Classifying relationships as direct variation, inverse variation or neither; writing an inverse variation equation from a single data pair; building inverse variation models and reading them from tables; testing data by checking whether the products are constant; and writing joint and combined variation equations from a sentence.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 8 — Rational Functions
Model Inverse and Joint Variation
Objectives
Five outcomes. One new form, and a habit of testing data before trusting it.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.1 Model Inverse and Joint Variation §8.1, pp. 551-555 — the lesson these objectives are drawn from
Warm-up
Lesson 2.4 gave you direct variation: y equals a times x, a line through the origin.
Discussion prompt
Six people share a job that takes 12 hours for one person. If three people work, it takes 4 hours; if six work, 2 hours. Is the number of hours a direct variation with the number of people?
Hint: Compute the quotient, then compute the product.
Answer:
No. In direct variation the QUOTIENT is constant, and here 4 over 3 is not 2 over 6. But look at the products: 3 times 4 is 12, and 6 times 2 is 12.
\[ 3 \cdot 4 = 12, \quad 6 \cdot 2 = 12, \quad 1 \cdot 12 = 12 \]
The product is constant. That is inverse variation, and it is the pattern behind sharing anything fixed among a varying number of takers.
Concept
Two variables show inverse variation when y equals a over x for some nonzero constant a. Equivalently, their product is constant. As one grows the other shrinks, and neither can ever be zero.
inverse variation — The relationship y equals a over x, with a a nonzero constant called the constant of variation. Equivalently, xy equals a.
\[ y = \frac{a}{x}, \; a \neq 0 \quad \Longleftrightarrow \quad xy = a \]
Direct variation and inverse variation differ by one operation in the test — divide for one, multiply for the other — and by everything in the graph.
Figure (svg): Two columns comparing direct variation with inverse variation
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.1 Model Inverse and Joint Variation §8.1, pp. 551-551
Section
Section 1
Concept
Two variables show direct variation if y equals a times x, and inverse variation if y equals a over x, for some nonzero constant a. Anything that cannot be rearranged into one of those two forms is neither.
\[ \text{direct: } y = ax; \qquad \text{inverse: } y = \frac{a}{x} \]
Adding a constant breaks both patterns. The equation y equals x plus 3 looks close to direct variation but is not, because doubling x does not double y.
Figure (svg): Three equations rewritten and classified as direct, inverse, or neither
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.1 Model Inverse and Joint Variation §8.1, pp. 551-551 — Classify direct and inverse variation
Picture it
Example 1, all three parts.
Figure (svg): Three equations rewritten and classified as direct, inverse, or neither
The first two were disguised. Solving each for y is what makes the form visible, and only then can the type be read off.
Worked example
Example 1.
\[ \text{Classify } xy = 7, \; y = x+3, \; \tfrac{y}{4} = x. \]
First: solve for y
Why: Dividing by x gives 7 over x, the inverse form.
\[ y = \frac{7}{x},\text{ inverse} \]
Second: it is already solved
Why: Y equals x plus 3 is neither a times x nor a over x.
Third: multiply by 4
Why: This gives y equals 4x, the direct form.
\[ y = 4 x,\text{ direct} \]
State the rule used
Why: Only y equals ax and y equals a over x count as variation.
Figure (svg): Three equations rewritten and classified as direct, inverse, or neither
\[ \text{inverse}, \; \text{neither}, \; \text{direct} \]
Verify: test the middle one numerically
Why: At x equal to 1, y is 4; at x equal to 2, y is 5. Doubling x did not double y, and the product went from 4 to 10 rather than staying constant. So it is neither kind of variation, which the algebra already said.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.1 Model Inverse and Joint Variation §8.1, pp. 551-551
Sorting
Rewrite each in the form solved for y.
Sort into buckets
Sort each equation.
Two of these were disguised by their arrangement. Solving for y before deciding turns a judgement call into a reading.
