in both ratios, the cross product is a quadratic equation. Collect the terms on one side and factor, then check both solutions.
Two solutions are normal here.
Figure (svg): A proportion whose cross product is a quadratic
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.1 Proportions §11.1, pp. 634-634 — Example 3 and its Study Tip on checking both solutions
Picture it
Cross, collect, factor.
Figure (svg): A proportion whose cross product is a quadratic
Everything after the cross product is Chapter 10 material. The proportion only supplied the equation; the factoring finished it.
Worked example
This is Example 3 from the textbook.
\[ \text{Solve } \dfrac{3}{x} = \dfrac{x + 1}{4}. \]
Cross multiply
Why: Three times four, x times the bracket.
\[ 12 = x ^{2} + x \]
Collect on one side
Why: Subtract twelve.
\[ 0 = x ^{2} + x - 12 \]
Factor
Why: Four and negative three.
\[ 0 = (x + 4) (x - 3) \]
Solve
Why: Set each factor to nought.
\[ x = -4 \text{ or } 3 \]
Figure (svg): A proportion whose cross product is a quadratic
\[ x = -4 \text{ and } x = 3 \]
Verify: check both solutions
Why: At three the left side is one and the right is four over four, also one. At negative four the left is negative three quarters and the right is negative three over four, also negative three quarters. Both work, and neither makes a denominator nought.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.1 Proportions §11.1, pp. 634-634
Faded example
Standard form before factoring.
Fill in the blanks
12 = x^2 + x \;\to\; 0 = x^2 + x - 12 \;\to\; 0 = (x + 4)(x - 3)
Why: The zero has to be on one side before the zero-product property applies, exactly as in Lesson 10.4. Factoring the quadratic is then ordinary Chapter 10 work.
Worked example
Guided Practice items of the same shape.
\[ \text{Solve } \dfrac{3}{x} = \dfrac{6}{x + 3} \text{ and } \dfrac{x}{2} = \dfrac{8}{x}. \]
Cross multiply the first
Why: Three times the bracket.
\[ 3 x + 9 = 6 x \]
Solve it
Why: This one stays linear.
\[ x = 3 \]
Cross multiply the second
Why: x times x.
\[ x ^{2} = 16 \]
Solve it
Why: Take square roots.
\[ x = \pm 4 \]
Figure (svg): A proportion whose cross product is a quadratic
\[ x = 3; \qquad x = 4 \text{ and } x = -4 \]
Verify: check the second's negative solution
Why: At negative four the left side is negative two and the right is eight over negative four, also negative two. A negative solution is perfectly valid in a pure proportion, though a situation involving lengths or counts would reject it.
Trap
\[ \dfrac{3}{x} = \dfrac{x+1}{4} \;\Longrightarrow\; 3 = x + 1 \]
Cancel the x with something on the other side
Why: The x looked like it could be removed.
There is no x on the right-hand side to cancel with, so nothing may be removed. This gives x equal to two, and checking shows three halves is not equal to three quarters.
\[ (3)(4) = (x)(x + 1) \]
Cross multiply first and simplify afterwards
Why: Cancelling across an equals sign is not a legal move.
Cancelling is for factors within a single fraction, which is the subject of Lesson 11.3.
Prediction
Some give one and some give two.
Predict first
What makes a proportion's cross product quadratic?
Correct: The unknown appearing in both ratios.
\[ \tfrac{3}{x} = \tfrac{x+1}{4} \to x^2 + x \qquad \tfrac{3}{y} = \tfrac{5}{8} \to 5y \]
Why: Cross multiplying pairs each side's numerator with the other side's denominator, so if the unknown appears in both ratios one of the products contains it twice. That produces a squared term and therefore up to two solutions. When the unknown is confined to one ratio, both products are linear in it and the equation stays linear.
Sorting
Look at where the unknown appears.
Sort into buckets
Sort each proportion by the kind of equation its cross product gives.
The fifth is worth noticing: the unknown appears in both ratios but only once in each product, so the cross product is three x plus nine equals six x, which is linear. Where the unknown sits matters as much as how often it appears.
