The direct variation equation y equals a x, the constant of variation, the family of lines through the origin, the constant-ratio test for deciding whether data shows direct variation, and building a direct variation model from one data pair.
Subject: Algebra 2 · 65 slides · symbolic lesson
Open the interactive version of this deck
Title
Algebra 2 · Chapter 2 — Linear Equations and Functions
Model Direct Variation
Objectives
Five outcomes. The fourth is the one that lets you decide whether the model fits at all.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.5 Model Direct Variation §2.5, pp. 107-109 — the lesson these objectives are drawn from
Warm-up
Lesson 2.4 built models with an intercept and a rate. This lesson is the special case where the intercept is zero.
Discussion prompt
A car travels at a steady 60 miles per hour. Write the equation for distance after t hours. What is the distance at t equal to zero, and what does that say about where the graph crosses the vertical axis?
Hint: Ask how far you have gone before you set off.
Answer:
\[ d = 60t \;\Longrightarrow\; d(0) = 0 \]
Zero miles at zero hours, so the graph passes through the origin and there is no constant term at all. Any situation where zero input must give zero output has this shape, and it is common enough to have its own name: direct variation.
Concept
When y equals a x for some nonzero constant a, we say y varies directly with x. Doubling x doubles y; tripling x triples y. The constant a is the fixed multiplier that turns one into the other.
constant of variation — The nonzero constant a in the equation y equals a x. It is the slope of the graph and the fixed ratio of y to x.
\[ y = ax, \quad a \neq 0 \]
Because the equation can be rearranged as y over x equals a, direct variation is equivalent to the ratio of the two quantities staying the same.
Figure (svg): The direct variation equation shown two ways: y equals a x, and y over x equals a
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.5 Model Direct Variation §2.5, pp. 107-107 — Direct Variation
Section
Section 1
Concept
A direct variation equation is a linear equation whose constant term is zero. Its slope is the constant of variation, and its graph passes through the origin because zero input gives zero output.
direct variation — The relationship y equals a x between two variables, in which y is a fixed multiple of x. The graph is a line through the origin.
\[ y = ax \quad \text{is } y = mx + b \text{ with } b = 0 \]
The requirement that a is nonzero rules out the horizontal line y equals zero, which would make y constant rather than varying with anything.
Figure (svg): Five lines through the origin with different slopes, showing the family of direct variation graphs
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.5 Model Direct Variation §2.5, pp. 107-107
Picture it
Four members, differing only in their slopes.
Figure (svg): Five lines through the origin with different slopes, showing the family of direct variation graphs
Every linear function has two dials, m and b. Direct variation is what you get when the b dial is locked at zero, so only steepness remains free.
Worked example
Reading a from three equations, including one that has to be rearranged first.
\[ \text{Find the constant of variation in } y = 3x, \; y = -\tfrac{2}{5}x, \; \text{and } 4y = 12x. \]
Read the first directly
Why: The coefficient of x is the constant of variation.
\[ a = 3 \]
Read the second directly
Why: Negative two fifths, sign included.
\[ a = -\frac{2}{5} \]
Rearrange the third into the form y equals a x
Why: Dividing both sides by four leaves y alone on the left.
\[ y = 3 x \]
Read the constant off the rearranged form
Why: Twelve over four is three, so this is the same relationship as the first.
\[ a = 3 \]
Figure (svg): The solution to Worked example identify the constant of variation shown as a ladder of expressions, one row per algebraic move
\[ a = 3, \quad a = -\tfrac{2}{5}, \quad a = 3 \]
Verify: test the ratio y over x at one point of each
Why: For the first, at x equal to 2 the value is 6, and 6 over 2 is 3. For the third, at x equal to 2 the equation gives 4y equal to 24, so y is 6 and the ratio is again 3. Both routes agree, and the first and third really are the same line.
Sorting
Solve for y in your head first if you need to.
Sort into buckets
Sort each equation.
The test is one substitution: put x equal to zero and see whether y comes out zero.
Worked example
The test is whether the constant term is zero once the equation is solved for y.
\[ \text{Which of } y = 5x, \; y = 5x + 2, \; 3y = -9x, \; y = \tfrac{x}{4} + 1 \text{ show direct variation?} \]
Check the first
Why: Already in the form y equals a x with no constant term.