Worked example
Guided Practice 1 to 3.
\[ \text{Classify } 3x = y, \; xy = 0.75, \; y = x-5. \]
First: already in direct form
Why: Three x equals y is y equals 3x with the sides swapped.
Second: solve for y
Why: Dividing by x gives 0.75 over x.
Third: a constant is subtracted
Why: Neither form allows an added or subtracted constant.
Note what the constant does
Why: It shifts the graph off the origin, which direct variation never does.
Figure (svg): The solution to Worked example three more to classify shown as a ladder of expressions, one row per algebraic move
\[ \text{direct}, \; \text{inverse}, \; \text{neither} \]
Verify: check the second by multiplying
Why: Take x equal to 3, giving y equal to 0.25; the product is 0.75. Take x equal to 0.5, giving y equal to 1.5; the product is again 0.75. A constant product confirms inverse variation, which is the test the next idea will make explicit.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.1 Model Inverse and Joint Variation §8.1, pp. 552-552
Trap
\[ y = \frac{x}{4} \]
Classify it as inverse variation
Why: A fraction bar with x in it is taken as the signal.
\[ \text{inverse} \quad \text{(wrong)} \]
Here x is on TOP. The equation is y equals one quarter times x, which is direct variation with a equal to one quarter.
\[ y = \frac{x}{4} = \tfrac{1}{4}x \;\Longrightarrow\; \text{direct} \]
Ask where the variable sits, not whether a fraction appears
Why: Inverse variation needs x in the DENOMINATOR.
\[ y = \frac{4}{x} \;\Longrightarrow\; \text{inverse} \]
The same reading decided between exponential and power functions in Lesson 7.7. Where the variable sits is the whole question.
Fill the middle
Example 1a.
Fill in the blanks
xy = 7 \;\Longrightarrow\; y = \fracx___}
Why: Dividing both sides by x puts x in the denominator, which is the inverse variation form. The constant of variation is 7, and it is also the constant product.
Matching
Two forms count, everything else does not.
Match the pairs
Why: The last two rows are the same relationship written differently, which is exactly why solving for y is the reliable first move. The third row is the one that fools people, since it looks almost like the first.
Prediction
Commit before reasoning.
Predict first
Compare y equals 4x and y equals 7 over x at x equal to 0.
Correct: The direct one gives 0; the inverse one is undefined.
\[ y = 4x \text{ at } x=0 \text{ gives } 0; \quad y = \tfrac{7}{x} \text{ at } x=0 \text{ is undefined} \]
Why: A direct variation graph passes through the origin by construction: a times 0 is 0 for any a. An inverse variation graph cannot, because dividing by zero is undefined — which is why its graph has two separate branches and never touches either axis. That single difference is visible immediately in the two pictures, and it is the reason the next lesson calls x equal to 0 an asymptote.
Section
Section 2
Concept
Write the general equation y equals a over x, substitute the given pair of values, and solve for a. The result is a specific equation that can then be used at any other value.
\[ 7 = \frac{a}{4} \;\Longrightarrow\; a = 28 \;\Longrightarrow\; y = \frac{28}{x} \]
Only one point is needed, because there is only one constant to find — the same counting argument that made two points enough for the two-constant families of Lesson 7.7.
Figure (svg): A direct variation graph beside an inverse variation graph
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.1 Model Inverse and Joint Variation §8.1, pp. 551-551 — Write an inverse variation equation
Picture it
Direct variation beside inverse variation, plotted on matching axes.
Figure (svg): A direct variation graph beside an inverse variation graph
The inverse graph has a branch in each of two quadrants and approaches both axes without ever reaching them. The sign of a decides which two quadrants.