Socratic
Both came from correct algebra.
Discussion prompt
Give two reasons for substituting both solutions of a quadratic proportion back into the original. Then say which reason is specific to proportions.
Hint: One reason applies to any equation.
Answer:
The first reason is general: a substitution catches arithmetic slips in the cross multiplying, the collecting or the factoring, and it is cheap. The second is specific to proportions — cross multiplying removes the denominators, and with them the restriction that they must not be nought, so a solution can emerge that the original equation never allowed.
That second reason is what the next section is entirely about. In Example 3 neither solution caused a problem, but in Example 4 one of the two must be discarded, and nothing in the algebra itself signals which.
Section
Section 4
Concept
Cross multiplying removes the denominators and the restrictions they carry. Any value that makes an original denominator nought must be discarded, however correctly it was derived.
Division by nought is undefined, so such a value was never allowed.
Figure (svg): A candidate solution rejected because it makes a denominator zero
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.1 Proportions §11.1, pp. 635-635 — Example 4 and the Exclude Zero Denominators note
Picture it
Both came from the same algebra.
Figure (svg): A candidate solution rejected because it makes a denominator zero
The rejection is not a correction of a mistake. The algebra was right and the equation simply does not admit that value, which is a distinction worth stating when writing the answer.
Worked example
This is Example 4 from the textbook.
\[ \text{Solve } \dfrac{y^2 - 9}{y - 3} = \dfrac{y + 3}{2}. \]
Cross multiply
Why: Both sides across.
\[ 2(y ^{2} - 9) = (y - 3) (y + 3) \]
Multiply out
Why: The right side is a difference of squares.
\[ 2 y ^{2} - 18 = y ^{2} - 9 \]
Isolate
Why: Subtract y squared and add nine.
\[ y ^{2} = 9 \]
Take roots and check
Why: Three makes the denominator nought.
\[ y = -3 \]
Figure (svg): A candidate solution rejected because it makes a denominator zero
\[ y = -3 \]
Verify: substitute the accepted solution
Why: At negative three the left side is nought over negative six, which is nought, and the right side is nought over two, also nought. At positive three the left side would be nought over nought, which is undefined, so that value could never have been a solution.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.1 Proportions §11.1, pp. 635-635
Faded example
Set the variable denominator to nought.
Fill in the blanks
\text3 y - 3: \quad y - 3 = 0 \;\Longrightarrow\; y = 3, \text___ y \ne ___
Why: The value that makes a denominator nought is exactly the value the expression forbids. Listing it before solving turns the final check into a comparison rather than a discovery.
Worked example
Deciding what to reject before solving anything.
\[ \text{Which values must be excluded from } \dfrac{y^2 - 9}{y - 3} = \dfrac{y + 3}{2}? \]
Look at each denominator
Why: One contains the variable.
\[ y - 3 \text{ and } 2 \]
Set the variable one to nought
Why: Find the forbidden value.
\[ y - 3 = 0 \]
Solve
Why: That value is excluded.
\[ y = 3 \]
Note the other
Why: Two is never nought.
Figure (svg): A candidate solution rejected because it makes a denominator zero
\[ y \ne 3 \]
Verify: see why doing this first helps
Why: Knowing in advance that three is forbidden means the rejection is a matter of reading a list rather than of noticing something at the last moment. It also makes the answer easier to write, since the restriction can be stated alongside the solution.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.1 Proportions §11.1, pp. 635-635
Trap
\[ y^2 = 9 \;\Longrightarrow\; y = 3 \text{ and } y = -3 \]
Report both square roots
Why: Lesson 9.2 always kept both.
At y equal to three the original left side is nought divided by nought, which is undefined, so three is not a solution of the original equation. The cleared equation has two roots and the original has one.
\[ y = -3 \quad \text{only} \]
Check each root against the original denominators
Why: Clearing fractions loses their restrictions.
This is the first place in the course where correct algebra can produce a value that is not a solution.
Elimination
It satisfies the cleared equation.