Check the second
Why: There is a constant term of 2, so at x equal to zero the value is 2 rather than 0.
Check the third
Why: Dividing by three gives y equals negative 3x, with no constant term.
Check the fourth
Why: There is a constant term of 1, so the graph misses the origin.
Figure (svg): The solution to Worked example which equations are direct variations shown as a ladder of expressions, one row per algebraic move
\[ y = 5x \;\text{ and }\; 3y = -9x \quad \text{are direct variations} \]
Verify: substitute x equal to zero into each
Why: The two direct variations give y equal to zero, so their graphs pass through the origin. The other two give 2 and 1, so they do not. Substituting zero is the fastest single test there is for this.
Trap
\[ y = 5x + 2 \]
Call this a direct variation because y depends on x
Why: Any dependence at all is read as direct variation.
Then doubling x should double y. At x equal to 1, y is 7; at x equal to 2, y is 12 — not 14.
\[ y = 5x + 2 \quad \text{is NOT a direct variation} \]
Check whether zero input gives zero output
Why: Direct variation requires the constant term to be zero, so that the graph passes through the origin.
\[ y(0) = 2 \neq 0 \]
Compare y equals 5x, where doubling x really does double y: 5 becomes 10. The constant term is what breaks the doubling property.
Prediction
Commit before you compute.
Predict first
In a direct variation, if x is multiplied by 3, what happens to y?
Correct: y is multiplied by 3, whatever the constant of variation is.
\[ y = ax \;\Longrightarrow\; a(3x) = 3(ax) = 3y \]
Why: Replacing x by 3x in y equals a x gives a times 3x, which is 3 times a x, or three times the original y. The constant a factors out and therefore cannot affect the result. This multiply-together property is what direct variation actually means, and it is what fails as soon as a constant term is added.
Explain it to yourself
The definition rules out one value.
\[ y = ax, \quad a \neq 0 \]
Discussion prompt
What relationship would y equals 0 times x describe, and why is it excluded from direct variation? What would happen to the ratio test on such data?
Hint: Write out what the equation says when a is zero.
Answer:
With a equal to zero the equation becomes y equals 0 for every x — a horizontal line along the axis. Nothing varies at all, so calling it a variation would be misleading.
The ratio test would give y over x equal to zero for every pair, which is technically constant. Excluding a equal to zero is what stops that degenerate case from counting, and it is the same instinct as excluding a vertical line from having a slope.
Matching
Some need rearranging first.
Match the pairs
Why: The constant of variation is only readable once y stands alone with coefficient one. Two of these need a division and one needs a term moved, and in each case the answer changes if that step is skipped: reading 3y equals -9x as a equal to -9 would be wrong by a factor of three.
Section
Section 2
Concept
An ordinary line needs two points, but a direct variation already passes through the origin — so one more point determines it completely. Substitute the pair, solve for a, and write the equation.
\[ y = ax \;\Longrightarrow\; a = \frac{y}{x} \]
That is why direct variation problems give you a single measurement where an ordinary linear model would need two.
Figure (svg): Five lines through the origin with different slopes, showing the family of direct variation graphs
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.5 Model Direct Variation §2.5, pp. 107-107 — Write and graph a direct variation equation
Picture it
Every line in the family already goes through it.
Figure (svg): Five lines through the origin with different slopes, showing the family of direct variation graphs
So a direct variation problem is really a two-point problem in which one of the points was given away for free by the assumption.
Worked example
Example 1. Substitute, solve, write, graph.
\[ \text{Write and graph a direct variation equation with } (-4, 8) \text{ as a solution.} \]
Start from the general form
Why: Direct variation is assumed, so the form is y equals a x with a unknown.
\[ y = a x \]
Substitute the ordered pair
Why: Eight for y and negative four for x.
\[ 8 = a(-4) \]
Solve for a
Why: Dividing both sides by negative four gives negative two.
\[ a = -2 \]
Write the equation and describe the graph
Why: A line through the origin with slope negative two.
\[ y = -2 x \]
Figure (svg): The solution to Worked example from one ordered pair shown as a ladder of expressions, one row per algebraic move
\[ y = -2x \]
Verify: substitute the given pair back in
Why: At x equal to negative four, negative two times negative four is eight — the given y value. And at x equal to zero the equation gives zero, so the line passes through the origin as every direct variation must.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.5 Model Direct Variation §2.5, pp. 107-107
Fill the middle
Guided Practice 3.