Worked example
Example 2.
\[ \text{The variables vary inversely, with } y = 7 \text{ when } x = 4. \text{ Find } y \text{ when } x = -2. \]
Write the general equation
Why: Every inverse variation has this form.
\[ y = \frac{a}{x} \]
Substitute the given pair
Why: Seven equals a over 4.
\[ 7 = \frac{a}{4} \]
Solve for the constant
Why: Multiplying both sides by 4 gives 28.
\[ a = 28 \]
Evaluate at the new input
Why: Twenty-eight over negative 2.
\[ y = -14 \]
Figure (svg): The solution to Worked example write and use the equation shown as a ladder of expressions, one row per algebraic move
\[ y = \frac{28}{x}; \quad y = -14 \text{ when } x = -2 \]
Verify: check the product both times
Why: At the given pair the product is 4 times 7, or 28. At the new pair it is negative 2 times negative 14, also 28. A constant product is what inverse variation means, so this check tests the relationship itself rather than just the arithmetic.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.1 Model Inverse and Joint Variation §8.1, pp. 551-551
Fill the middle
Example 2.
Fill in the blanks
7 = \frac28___ \;\Longrightarrow\; a = ___
Why: Multiplying both sides by 4 gives 28, which is also the product of the given pair. Recognising that shortcut turns a two-line solve into a single multiplication.
Worked example
Guided Practice 4 to 6, each asking for y at x equal to 2.
\[ \text{Inverse variation with } (4,3), \; (8,-1), \; \left(\tfrac{1}{2},12\right). \text{ Find } y \text{ at } x = 2 \text{ each time.} \]
First: the constant is the product
Why: Four times 3 is 12, so y equals 12 over x.
\[ \text{at } x = 2, y = 6 \]
Second: a negative constant
Why: Eight times negative 1 is negative 8.
\[ \text{at } x = 2, y = -4 \]
Third: a fractional input
Why: One half times 12 is 6.
\[ \text{at } x = 2, y = 3 \]
Note the shortcut
Why: The constant of variation is simply the product of any given pair.
\[ a = x y \]
Figure (svg): The solution to Worked example three more equations shown as a ladder of expressions, one row per algebraic move
\[ y = 6, \quad y = -4, \quad y = 3 \]
Verify: check the second's sign
Why: A negative constant means the two branches sit in the second and fourth quadrants, so a positive x gives a negative y. At x equal to 2 the answer negative 4 has the right sign, and 2 times negative 4 is negative 8, the constant. Sign and product both check.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.1 Model Inverse and Joint Variation §8.1, pp. 552-552
Error analysis
A student is told that y varies inversely with x and that y is 7 when x is 4.
Annotate
On: \( y = ax \;\Longrightarrow\; 7 = 4a \;\Longrightarrow\; a = 1.75, \; y = 1.75x \)
The two models predict opposite things. At x equal to 8 the wrong one gives 14, while the right one gives 3.5 — one rises, the other falls.
Matching
The constant is the product.
Match the pairs
Why: Every constant here is just the product of the pair given. The third is negative, which puts its branches in the second and fourth quadrants rather than the first and third.
Prediction
Commit before reasoning.
Predict first
Compare y equals 8 over x with y equals negative 8 over x.
Correct: The branches move from the first and third quadrants to the second and fourth.
\[ xy = 8 > 0: \text{ same signs}; \qquad xy = -8 < 0: \text{ opposite signs} \]
Why: With a positive constant, x and y always share a sign, so the branches sit where both coordinates agree. With a negative constant they always differ, putting the branches in the other two quadrants. The shape is identical; it is reflected across the vertical axis. Neither version ever crosses an axis, because the product would have to be zero.
Ranking
Writing an inverse variation equation from one pair.
Put in order
Why: Step one is the one that decides everything: writing the direct form here instead would produce a model that rises where the real one falls. Read the word inversely before writing anything down.
Section
Section 3
Concept
Inverse variation describes anything fixed being shared out: a storage capacity among songs, a job among workers, a distance among speeds. The constant of variation is the fixed total, which gives it a physical meaning.
\[ n = \frac{10{,}000}{s} \]
That is why the model can be trusted beyond the data. The constant is not a fitted number but a real quantity — here, the player's storage in megabytes.
