Two quantities that vary directly, meaning their ratio is a constant k called the constant of variation, so that y equals kx. Includes finding k from one pair of values, graphing a direct variation model as a line through the origin whose slope is k, and fitting an approximate model to real data.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 4 — Graphing Linear Equations and Functions
Direct Variation
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. 4 Graphing Linear Equations and Functions — Lesson 4.6 Direct Variation §4.6, pp. 236-242 — the lesson these objectives are drawn from
Warm-up
Lesson 4.5 found the slope of a line. This lesson looks at the family of lines whose slope is the whole story.
Discussion prompt
For the equation y equals 4x, work out y at x equal to 1, 2, 3 and 5, then divide each y by its x. What do you notice, and what does it say about where the graph must pass?
Hint: Compute the four ratios and compare them.
Answer:
\[ (1, 4), \; (2, 8), \; (3, 12), \; (5, 20) \quad \dfrac{y}{x} = 4 \text{ every time} \]
The ratio is four in every case. And when x is zero, y is four times zero, which is zero — so the point (0, 0) is a solution too, and the graph must pass through the origin.
Concept
When two quantities y and x have a constant ratio k, they are said to have direct variation, and k is called the constant of variation. The relationship is written y equals kx, with k not zero.
direct variation — A relationship between two quantities whose ratio is a constant. If y varies directly with x then y equals kx, where the constant k is the constant of variation.
The model y equals kx is read as y varies directly with x.
Figure (svg): A table of paired values whose ratios are all the same number
McDougal Littell Algebra 1: Concepts and Skills, Ch. 4 Graphing Linear Equations and Functions — Lesson 4.6 Direct Variation §4.6, pp. 236-236
Section
Section 1
Concept
Two quantities vary directly when dividing one by the other always gives the same number. That number is the constant of variation, and it converts the statement about a ratio into an equation.
\[ \dfrac{y}{x} = k \;\Longleftrightarrow\; y = kx, \quad k \neq 0 \]
Either form says the same thing; the second is easier to substitute into.
Figure (svg): A table of paired values whose ratios are all the same number
McDougal Littell Algebra 1: Concepts and Skills, Ch. 4 Graphing Linear Equations and Functions — Lesson 4.6 Direct Variation §4.6, pp. 236-236 — the Model for Direct Variation box
Picture it
The bottom row never changes.
Figure (svg): A table of paired values whose ratios are all the same number
The x-values and y-values both grow, and the row that matters is the third one. Direct variation is a statement about that row rather than about either of the first two.
Worked example
The test is to compute every ratio and compare them.
\[ \text{Do these pairs show direct variation? } \; (2, 6), \; (3, 9), \; (5, 15). \]
Compute the first ratio
Why: Six divided by two is three.
\[ 3 \]
Compute the second
Why: Nine divided by three is three.
\[ 3 \]
Compute the third
Why: Fifteen divided by five is three.
\[ 3 \]
Draw the conclusion
Why: All three ratios agree, so k is three and the model is y equals 3x.
\[ y = 3 x \]
Figure (svg): The solution to Worked example check a table for direct variation shown as a ladder of expressions, one row per algebraic move
\[ \dfrac{6}{2} = \dfrac{9}{3} = \dfrac{15}{5} = 3 \;\Longrightarrow\; y = 3x \]
Verify: test the model on a pair it was not built from
Why: At x equal to four the model predicts twelve, and twelve over four is three, consistent with the others. Every pair the model produces has the same ratio by construction, which is what makes it the right description of the table.
Sorting
Test whether the ratio of the two quantities is constant.
Sort into buckets
Sort each relationship by whether it is direct variation.
Exercises 25 and 26 in the textbook ask exactly this about bicycling and about a circle's circumference. Both are direct variation, and the constant in the second one is a number you have known for years.