Eliminate the wrong options
What is the reason for discarding it?
Survives elimination: A
Why: Division by nought is undefined, so three was never permitted in the original equation whatever the cleared version says. Option C is worth rejecting explicitly: nothing was done wrongly, and the value still has to go.
Hypothesis
The steps are all reversible-looking.
Predict first
Why can cross multiplying introduce a value that is not a solution?
Correct: It removes the denominators, and with them their restrictions.
Every method in Chapter 11 that clears denominators needs this same final check.
Why: The original equation is only defined where its denominators are non-zero, so it carries a restriction that the cleared polynomial equation does not. Multiplying both sides by a quantity that happens to be nought is not a valid step, and cross multiplying does exactly that when the variable takes the forbidden value. The cleared equation is therefore a slightly larger problem than the original, and its extra root has to be filtered out by hand.
Socratic
y squared minus nine over y minus three looks reducible.
Discussion prompt
The left side factors as y plus three times y minus three, over y minus three. What happens if you cancel, and why must the restriction survive?
Hint: What does the expression become, and where is it still undefined?
Answer:
Cancelling gives y plus three, so the equation becomes y plus three equals y plus three over two — which is a perfectly ordinary equation with the single solution negative three. The cancellation is legitimate and considerably quicker than cross multiplying.
But the restriction must be carried along: the original expression is undefined at three, and the cancelled version is not, so they are equal everywhere except at that one point. Writing the simplified form without noting that y cannot be three would silently change the problem. Lesson 11.3 makes this the central issue of simplifying rational expressions.
Section
Section 5
Concept
When one part of something has been counted and is thought to represent the rest, a proportion scales the count up. Parts counted directly are added at the end rather than scaled.
The two sides must use the same units.
Figure (svg): A long pit with two end sections and a sampled strip in the middle
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.1 Proportions §11.1, pp. 635-635 — Example 5, Write and Use a Proportion, on the clay warriors
Picture it
One part scaled, two counted.
Figure (svg): A long pit with two end sections and a sampled strip in the middle
Only the central region is estimated by scaling. The two end sites were counted, so scaling them as well would count them twice over.
Worked example
This is Example 5 from the textbook.
\[ \text{A } 10 \text{ m strip holds } 282 \text{ warriors and represents a } 200 \text{ m region. The ends hold } 450. \text{ Estimate the total.} \]
Write the proportion
Why: Sampled count over sampled width.
\[ \tfrac{282}{10} = \tfrac{w}{200} \]
Cross multiply
Why: Two hundred times two hundred and eighty-two.
\[ 10 w = 56 \, 400 \]
Solve
Why: Divide by ten.
\[ w = 5640 \]
Add the ends
Why: Those were counted directly.
\[ 5640 + 450 = 6090 \]
Figure (svg): A long pit with two end sections and a sampled strip in the middle
\[ \text{about } 6090 \]
Verify: check the scaling factor
Why: Two hundred metres is twenty times ten metres, and twenty times two hundred and eighty-two is five thousand six hundred and forty. Scaling by the ratio of the widths is the same calculation the proportion performs, which confirms the setup.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.1 Proportions §11.1, pp. 635-635
Faded example
Count over width, both sides.
Fill in the blanks
\dfrac2005640 = \dfrac______} \;\Longrightarrow\; w = ___
Why: Both sides are warriors per metre, so the units match and the proportion is meaningful. Mixing a count with a width on the same side would make the equation say nothing.
Worked example
Reading the assumption out of the model.
\[ \text{What must be true for the } 6090 \text{ estimate to be reliable?} \]
Identify the scaling step
Why: The sample stood for the region.
\[ 282 \text{ in } 10 \text{ m} \]
State the assumption
Why: The density is the same throughout.
Consider a failure
Why: If the middle were sparser.
Note what is not assumed
Why: The ends were counted, not scaled.