Figure (svg): The number of songs an MP3 player holds against the average size of a song
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.1 Model Inverse and Joint Variation §8.1, pp. 552-552 — Write an inverse variation model
Picture it
Example 3: a player that holds 2500 songs of 4 megabytes each.
Figure (svg): The number of songs an MP3 player holds against the average size of a song
Every point on the curve has the same product, 10,000. The curve falls steeply at first and then flattens, but never reaches zero.
Worked example
Example 3.
\[ \text{A player holds } 2500 \text{ songs at } 4 \text{ MB each. Find } n \text{ at } s = 2, 2.5, 3, 5. \]
Write the general equation and substitute
Why: Twenty-five hundred equals a over 4.
\[ a = 10, 000 \]
State the model
Why: Ten thousand megabytes shared among songs of size s.
\[ n = 10, \frac{000}{s} \]
Evaluate at each size
Why: Ten thousand divided by 2, 2.5, 3 and 5.
\[ 5000, 4000, 3333, 2000 \]
Describe the trend
Why: As the average size rises, the count falls.
Figure (svg): The number of songs an MP3 player holds against the average size of a song
\[ n = \frac{10{,}000}{s} \]
Verify: interpret the constant
Why: Ten thousand is the player's storage in megabytes: 2500 songs of 4 megabytes each is 10,000 megabytes. That is why the model holds at song sizes never tested — the constant is a fact about the hardware, not a number chosen to fit a curve.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.1 Model Inverse and Joint Variation §8.1, pp. 552-552
Fill the middle
Example 3, Step 1.
Fill in the blanks
2500 = \frac10000___ \;\Longrightarrow\; a = ___
Why: The constant is 10,000, and it is the total storage in megabytes. Reading a constant of variation as a physical quantity is what turns a formula into a model.
Worked example
Guided Practice 7.
\[ \text{A player holds } 3000 \text{ songs at } 5 \text{ MB each. Write its model and compare.} \]
Find the constant
Why: Three thousand times 5.
\[ a = 15, 000 \]
State the model
Why: Fifteen thousand megabytes of storage.
\[ n = 15, \frac{000}{s} \]
Compare with the first player
Why: Fifteen thousand against 10,000 megabytes.
\[ 50 \%\text{ more storage} \]
Check at a common size
Why: At 4 megabytes this player holds 3750 songs.
\[ \text{against } 2500 \]
Figure (svg): The solution to Worked example a different player shown as a ladder of expressions, one row per algebraic move
\[ n = \frac{15{,}000}{s} \]
Verify: compare the two at one size
Why: At 4 megabytes per song the first player holds 2500 and the second 3750, a ratio of 1.5 — exactly the ratio of the two constants. In inverse variation the constant scales every output by the same factor, so comparing constants compares the devices directly.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.1 Model Inverse and Joint Variation §8.1, pp. 552-552
Trap
\[ n = \frac{10{,}000}{s}, \; n = 2500 \text{ at } s = 4 \]
At s equal to 2, expect half as many songs
Why: The linear instinct is applied to a non-linear model.
\[ n = 1250 \quad \text{(wrong)} \]
Halving the song size DOUBLES the count, to 5000. Smaller songs mean more of them fit.
\[ n = \frac{10{,}000}{2} = 5000 \]
Halve the input and the output doubles
Why: In inverse variation the two move in opposite directions, and by reciprocal factors.
\[ s \to \tfrac{s}{2} \;\Longrightarrow\; n \to 2n \]
The product stays fixed, so whatever multiplies one variable divides the other. That reciprocal relationship is the single most useful thing to know about these models.
Comparison
Fill the blanks. Multiply one, divide the other.