Fill in the blanks
3 = a(5) \;\Longrightarrow\; a = \frac{3}{5} \;\Longrightarrow\; y = \tfrac______x
Why: Dividing both sides by 5 gives a equal to three fifths. Substituting back: three fifths of five is three, which is the given y value. Note the constant is positive here because x and y share a sign, in contrast with the two negative constants from the previous example.
Worked example
Guided Practice 2 and 4. The constant does not have to be a whole number.
\[ \text{Write direct variation equations through } (-7, 4) \text{ and through } (6, -2). \]
Substitute the first pair
Why: Four for y and negative seven for x.
\[ 4 = a(-7) \]
Solve for a
Why: Four divided by negative seven is negative four sevenths.
\[ a = -\frac{4}{7} \]
Substitute the second pair
Why: Negative two for y and six for x.
\[ -2 = a(6) \]
Solve and reduce
Why: Negative two over six reduces to negative one third.
\[ a = -\frac{1}{3} \]
Figure (svg): The solution to Worked example a fractional constant shown as a ladder of expressions, one row per algebraic move
\[ y = -\tfrac{4}{7}x \qquad \text{and} \qquad y = -\tfrac{1}{3}x \]
Verify: substitute each given pair
Why: Negative four sevenths of negative seven is four, and negative one third of six is negative two. Both original pairs are recovered. Note that both constants are negative because in each pair x and y have opposite signs — a fact you could have predicted before dividing.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.5 Model Direct Variation §2.5, pp. 107-107
Trap
\[ \text{Direct variation through } (6, -2). \]
Compute the constant as x over y
Why: The two numbers are divided in the order they appear in the ordered pair.
\[ a = \frac{6}{-2} = -3 \;\Longrightarrow\; y = -3x \]
Substituting x equal to 6 gives negative eighteen, not negative two.
\[ \text{Direct variation through } (6, -2). \]
Solve y equals a x for a, which puts y on top
Why: The constant of variation is the OUTPUT divided by the input, because that is what dividing both sides by x leaves.
\[ a = \frac{y}{x} = \frac{-2}{6} = -\tfrac{1}{3} \]
\[ y = -\tfrac{1}{3}x \;\Longrightarrow\; y(6) = -2 \quad \checkmark \]
Elimination
A direct variation passes through (-3, 12).
Eliminate the wrong options
What is its equation?
Survives elimination: A
Why: Substituting the pair into y equals a x gives 12 equal to a times -3, so a is -4. Checking: negative four times negative three is twelve. A quick sanity check on the sign comes first — opposite signs in the pair force a negative constant.
Discrimination
Do not divide anything. Decide the sign from the pair alone.
Sort into buckets
Sort each ordered pair by the sign of its constant of variation.
Reverse engineer
Work in the other direction.
Fill in the blanks
y = -\tfrac-21___x \;\Longrightarrow\; \text___ (6, ___)
Why: Negative seven halves of six is negative twenty-one, so the point (6, -21) lies on the line. This is the reverse of the writing task, and it is worth practising because it is how you generate your own check points: pick any x, compute y, and confirm the pair satisfies the original relationship.
Section
Section 3
Concept
Graphing a direct variation needs no intercept calculation: the line goes through the origin. Plot that, step off the constant of variation, and draw.
\[ y = ax \;\Longrightarrow\; \text{slope } a, \; y\text{-intercept } 0 \]
That the whole family shares one point is unusual and useful. It means two direct variations can only ever intersect at the origin, unless they are the same line.
Figure (svg): Five lines through the origin with different slopes, showing the family of direct variation graphs
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.5 Model Direct Variation §2.5, pp. 107-107
Picture it
The family of direct variation graphs.
Figure (svg): Five lines through the origin with different slopes, showing the family of direct variation graphs
A positive constant puts the line in the first and third quadrants; a negative one puts it in the second and fourth. The size of the constant sets the steepness.
Worked example
Example 1's graph, drawn step by step.
\[ \text{Graph } y = -2x. \]
Plot the origin
Why: Every direct variation passes through it, so no calculation is needed for this point.
\[ \text{plot } (0, 0) \]
Step off the constant of variation
Why: Negative two means down two for every one across.
\[ \text{from } (0, 0)\text{ to } (1, -2) \]
Plot the given point as a third check
Why: The pair (-4, 8) was the one the equation was built from, and it should lie on the line.
\[ (-4, 8)\text{ lies on it} \]
Draw the line through them
Why: Three collinear points is a stronger draw than two.