Figure (svg): A long pit with two end sections and a sampled strip in the middle
\[ \tfrac{282}{10} \text{ per metre throughout} \]
Verify: test the sensitivity
Why: If the true average were ten per cent lower, the central estimate would fall to about five thousand and eighty and the total to about five and a half thousand. A ten per cent error in the sample becomes a ten per cent error in the scaled part, so the estimate is only as good as the sampling.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.1 Proportions §11.1, pp. 635-635
Trap
\[ \tfrac{282 + 450}{10} = \tfrac{t}{200} \]
Add all the known warriors, then scale
Why: All the counts were put together before scaling.
The four hundred and fifty were counted in the end sites, not in the ten-metre strip, so they must not be scaled by twenty. This gives fourteen thousand six hundred and forty, more than twice the correct estimate.
\[ \tfrac{282}{10} = \tfrac{w}{200}, \quad \text{then add } 450 \]
Scale only what was sampled, and add the rest at the end
Why: Two different kinds of number.
Keeping the sampled and counted parts separate until the last step is what prevents this.
Sorting
Sampled parts are scaled; counted parts are not.
Sort into buckets
Sort each quantity by how it enters the total.
Three quantities build the proportion and three are simply added. Sorting them before writing anything is what keeps the two kinds from being mixed.
Prediction
Containing twice as many warriors.
Predict first
How would the estimate change?
Correct: Not at all, since the density is the same.
\[ \tfrac{564}{20} = \tfrac{w}{200} \;\Longrightarrow\; w = 5640 \]
Why: A twenty-metre strip with five hundred and sixty-four warriors has the same warriors-per-metre as a ten-metre strip with two hundred and eighty-two, so the proportion gives the same total. What a wider strip would improve is the reliability of the estimate rather than its value, since a larger sample is less likely to be unrepresentative of the whole. That is a statistical point rather than an algebraic one, and it is why archaeologists sample as much as they can.
Socratic
It is reported as about six thousand.
Discussion prompt
Say what the estimate depends on and how precisely it should be reported. Then say what a second sample would add.
Hint: Think about what was assumed.
Answer:
It depends entirely on the sampled strip having the same density as the rest of the central region, which cannot be known from one sample. Reporting six thousand and ninety to the nearest warrior implies a precision the method does not have; about six thousand is more honest, since the last two digits are an artefact of the arithmetic rather than of the evidence.
A second sample from a different part of the region would show whether the density is roughly constant. If two widely separated strips agreed, the assumption would be supported; if they differed, the single-sample estimate would be exposed as unreliable and an average or a more careful survey would be needed.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| The unknown | The cross product | Solutions |
|---|---|---|
| in one ratio only | linear | one |
| in both ratios | often quadratic | up to two |
| in a denominator | may be cleared away | check for excluded values |
The last row is the one that is new in this chapter. Wherever a variable sits in a denominator, the final answer has a restriction attached to it.
Pattern
To solve any proportion, these five moves cover it.
Step one takes a few seconds and turns step five into a comparison rather than a discovery, which matters because nothing in the algebra itself flags a forbidden value.
OpenStax Elementary Algebra 2e, §8.7 Solve Proportion and Similar Figure Applications §8.7
Check
Multiply across.
Check your understanding
Solve 3/y = 5/8.
Answer: A
Why: The extremes are three and eight, giving twenty-four, and the means are y and five, giving five y. So five y equals twenty-four.
Check
The unknown is in both ratios.
Check your understanding
Solve 3/x = (x + 1)/4.
Answer: A
Why: Cross multiplying gives twelve equals x squared plus x, so x squared plus x minus twelve is nought, which factors as x plus four times x minus three.
Check
Watch the denominators.
Check your understanding
Solving a proportion gives y = 3 and y = -3, but one denominator is y - 3. What is the solution?
Answer: A
Why: At y equal to three the denominator y minus three is nought, and division by nought is undefined, so that value must be discarded.
Real world
This is the clay warriors question from the lesson opener. Pit 1 of the tomb has two end sites containing 450 warriors between them, and a 200 metre central region. A 10 metre strip of that region contains 282 warriors and is thought to be representative.