Comparison matrix
| Change to s | Effect on n | Example from the model |
|---|---|---|
| halve it | n doubles | s from 4 to 2 gives 2500 to 5000 |
| double it | n halves | s from 2.5 to 5 gives 4000 to 2000 |
| multiply by 10 | n divides by 10 | s from 0.5 to 5 gives 20000 to 2000 |
| leave it alone | n is unchanged | the product stays 10,000 |
Every row is the same statement: the product is fixed, so a factor applied to one variable is a divisor applied to the other.
Sorting
Ask whether something fixed is being shared.
Sort into buckets
Sort each situation.
The question to ask is whether a total is being divided up or built up. Divided means inverse; built up means direct.
Prediction
Commit before reasoning.
Predict first
In the model n equals 10,000 over s, is there a song size for which no songs fit?
Correct: No — the count gets small but never reaches zero.
\[ 0 = \frac{10{,}000}{s} \text{ has no solution} \]
Why: Setting n equal to 0 gives 10,000 equal to 0, which is impossible, so the equation has no solution. At a song size of 20,000 megabytes the model gives half a song, which in context means none fits — the mathematics and the physical situation part company there. Recognising where a model stops describing reality is part of using it, exactly as the extrapolation limit was in Lesson 7.7.
Section
Section 4
Concept
The equation y equals a over x rearranges to xy equals a. So a set of data pairs shows inverse variation exactly when their products are constant, or close to constant for real measurements.
\[ y = \frac{a}{x} \;\Longleftrightarrow\; xy = a \]
For direct variation you would compute the quotients instead. Reaching for the wrong operation is the error the book flags beside this example.
Figure (svg): Four data pairs with their products computed to test for inverse variation
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.1 Model Inverse and Joint Variation §8.1, pp. 553-553 — Check data for inverse variation
Picture it
Example 4: chip area against chips obtained from one wafer.
Figure (svg): Four data pairs with their products computed to test for inverse variation
The four products range from 25,872 to 26,320 — close enough that 26,000 serves as the constant and the model is worth trusting.
Worked example
Example 4.
\[ \text{Areas } 58, 62, 66, 70 \text{ give } 448, 424, 392, 376 \text{ chips. Model and predict at } A = 81. \]
Compute every product
Why: Fifty-eight times 448, and so on for all four pairs.
\[ 25, 984; 26, 288; 25, 872; 26, 320 \]
Judge the constancy
Why: All four are within about 1 percent of 26,000.
Write the model
Why: The constant product becomes the constant of variation.
\[ c = 26, \frac{000}{A} \]
Predict at the new area
Why: Twenty-six thousand divided by 81.
\[ \text{about } 321\text{ chips} \]
Figure (svg): Four data pairs with their products computed to test for inverse variation
\[ c = \frac{26{,}000}{A} \;\Longrightarrow\; c \approx 321 \]
Verify: sanity-check the prediction
Why: Eighty-one is larger than every area in the table, so the count should be smaller than every count in the table — and 321 is below 376. The constant has a meaning too: 26,000 square millimeters is roughly the usable area of the wafer, which is why the same constant governs a chip size never tested.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.1 Model Inverse and Joint Variation §8.1, pp. 553-553
Fill the middle
Example 4, Step 1.
Fill in the blanks
58 \cdot 448 = 25984
Why: The product is 25,984, close to 26,000. Computing all four and seeing them agree is what licenses writing the model at all.
Worked example
Guided Practice 8.
\[ \text{Predict the number of chips per wafer when } A = 79. \]
Use the same model
Why: The constant does not change with the chip size.
\[ c = 26, \frac{000}{A} \]
Substitute the new area
Why: Twenty-six thousand over 79.
\[ 329.1 \]
Round in context
Why: Chips come in whole numbers.
\[ \text{about } 329 \]
Compare with the earlier prediction
Why: A smaller chip gives more chips per wafer.
\[ 329\text{ against } 321 \]
Figure (svg): The solution to Worked example a second prediction shown as a ladder of expressions, one row per algebraic move
\[ c = \frac{26{,}000}{79} \approx 329 \]
Verify: check the direction of the change
Why: Seventy-nine is smaller than 81, and 329 is larger than 321 — the two move in opposite directions, as inverse variation requires. A prediction that moved the same way as the input would signal a model set up backwards.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.1 Model Inverse and Joint Variation §8.1, pp. 553-553
Error analysis
A student tests the chip data for inverse variation.