Worked example
A relationship can be perfectly regular and still not be direct variation.
\[ \text{Do these pairs show direct variation? } \; (1, 7), \; (2, 11), \; (3, 15). \]
Compute the first ratio
Why: Seven divided by one is seven.
\[ 7 \]
Compute the second
Why: Eleven divided by two is five and a half.
\[ 5.5 \]
Compare
Why: The ratios differ, so there is no constant of variation.
Notice what is constant instead
Why: Each y is four more than the last as x rises by one, so the differences are constant even though the ratios are not.
Figure (svg): The solution to Worked example a table that fails the test shown as a ladder of expressions, one row per algebraic move
\[ \dfrac{7}{1} = 7, \quad \dfrac{11}{2} = 5.5 \quad \text{not equal} \]
Verify: find the equation and see why it fails
Why: The constant difference of four gives y equals 4x plus 3, which is a line but not through the origin. At x equal to zero it gives y equal to three rather than zero, and that stray constant is exactly what breaks the ratio.
Trap
A taxi charges 3 dollars to board plus 2 dollars a mile. Does the fare vary directly with distance?
Say yes: further means more expensive, so they go up together
Why: Both quantities increase, which is what direct variation feels like it should mean.
One mile costs five dollars and two miles cost seven, so the ratios are five and three and a half. They are not equal, so this is not direct variation.
\[ f = 2m + 3 \quad \text{not of the form } y = kx \]
Test the ratio rather than the direction
Why: Direct variation requires a constant quotient, which is a stronger condition than both quantities rising together.
The boarding charge is what breaks it: at zero miles the fare is three dollars rather than nothing, so the graph misses the origin.
Faded example
Divide each y by its x and compare.
Fill in the blanks
(2, 6), (3, 9), (5, 15): \quad \dfrac33 = 3, \; \dfrac______ = ___, \; \dfrac______ = ___
Why: All three ratios are three, so the quantities vary directly with a constant of variation of three and the model is y equals 3x. Had any one of them differed, the answer would have been that there is no direct variation at all.
Elimination
Several statements sound like direct variation.
Eliminate the wrong options
Which one actually defines it?
Survives elimination: A
Why: The constant ratio is the definition, and everything else in the lesson follows from it. Option D is worth dwelling on: it names a true consequence rather than the definition, and mistaking a consequence for a definition is what lets non-examples like y equals 4x plus 3 slip through.
Socratic
The textbook states this and it is worth deriving.
Discussion prompt
Explain why every direct variation graph contains the point (0, 0). Then say what it would mean physically if a real relationship's graph missed the origin.
Hint: Substitute zero into the model.
Answer:
Substituting x equal to zero into y equals kx gives y equal to k times zero, which is zero whatever k is. So (0, 0) satisfies every direct variation equation, and the graph must contain it.
A graph missing the origin means that when one quantity is zero the other is not — a boarding charge, a starting balance, a fixed fee. That stray amount is precisely what stops the ratio being constant, so checking whether zero maps to zero is a fast physical test for direct variation before any arithmetic.
Section
Section 2
Concept
Since the model y equals kx has only one unknown, substituting a single known pair and solving gives k. The whole relationship follows from one data point.
This is a one-step equation of exactly the kind Lesson 3.1 solved.
Figure (svg): Finding the constant of variation from one pair of values
McDougal Littell Algebra 1: Concepts and Skills, Ch. 4 Graphing Linear Equations and Functions — Lesson 4.6 Direct Variation §4.6, pp. 236-236 — Example 1, Write a Direct Variation Model
Picture it
Substitute, then divide.
Figure (svg): Finding the constant of variation from one pair of values
The unknown here is the constant rather than a variable, which is a small shift in perspective. Once k is found it stops being unknown and the equation becomes usable.