Discussion prompt
Estimate the total number of warriors in Pit 1, say what assumption the estimate rests on, and say how precisely you would report it.
Hint: Scale the strip, then add the ends.
Answer:
\[ \dfrac{282}{10} = \dfrac{w}{200} \;\Longrightarrow\; w = 5640, \quad \text{total } 5640 + 450 = 6090 \]
The two end sites were counted rather than sampled, so their four hundred and fifty warriors are added at the end rather than scaled by twenty.
The whole estimate rests on the sampled strip having the same density of warriors as the rest of the central region, which one sample cannot establish. Reporting six thousand and ninety to the nearest warrior claims a precision the method does not have — about six thousand is the honest figure, and a second sample from elsewhere in the region would be worth far more than any refinement of the arithmetic.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Cross multiplying gives y squared = 9, and one denominator was y - 3. What are the solutions?
Correct: y = -3 only.
\[ y = 3: \; \tfrac{0}{0} \text{ undefined} \qquad y = -3: \; \tfrac{0}{-6} = 0 = \tfrac{0}{2} \]
Why: Both values satisfy the cleared equation, and the algebra producing them was correct — but at y equal to three the original denominator y minus three becomes nought, and division by nought is undefined, so three was never a permitted value. Cross multiplying removes the denominators and with them the restriction they carried, which means the cleared equation is a slightly larger problem than the original and can have roots the original forbids. Nothing in the algebra flags this, which is why the check against the original denominators is not optional. Negative three causes no such problem: it makes the denominator negative six, and substituting gives nought on both sides.
Explain it
They multiplied the two numerators together and the two denominators together.
Discussion prompt
In no more than four sentences, explain the correct pairing and where the rule comes from. Then give them a check.
Hint: Which numbers are the extremes?
Answer:
A usable answer: the rule pairs each numerator with the denominator on the other side, so for three over y equals five over eight you get three times eight equals y times five. It comes from writing both fractions over the common denominator and comparing numerators, which is why the pairing crosses over.
The check is to substitute your answer back into the original proportion and see whether the two fractions really are equal. Their version usually gives the reciprocal of the right answer, which the check exposes immediately.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: The pairing is fixed by drawing the two diagonal arcs before multiplying anything. Quadratic cross products are fixed by putting the equation in standard form and factoring as in Chapter 10. Exclusions are fixed by listing the forbidden values before you start solving. Setting up is fixed by checking that both sides carry the same units. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of a page write a general proportion, label the extremes and the means, and note the restriction on the denominators. Underneath, derive the cross product property by writing both fractions over a common denominator, so the rule is built rather than quoted, and draw the two diagonal arcs beside it. In the middle, solve one proportion by the reciprocal property and the same one by cross products, side by side, and note which was shorter and why. Beneath that, solve a proportion whose cross product is quadratic, writing every line from the cross multiplication through the factoring, and check both solutions in the original. In the lower half, take a proportion with a variable denominator: list the excluded values first, solve it, and strike through any solution that appears on your list, writing one sentence saying why the algebra alone could not have told you. Finally, in the margin, sketch the sampling problem with the sampled part and the counted parts marked differently.
Your excluded-value list should be written before you solve, not after. If you find yourself discovering the exclusion only at the checking stage, the habit has not formed yet and the next few lessons will punish it.
Recap
Five things, and the fourth is new to this chapter.
| If the question says | Your first move is |
|---|---|
| Solve a proportion | List the excluded values, then cross multiply |
| The unknown is in a denominator | Consider flipping both sides |
| The unknown is in both ratios | Expect a quadratic and factor it |
| A denominator contains the variable | Note what value it forbids |
| A sample represents a whole | Scale the sample; add counted parts separately |
Lesson 11.2 looks at two particular kinds of relationship between quantities. Direct variation is a proportion in disguise, and inverse variation is what happens when one quantity rises as the other falls.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.1 Proportions §11.1, pp. 633-638 — everything on these slides traces back here
Want this taught 1-on-1? Alexander tutors Algebra 1 — $55/session, free consultation.