Figure (svg): The solution to Worked example graph a direct variation shown as a ladder of expressions, one row per algebraic move
\[ y = -2x \]
Verify: check the quadrants the line occupies
Why: A negative constant sends the line through the second quadrant, where x is negative and y positive, and the fourth, where the signs are reversed. The given point (-4, 8) is in the second quadrant, which agrees. A line drawn through the first and third quadrants would signal a lost minus sign.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.5 Model Direct Variation §2.5, pp. 107-107
Sorting
Decide from the sign of the constant alone.
Sort into buckets
Sort each direct variation by which pair of quadrants its graph runs through.
The steepness varies within each group, but the pair of quadrants depends only on the sign. That makes it a quick check on a sketch.
Worked example
Guided Practice 1 and 3 on the same axes, to see the effect of the constant.
\[ \text{Graph } y = -3x \text{ and } y = \tfrac{3}{5}x \text{ on the same axes.} \]
Note what they share
Why: Both pass through the origin, so they meet there and nowhere else.
\[ \text{both through } (0, 0) \]
Step off the first constant
Why: Negative three: down three, right one. Steep and falling.
\[ (1, -3) \]
Step off the second constant
Why: Three fifths: up three, right five. Shallow and rising.
\[ (5, 3) \]
Compare their steepness
Why: Three is larger than three fifths, so the first is the steeper of the two.
Figure (svg): The solution to Worked example two graphs compared shown as a ladder of expressions, one row per algebraic move
\[ y = -3x \;\text{ and }\; y = \tfrac{3}{5}x \]
Verify: confirm they meet only at the origin
Why: Setting the two expressions equal gives negative 3x equal to three fifths x, so eighteen fifths x equals zero and x is zero. Two direct variations with different constants can only cross at the origin, which the algebra confirms.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.5 Model Direct Variation §2.5, pp. 107-107
Error analysis
A student graphs y equals 2x by treating the given point as an intercept.
Annotate
On: \( y = 2x \text{ through } (3, 6) \;\Longrightarrow\; \text{plot } (0, 6), \text{ then step up 2 right 1} \)
For a direct variation the intercept is never something you look up. It is zero by definition, and the given point is a check rather than a starting place.
Tweak it
Only one dial exists in this family.
Parameter explorer
What stays fixed as the constant of variation changes, and what moves?
\[ y = {a}x \]
Comparison
Fill the blanks. One row is the whole difference.
Comparison matrix
| Feature | Direct variation | General linear function |
|---|---|---|
| Form | y = ax | y = mx + b |
| y-intercept | always 0 | any value b |
| Points needed to determine it | one (plus the origin) | two |
| Doubling x doubles y? | yes | no, unless b = 0 |
| Ratio y over x | constant | not constant |
The last row is the operational difference and the one Section 4 turns into a test you can run on data.
Counterexample
A classmate proposes a shortcut.
\[ \text{if y increases when x increases, then y varies directly with x} \]
Discussion prompt
Find a linear equation where y increases with x but the relationship is not a direct variation. Then say what extra condition the definition requires beyond increasing together.
Hint: Add a constant to a direct variation and check the origin.
Answer:
\[ y = 5x + 2 \quad \text{increases, but } y(0) = 2 \neq 0 \]
Increasing together is necessary for a positive constant but nowhere near sufficient. The extra condition is proportionality: doubling x must double y, which requires the graph to pass through the origin.
Testing it on the counterexample: at x equal to 1, y is 7; doubling x to 2 gives 12, not 14. The constant term is exactly what breaks the doubling.
Section
Section 4
Concept
Because y equals a x can be written as y over x equals a, a table of data pairs shows direct variation exactly when the ratio of y to x is the same for every pair. That ratio is the constant of variation.
\[ \frac{y}{x} = a \text{ for every pair} \]
For real data the ratios never come out exactly equal. Approximately equal is what makes direct variation a plausible model, and the textbook says so explicitly.