Worked example
This is Example 1 from the textbook, both parts.
\[ x \text{ and } y \text{ vary directly, and } y = 20 \text{ when } x = 5. \text{ Write the equation, then find } y \text{ when } x = 12. \]
Write the model
Why: Direct variation means y equals kx.
\[ y = k x \]
Substitute the known pair
Why: Twenty for y and five for x.
\[ 20 = k(5) \]
Solve for k
Why: Divide each side by five.
\[ k = 4 \]
Use the completed model
Why: Substitute twelve for x in y equals 4x.
\[ y = 48 \]
Figure (svg): Finding the constant of variation from one pair of values
\[ y = 4x \qquad y(12) = 48 \]
Verify: check the ratio of the new pair
Why: Forty-eight divided by twelve is four, the same constant as twenty over five. Every pair the model produces has that ratio, so checking it confirms the model rather than the arithmetic alone.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 4 Graphing Linear Equations and Functions — Lesson 4.6 Direct Variation §4.6, pp. 236-236
Translation
Divide y by x to find the constant.
Match the pairs
Why: In each case the constant is the y-value divided by the x-value. The third and fourth are worth comparing: a k of twelve and a y of twelve are different things, and reading the model back at x equal to one distinguishes them.
Worked example
Guided Practice 1 to 3. Same three steps each time.
\[ \text{Write the model for } \; (2, 6), \quad (3, 21), \quad (8, 96). \]
Take the first pair
Why: Six equals k times two, so k is three.
\[ y = 3 x \]
Take the second
Why: Twenty-one equals k times three, so k is seven.
\[ y = 7 x \]
Take the third
Why: Ninety-six equals k times eight, so k is twelve.
\[ y = 12 x \]
Notice the pattern
Why: In every case k is simply y divided by x.
\[ k = \frac{y}{x} \]
Figure (svg): The solution to Worked example three from guided practice shown as a ladder of expressions, one row per algebraic move
\[ y = 3x, \quad y = 7x, \quad y = 12x \]
Verify: confirm the shortcut against the algebra
Why: Solving k times x equals y for k gives k equal to y over x, which is the ratio from the definition. The substitution and the division are the same computation, so the shortcut is safe — though writing the model first keeps the meaning visible.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 4 Graphing Linear Equations and Functions — Lesson 4.6 Direct Variation §4.6, pp. 236-236
Error analysis
The student was told that y varies directly with x, and that y is 20 when x is 5.
Annotate
On: \( \begin{aligned} y &= kx \\ 20 &= k(5) \\ k &= 20 \times 5 \\ k &= 100 \\ y &= 100x \end{aligned} \)
Every direct variation model can be tested against the pair that produced it. That check costs one substitution and catches the whole class of undoing errors.
Faded example
The substitution is done. Finish it.
Fill in the blanks
20 = k(5) \;\Longrightarrow\; k = \dfrac54} = 4 \;\Longrightarrow\; y = ___x
Why: Dividing both sides by five isolates k, giving four, and that constant goes straight into the model. Substituting the original pair back in gives twenty, which confirms the whole thing in one line.
Prediction
The quantities vary directly.
Predict first
If y varies directly with x and x is tripled, what happens to y?
Correct: It is tripled.
\[ y = kx \;\Longrightarrow\; k(3x) = 3(kx) = 3y \]
Why: In y equals kx, replacing x by three x gives three times k x, which is three times the old y. The constant k stays fixed and the output scales with the input, which is the whole practical content of direct variation — and it holds whatever k happens to be, so no data is needed to answer.
Socratic
Most relationships need more than one observation to pin down.
Discussion prompt
Explain why a single pair of values determines a direct variation model completely, and say how many points would be needed for a line that is not through the origin.
Hint: Count the unknowns in each equation.
Answer:
The model y equals kx has exactly one unknown, so one equation determines it. Substituting the pair gives a one-step equation, and solving it fixes k for good — the assumption that the relationship is direct variation is doing the work that a second data point would otherwise have to do.
A general line has the form y equals mx plus b, with two unknowns, so it needs two points. That is the same count as the two points needed to draw a line, and Lesson 5.3 will use exactly this idea to find the equation of a line from two of its points.
Section
Section 3
Concept
Because the origin is always a solution, graphing a direct variation model needs only one computed point. Plot the origin, find a second point by substituting any convenient value of x, and draw the line.