Annotate
On: \( \frac{448}{58} \approx 7.7, \; \frac{424}{62} \approx 6.8, \; \frac{392}{66} \approx 5.9 \)
The book prints this caution beside the example: quotients for direct, products for inverse. Applying the wrong test hides a real pattern.
Sorting
One operation for each kind of variation.
Sort into buckets
Sort each question.
Each test is asking whether a particular combination stays fixed. Naming what that combination means physically makes the choice obvious rather than memorised.
Two truths and a lie
All three are about testing data.
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. Non-constant quotients rule out DIRECT variation only. The chip data has quotients running from 5.4 to 7.7 and is nevertheless a textbook case of inverse variation. Ruling out one pattern is not the same as ruling out all of them, exactly as a curved semi-log plot in Lesson 7.7 ruled out only the exponential family.
Prediction
Commit before reasoning.
Predict first
The four products are 25,984, 26,288, 25,872 and 26,320. Is that constant?
Correct: Close enough — they agree to about 1 percent, which is normal for measurements.
\[ \text{spread} = 26{,}320 - 25{,}872 = 448 \approx 1.7\% \text{ of } 26{,}000 \]
Why: The spread from 25,872 to 26,320 is 448, less than 2 percent of the mean. Real measurements never agree exactly, so the standard is whether the variation is small relative to the values — the same judgement made about straightness in Lesson 7.7's transformed plots. Demanding exact equality would reject every real data set ever collected.
Section
Section 5
Concept
Joint variation occurs when a quantity varies directly with the product of two or more others. Combining direct and inverse relationships puts some variables in the numerator and others in the denominator, with one constant out front.
\[ z = axy; \qquad z = \frac{ay}{x}; \qquad x = \frac{atr}{s} \]
Reading the sentence is the whole task. Directly and jointly send a variable up top; inversely sends it underneath; the constant a always stays in front.
Figure (svg): Five statements in words translated into variation equations
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.1 Model Inverse and Joint Variation §8.1, pp. 553-554 — Joint Variation
Picture it
Example 6, translated one line at a time.
Figure (svg): Five statements in words translated into variation equations
Every named quantity appears exactly once, above or below the bar, and nothing is ever added. That is the whole grammar of variation statements.
Worked example
Example 5.
\[ \text{z varies jointly with x and y, and } z = -75 \text{ when } x = 3, y = -5. \text{ Find } z \text{ at } x=2, y=6. \]
Write the general equation
Why: Joint variation with two quantities.
\[ z = a x y \]
Substitute the given values
Why: Negative 75 equals a times 3 times negative 5.
\[ -75 = -15 a \]
Solve for the constant
Why: Dividing both sides by negative 15.
\[ a = 5 \]
Evaluate at the new inputs
Why: Five times 2 times 6.
\[ z = 60 \]
Figure (svg): The solution to Worked example write a joint variation equation shown as a ladder of expressions, one row per algebraic move
\[ z = 5xy; \quad z = 60 \]
Verify: check the constant against the given data
Why: Five times 3 times negative 5 is negative 75, the value given. And the new answer is positive because both new inputs are positive, whereas the original pair included a negative — a sign check that confirms the substitution went where it should.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.1 Model Inverse and Joint Variation §8.1, pp. 554-554
Matching
Above the bar or below it.
Match the pairs
Why: In every row the constant a sits in front and each named quantity appears exactly once. Directly and jointly place a variable in the numerator; inversely places it in the denominator. Nothing is ever added.