Figure (svg): A table of shark tooth lengths and body lengths with the ratio computed under each column, all near 120
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.5 Model Direct Variation §2.5, pp. 108-108 — Use ratios to identify direct variation, and its Avoid Errors note
Picture it
Example 3: tooth length against body length for six great white sharks.
Figure (svg): A table of shark tooth lengths and body lengths with the ratio computed under each column, all near 120
The ratios run 119, 121, 121, 119, 120, 120 — close enough that a single constant of 120 describes them all. Nothing biological gives exact ratios.
Worked example
Example 3. Six pairs, six divisions, one verdict.
\[ \text{Tooth lengths } 1.8, 2.4, 2.9, 3.6, 4.7, 5.8 \text{ cm; body lengths } 215, 290, 350, 430, 565, 695 \text{ cm.} \]
Divide body length by tooth length for the first pair
Why: Two hundred and fifteen over 1.8 is about 119.
\[ \text{about } 119 \]
Repeat for every pair
Why: The six ratios come out at about 119, 121, 121, 119, 120 and 120.
\[ 119\text{ to } 121 \]
Judge whether they are close enough
Why: They differ by less than two percent across a threefold range of sizes, which for biological data is very close.
Write the model using a representative ratio
Why: One hundred and twenty is a sensible round value in the middle of the six.
\[ b = 120 t \]
Figure (svg): The solution to Worked example does the data show direct variation shown as a ladder of expressions, one row per algebraic move
\[ \frac{b}{t} \approx 120 \;\Longrightarrow\; b = 120t \]
Verify: test the model against the extreme data points
Why: At t equal to 1.8 the model gives 216 cm against the measured 215; at t equal to 5.8 it gives 696 against the measured 695. Both are within about half a percent, and the errors do not grow with size — which is what confirms a proportional model rather than merely a coincidence at one end.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.5 Model Direct Variation §2.5, pp. 108-108
Definition probe
Each item gives three ratios computed from a data set.
Sort into buckets
Sort each data set by whether direct variation is a plausible model.
Worked example
Guided Practice 6. The same six sharks, but body MASS instead of body length.
\[ \text{Masses } 80, 220, 375, 730, 1690, 3195 \text{ kg against the same tooth lengths.} \]
Divide mass by tooth length for the first pair
Why: Eighty over 1.8 is about 44.
\[ \text{about } 44 \]
Compute a middle ratio
Why: Seven hundred and thirty over 3.6 is about 203.
\[ \text{about } 203 \]
Compute the last ratio
Why: Three thousand one hundred and ninety-five over 5.8 is about 551.
\[ \text{about } 551 \]
Judge the spread
Why: The ratios climb from 44 to 551, more than a twelvefold increase. That is not approximately constant by any standard.
Figure (svg): The solution to Worked example data that fails the test shown as a ladder of expressions, one row per algebraic move
\[ \frac{m}{t}: \; 44, \; 92, \; 129, \; 203, \; 360, \; 551 \quad \text{not constant} \]
Verify: say why this is what biology predicts
Why: Length is a one-dimensional measurement and mass grows roughly with volume, which is three-dimensional. So mass should grow roughly with the CUBE of tooth length, not in proportion to it. Checking: mass over tooth length cubed gives about 13.7, 15.9, 15.4, 15.6, 16.3 and 16.4 — far closer to constant, which is the model the data actually supports.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.5 Model Direct Variation §2.5, pp. 109-109
Error analysis
A student checks the shark mass data and concludes it shows direct variation.
Annotate
On: \( \frac{375}{2.9} \approx 129, \quad \frac{730}{3.6} \approx 203 \;\Longrightarrow\; \text{close enough; direct variation} \)
Compute every ratio, not two, and look at whether they scatter around a value or march in one direction. A trend is different from noise.
Fill the middle
One of the shark pairs from Example 3.
Fill in the blanks
\frac119___ \approx ___ \quad \text___
Why: Four hundred and thirty divided by 3.6 is about 119.4, which rounds to 119. It sits inside the 119-to-121 band the other five pairs produced, which is what makes the pooled model b equals 120t reasonable. Note the unit of the ratio: centimetres of body per centimetre of tooth, which is dimensionless but still worth stating.
Estimation
A shark with a 4.7 cm tooth measured 565 cm long.
Predict first
Roughly what is the body-to-tooth ratio?
Correct: About 120.
\[ \frac{565}{4.7} \approx 120.2 \]
Why: Four point seven times a hundred is 470 and times 120 is 564, so the ratio is essentially 120. Estimating first catches a slipped decimal, which is the main risk when dividing by a number between 1 and 10. The fourth option is the reciprocal, which is what you get from dividing the wrong way round.
Socratic
One question, and nothing else on this slide.
\[ 119, \; 121, \; 121, \; 119, \; 120, \; 120 \]
Discussion prompt
The textbook says the ratios do not have to be exactly equal for direct variation to be plausible. Where would you draw the line? What would you want to know about how the data was measured before deciding, and does the number of data points change your answer?
Hint: Think about the precision of the measurements themselves.
Answer:
Any measurement has a precision, and ratios cannot be more consistent than the numbers they come from. Tooth lengths given to one decimal place already carry about a one percent uncertainty, which alone explains the spread from 119 to 121.
More data points make a verdict safer in both directions: six consistent ratios is decent evidence, and two would be almost none. What matters most is whether the deviations have a direction — random scatter supports the model, a steady climb refutes it however small the steps.
Lesson 2.6 and Chapter 11 make this precise. For now the honest phrase is the textbook's: direct variation is a plausible model, not a proven fact.
Section
Section 5
Concept
When a situation is known to be proportional, a single measurement fixes the constant of variation, and the constant then answers every other question. The constant carries units: output units per input unit.
\[ d = 0.0625t \]
The assumption of proportionality is doing real work. It is what lets one measurement replace the two an ordinary linear model would need.
Figure (svg): A hailstone growing from 0.75 inches in 12 minutes, with the constant of variation computed and used to predict 20 minutes
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.5 Model Direct Variation §2.5, pp. 108-108 — Write and apply a model for direct variation
Picture it
Example 2: a hailstone 0.75 inches across after 12 minutes.
Figure (svg): A hailstone growing from 0.75 inches in 12 minutes, with the constant of variation computed and used to predict 20 minutes
The constant 0.0625 carries the units inches per minute, which is what makes the prediction at twenty minutes a length rather than a bare number.
Worked example
Example 2, both parts.
\[ \text{A hailstone is } 0.75 \text{ in across after } 12 \text{ min. Write the model and predict at } 20 \text{ min.} \]
Write the direct variation form with the right letters
Why: Diameter varies directly with time, so d equals a times t.
\[ d =\text{ at} \]
Substitute the measurement
Why: Nought point seven five for d and twelve for t.
\[ 0.75 = a(12) \]
Solve for the constant and attach its units
Why: Nought point seven five over twelve is 0.0625, in inches per minute.
Predict at twenty minutes
Why: Nought point 0625 times twenty is 1.25 inches.
\[ d = 1.25\text{ in} \]
Figure (svg): The solution to Worked example the hailstone shown as a ladder of expressions, one row per algebraic move
\[ d = 0.0625t \qquad \text{and at } t = 20, \; d = 1.25 \text{ in} \]
Verify: check the proportionality directly
Why: Twenty minutes is five thirds of twelve minutes, so the diameter should be five thirds of 0.75, which is 1.25 inches. Reaching the answer by scaling rather than by the equation confirms both the model and the arithmetic, and it uses the doubling property that defines direct variation.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.5 Model Direct Variation §2.5, pp. 108-108
Real world
Three metres of a certain fabric cost 22.50 dollars, and the shop charges by the metre with no fixed fee.
Discussion prompt
Write the direct variation model, say what its constant means with units, and find the cost of 7 metres and the length you could buy for 60 dollars. Then say what would break the model.
Hint: No fixed fee is exactly what makes it a direct variation.
Answer:
\[ c = at, \; 22.50 = a(3) \;\Longrightarrow\; a = 7.50 \text{ dollars per metre} \]
\[ c(7) = 52.50 \qquad t = \frac{60}{7.50} = 8 \text{ metres} \]
A minimum cutting charge, or a discount for buying a whole roll, would break it — either adds a constant term or changes the rate, and both destroy the proportionality that made one measurement sufficient.