Worked example
Guided Practice 9 to 14.
\[ \text{Joint variation from } (1,2,7), (4,-3,-24), (-2,6,18), (-6,-4,56). \text{ Find } z \text{ at } x=-2, y=5. \]
First: the constant is z over the product
Why: Seven over 2 is 3.5, so z is 3.5xy.
\[ z = 3.5(-10) = -35 \]
Second and third
Why: Negative 24 over negative 12 is 2; 18 over negative 12 is negative 1.5.
\[ z = -20; z = 15 \]
Fourth
Why: Fifty-six over 24 is seven thirds.
\[ z = -\frac{70}{3} \]
The two translations
Why: Inversely puts a variable below the bar, directly and jointly above it.
\[ x = a w / y; p = a q r / s \]
Figure (svg): The solution to Worked example four more, then two translations shown as a ladder of expressions, one row per algebraic move
\[ -35, \; -20, \; 15, \; -\tfrac{70}{3} \]
Verify: check the third's sign
Why: The constant is negative 1.5, and the new inputs are negative 2 and 5, whose product is negative 10. A negative constant times a negative product gives a positive result, 15. Tracking signs through both the constant and the product is where these problems most often go wrong.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.1 Model Inverse and Joint Variation §8.1, pp. 554-554
Trap
\[ \text{z varies jointly with x and y} \]
Write z as a times the sum
Why: Jointly is read as involving both, so both are added in.
\[ z = a(x+y) \quad \text{(wrong)} \]
Jointly means varying with the PRODUCT. Doubling x alone should double z, and with a sum it would not.
\[ z = axy \]
Multiply the quantities named
Why: Joint variation is direct variation with a product in place of a single variable.
\[ x \to 2x \;\Longrightarrow\; z \to 2z; \quad \text{both doubled} \;\Longrightarrow\; z \to 4z \]
Doubling both variables quadruples z, which is what jointly promises and what a sum would never deliver.
Fill the middle
Example 5.
Fill in the blanks
-75 = a(3)(-5) = -15a \;\Longrightarrow\; a = 5
Why: Three times negative 5 is negative 15, and negative 75 over negative 15 is 5. Simplifying the product before dividing keeps the sign work in one place.
Sorting
One word decides each variable's position.
Sort into buckets
Sort each phrase by where the named variable goes.
Powers ride along with their variable: the square of x inversely means x squared in the denominator, not the whole expression squared.
Prediction
Commit before reasoning.
Predict first
In z equals 5xy, both x and y are doubled. What happens to z?
Correct: It quadruples.
\[ z = 5(2x)(2y) = 4 \cdot 5xy = 4z \]
Why: Each doubling multiplies z by 2, so doing both multiplies it by 4. Starting from x equal to 2 and y equal to 6, z is 60; at x equal to 4 and y equal to 12 it is 240. In a combined relationship the effects compound: if z varied directly with y and inversely with x, doubling both would leave z unchanged, since the two factors of 2 would cancel.
Comparison
Fill the blanks. One constant, three arrangements.
Comparison matrix
| Kind | Equation | Test on data |
|---|---|---|
| Direct | y = ax | the quotients y over x are constant |
| Inverse | y = a/x | the products xy are constant |
| Joint | z = axy | z divided by xy is constant |
| Combined | z = ay/x | z times x over y is constant |
Every test is the same move: isolate the constant a and check whether it really stays constant across the data.
Pattern
One routine for equations, one for data, one for sentences.
Quotients test direct variation and products test inverse variation. Using the wrong one hides a real pattern rather than revealing its absence.
OpenStax Algebra and Trigonometry 2e, §5.8 Modeling Using Variation §5.8
Check
Classifying. Solve for y first.
Check your understanding
Which of these shows inverse variation?
Answer: A
Why: Solving for y gives 0.75 over x, so the product is constant.
Check
One data pair, one constant.
Check your understanding
The variables vary inversely and y = 7 when x = 4. What is y when x = -2?
Answer: A
Why: The constant is 28, and 28 divided by -2 is -14.
Check
Translating a sentence.