Worked example
Guided Practice 5. Given a size, find the time — and watch the radius.
\[ \text{A hailstone has radius } 0.6 \text{ in. How long has it been forming?} \]
Convert the radius to a diameter
Why: The model is stated in terms of diameter, and the diameter is twice the radius.
\[ d = 1.2\text{ in} \]
Substitute into the model
Why: One point two equals 0.0625 times t.
\[ 1.2 = 0.0625 t \]
Solve for t
Why: One point two divided by 0.0625 is 19.2.
\[ t = 19.2 \]
State the answer with its unit
Why: About nineteen minutes.
\[ \text{about } 19.2\text{ minutes} \]
Figure (svg): The solution to Worked example run the model backwards shown as a ladder of expressions, one row per algebraic move
\[ t = \frac{1.2}{0.0625} = 19.2 \text{ minutes} \]
Verify: substitute the time back into the model
Why: Nought point 0625 times 19.2 is 1.2 inches of diameter, which is a radius of 0.6 inches as given. The radius-to-diameter step is the one this problem is really testing; skipping it would give 9.6 minutes, half the right answer.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.5 Model Direct Variation §2.5, pp. 109-109
Error analysis
A student answers Guided Practice 5 without converting.
Annotate
On: \( 0.6 = 0.0625t \;\Longrightarrow\; t = 9.6 \text{ minutes} \)
Before substituting a measurement into a model, check that it is the same quantity the model's variable was defined as. Radius and diameter are the classic pair.
Matching
Each model came from one measurement.
Match the pairs
Why: In every case the constant is the output divided by the input, and its unit is the output unit per input unit. That unit is what makes the constant interpretable: 60 miles per hour is a speed, 7.50 dollars per metre is a price, and 0.0625 inches per minute is a growth rate. A constant without its unit is just a number.
Commit first
Answer, then rate your confidence honestly.
Predict first
A direct variation model is built from one measurement. How much data would an ordinary linear model need for the same situation?
Correct: Two measurements.
\[ y = mx + b: \; \text{2 unknowns} \qquad y = ax: \; \text{1 unknown} \]
Why: A general line has two unknowns, the slope and the intercept, so two points are needed to pin them down. A direct variation has only one unknown, the constant of variation, because the assumption of proportionality has already supplied the intercept as zero. The assumption is doing the work of a whole extra data point — which is worth remembering, since an unjustified assumption costs you exactly that much accuracy.
Edge cases
The hailstone model, d equals 0.0625t.
Discussion prompt
What does the model predict at t equal to 0, and is that sensible? What about at t equal to 480 minutes, which is eight hours? Say what physical limit the model does not know about.
Hint: Think about how big a hailstone can actually get.
Answer:
At t equal to zero it predicts a diameter of zero, which is exactly right: the hailstone has not started forming. That is the one prediction direct variation always gets right by construction.
\[ d(480) = 0.0625(480) = 30 \text{ inches} \]
Thirty inches across is absurd — the largest hailstone ever recorded is about eight inches. The model does not know that an updraft can only support a stone up to some weight before it falls, so its useful domain ends long before eight hours.
Comparison
Fill the blanks. Every row follows from the intercept being zero.
Comparison matrix
| Question | Answer for y = ax | Why |
|---|---|---|
| Where does the graph cross the vertical axis? | at the origin | x = 0 gives y = 0 |
| How many data points determine it? | one | the origin is already known |
| What happens to y when x doubles? | y doubles | a factors out of a(2x) |
| How do you test data for it? | check the ratio y over x is constant | y = ax rearranges to y/x = a |
| What are the units of a? | output unit per input unit | it is a quotient |
Five different-looking facts, all consequences of one thing: there is no constant term.
Pattern
One routine for every direct variation question.
Step one is the step people skip, and it is the one that decides whether anything after it means anything.
OpenStax Algebra and Trigonometry 2e, §5.8 Modeling Using Variation §5.8
Check
Write the equation from one pair.
Check your understanding
Write a direct variation equation that has (6, -21) as a solution.
Answer: A
Why: Substituting gives -21 = a times 6, so a = -21/6, which reduces to -7/2. Checking: negative seven halves of six is -21.
Check
The ratio test on data.
Check your understanding
A table gives ratios of y to x of 44, 92, 129, 203, 360 and 551. Does the data show direct variation?
Answer: A
Why: Direct variation requires the ratio to be approximately constant. These climb from 44 to 551, a more than twelvefold increase with a clear direction, which is a systematic trend rather than measurement noise.
Check
A model, run backwards. Watch what the variable stands for.