Check your understanding
Which equation says that p varies jointly with q and r and inversely with s?
Answer: A
Why: Jointly puts q and r in the numerator as a product; inversely puts s underneath.
Real world
Boyle's law says that for a fixed amount of gas at a fixed temperature, pressure varies inversely with volume. A diver's lungs hold 6 litres of air at the surface, where the pressure is 1 atmosphere.
Discussion prompt
At 30 metres depth the pressure is 4 atmospheres. Find the volume, then explain why divers are told never to hold their breath while ascending.
Hint: Pressure times volume is constant.
Answer:
\[ PV = a \;\Longrightarrow\; (1)(6) = 6 \;\Longrightarrow\; V = \frac{6}{P} \]
\[ P = 4 \;\Longrightarrow\; V = \frac{6}{4} = 1.5 \text{ litres} \]
The air compresses to 1.5 litres at depth. Ascending reverses it: that same air expands back to 6 litres, four times its compressed volume.
A diver who holds their breath while rising traps air that is trying to quadruple in size, and lung tissue cannot stretch that far — the injury is called pulmonary barotrauma and it can be fatal from as little as a few metres. The inverse relationship is not a curiosity here; it is the reason the first rule of scuba training is to keep breathing. Notice too that the danger is worst near the surface, where the same change in depth causes the largest change in volume, because the curve is steepest there.
Commit first
Answer, then rate your confidence honestly.
Predict first
Does the graph of an inverse variation ever cross the horizontal axis?
Correct: No — that would require the product to be zero, but the product is a nonzero constant.
\[ y = 0 \;\Longrightarrow\; xy = 0 \neq a \]
Why: A crossing of the horizontal axis means y equal to 0, which makes the product xy equal to 0. But the product is a, and a is required to be nonzero. The same argument rules out crossing the vertical axis, since x equal to 0 makes the product zero too. So both axes are approached and never met, which is what Lesson 8.2 will call asymptotes. A direct variation graph, by contrast, passes right through the origin.
Explain it
They know direct variation and think inverse variation is just the opposite word.
Discussion prompt
In four sentences or fewer, explain what makes a relationship inverse variation, using an everyday example.
Hint: Think about sharing something fixed.
Answer:
Inverse variation means the PRODUCT of the two quantities stays the same. Think of a pizza cut into slices: more people means smaller slices, and the total pizza never changes.
So if you double the number of people, each share halves. In symbols that is y equal to a over x, where a is the fixed total being shared out.
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 classifying, solve for y before deciding anything. For the constant, remember it is just the product of the pair. For testing a table, multiply rather than divide. For translating, underline every quantity named and place each one above or below the bar before writing the equation out.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a variation page. Top left: write the three general equations — direct, inverse, joint — and beside each the test you would apply to a table of data. Top right: sketch a direct variation graph and an inverse variation graph on the same axes, marking where each meets an axis and where one does not, and write one sentence about why. Middle: work Example 3 in full, from the given pair to the model to the table of four values, and write what the constant means physically. Bottom left: take the chip data, compute all four products, and write the model and one prediction. Bottom right: write five variation sentences of your own, each mixing directly, jointly and inversely, and translate each into an equation, underlining the word that placed each variable above or below the bar.
If any of your five equations has an addition in it, rewrite it: variation statements only ever multiply and divide.
Recap
Five things, and one new form to add to direct variation.
| If you see | Then |
|---|---|
| y = ax | Direct variation; the quotient is constant |
| y = a/x or xy = a | Inverse variation; the product is constant |
| A constant added anywhere | Neither kind of variation |
| Jointly | Multiply those variables in the numerator |
| Inversely | Put that variable in the denominator |
| A table to test | Products for inverse, quotients for direct |
Lesson 8.2 graphs y equals a over x and its translations, giving a name to the two lines the branches approach: asymptotes.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.1 Model Inverse and Joint Variation §8.1, pp. 551-555 — everything on these slides traces back here
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