Check your understanding
A hailstone's diameter follows d = 0.0625t, with d in inches and t in minutes. A hailstone has a radius of 0.6 inches. How long has it been forming?
Answer: A
Why: A radius of 0.6 inches is a diameter of 1.2 inches, and 1.2 divided by 0.0625 is 19.2 minutes. Substituting back: 0.0625 times 19.2 gives 1.2 inches of diameter.
Real world
A recipe for four people uses 300 g of rice. You are cooking for seven.
Discussion prompt
Write the direct variation model, find how much rice you need, and say what the constant of variation means. Then say why a recipe's SPICE quantities often do not scale by direct variation even when the rice does.
Hint: Ask whether zero people would need zero rice.
Answer:
\[ r = ap, \; 300 = a(4) \;\Longrightarrow\; a = 75 \text{ g per person} \]
\[ r(7) = 75(7) = 525 \text{ g} \]
Seventy-five grams per person is the constant, and it is exactly the per-serving amount. Zero people needing zero rice is what makes the model proportional in the first place.
Spice often does not scale proportionally because perceived heat is not proportional to quantity, and because a larger pot loses less aroma per unit volume. Cooks scale it sub-proportionally by instinct — which is to say the real relationship has a smaller exponent than one, the kind of model Chapter 7 handles.
Commit first
Answer, then rate your confidence honestly.
Predict first
The perimeter of a square varies with its side length. Does its AREA vary directly with its side length too?
Correct: No — the ratio of area to side is the side itself, which is not constant.
\[ \frac{P}{s} = 4 \quad \text{constant} \qquad \frac{A}{s} = s \quad \text{not constant} \]
Why: The perimeter is 4s, so perimeter over side is always 4 and that is a genuine direct variation. But area is s squared, so area over side is s, which changes as the square grows: a side of 2 gives a ratio of 2, and a side of 5 gives a ratio of 5. Being determined by the side is not the same as varying directly with it, and this is the same distinction the shark mass data made.
Explain it
They have met y equals m x plus b but not direct variation.
Discussion prompt
In four sentences or fewer, explain what makes a relationship a direct variation, why one data point is enough to pin it down, and how they would test a table of numbers for it.
Hint: Lead with the origin.
Answer:
A direct variation is a line whose constant term is zero, so it passes through the origin: zero input gives zero output, and doubling the input doubles the output. One data point is enough because the origin is already known, and two points determine a line.
To test a table, divide the output by the input for every row. If those quotients are all about the same number, that number is the constant of variation and the model fits.
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 deciding, substitute zero and see whether the output is zero. For the ratio, remember it is output over input, because that is what dividing y equals a x by x leaves. For closeness, look for a direction in the variation rather than its size — scatter is fine, a trend is not. For running backwards, check first that your measurement is the same quantity the model's variable names. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Draw one coordinate plane in the middle of a page and graph four direct variations on it with different constants, two positive and two negative, labelling each with its equation. Circle the single point all four share and write beside it why it is shared. To the left, write the two forms of the direct variation equation — y equals a x and y over x equals a — and draw an arrow between them showing the division that connects them. To the right, make a small table of four data pairs of your own that DOES show direct variation, compute all four ratios, and beneath it a second table of four pairs that does not, computing those ratios too and writing one sentence about how the two sets of ratios differ. At the bottom, take one real proportional situation, build the model from a single measurement, write the constant with its units, and make one prediction.
If the two ratio tables look similar, make the second one worse: the point of the contrast is that a failing set of ratios trends in one direction rather than scattering.
Recap
Five things, and all of them follow from the intercept being zero.
| If you see | Then |
|---|---|
| y = ax with a not zero | Direct variation |
| A constant term in the equation | Linear, but not direct variation |
| One ordered pair and proportionality | Divide y by x to get a |
| A table of data | Test whether y over x is constant |
| Ratios that climb steadily | Some other model, not direct variation |
Lesson 2.6 drops the assumption that any single line fits exactly, and asks instead for the line that fits a cloud of points best.
McDougal Littell Algebra 2 (Texas Edition), Ch. 2 Linear Equations and Functions — Lesson 2.5 Model Direct Variation §2.5, pp. 107-109 — everything on these slides traces back here
Want this taught 1-on-1? Alexander tutors Algebra 2 — $55/session, free consultation.