The value x equal to one is usually the most convenient, since it makes y equal to k.
Figure (svg): The graph of y equals 2x, a line through the origin
McDougal Littell Algebra 1: Concepts and Skills, Ch. 4 Graphing Linear Equations and Functions — Lesson 4.6 Direct Variation §4.6, pp. 237-237 — Example 2, Graph a Direct Variation Model
Picture it
Two points, one of them free.
Figure (svg): The graph of y equals 2x, a line through the origin
Compare this with Lesson 4.4, where the intercept method failed exactly for lines through the origin because both intercepts gave the same point. Here the missing second point is supplied by one substitution.
Worked example
This is Example 2 from the textbook.
\[ \text{Graph the equation } \; y = 2x. \]
Plot the origin
Why: It is a solution of every direct variation equation.
\[ (0, 0) \]
Choose a value for x
Why: One is convenient.
\[ x = 1 \]
Substitute to find y
Why: Two times one is two.
\[ y = 2 \]
Plot and draw
Why: Plot (1, 2) and draw the line through both points.
Figure (svg): The graph of y equals 2x, a line through the origin
\[ \text{the line through } (0, 0) \text{ and } (1, 2) \]
Verify: check a third point
Why: At x equal to negative one the model gives y equal to negative two, and the point (-1, -2) does lie on the drawn line. A direct variation graph extends into the third quadrant as well as the first, which the two plotted points alone would not have shown.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 4 Graphing Linear Equations and Functions — Lesson 4.6 Direct Variation §4.6, pp. 237-237
Elimination
The equation is y equals 3x.
Eliminate the wrong options
Which point is NOT on the graph?
Survives elimination: A
Why: The point (3, 1) has a ratio of one third rather than three, so it fails the model — the coordinates have been swapped. Option D is worth noticing: a direct variation graph passes through the third quadrant as well as the first whenever k is positive, so negative coordinates are perfectly ordinary here.
Worked example
Guided Practice 5. The method does not change.
\[ \text{Graph } \; y = -2x. \]
Plot the origin
Why: As always.
\[ (0, 0) \]
Substitute x equal to 1
Why: Negative two times one is negative two.
\[ (1, -2) \]
Plot and draw
Why: The line falls from left to right.
Read the sign
Why: A negative constant of variation gives a falling line.
\[ k < 0 \]
Figure (svg): Two direct variation lines, one with a positive constant and one with a negative constant
\[ \text{the line through } (0, 0) \text{ and } (1, -2) \]
Verify: check the ratio at a second point
Why: At x equal to negative three the model gives six, and six over negative three is negative two — the same constant. The ratio is negative for every pair, which is exactly what a falling line through the origin looks like.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 4 Graphing Linear Equations and Functions — Lesson 4.6 Direct Variation §4.6, pp. 237-237
Trap
\[ y = 2x \quad \text{graph using intercepts} \]
Set y to zero to get the x-intercept, then x to zero for the y-intercept
Why: The quick graph from Lesson 4.4 is a reliable habit.
\[ x\text{-intercept } 0, \quad y\text{-intercept } 0 \]
Both intercepts give the same point, the origin, and one point does not determine a line.
Plot the origin, then substitute any non-zero value of x
Why: One substitution supplies the second point the intercepts could not.
A direct variation model is the case where the intercept method breaks down, and knowing why is more useful than remembering the exception.
Faded example
The origin is already plotted.
Fill in the blanks
y = 2x \text1 x = 1: \quad y = 2(2) = 2, \text___ (1, ___)
Why: Choosing x equal to one makes the arithmetic trivial and gives a y-value equal to k itself. That is worth noticing: the point (1, k) is on every direct variation graph, so it can be written down without any computation at all.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Table (4.2) | Intercepts (4.4) | Direct variation (4.6) | |
|---|---|---|---|
| Points computed | about five | two | one |
| Fails when | outputs are fractional | the line passes through the origin | the line misses the origin |
| Best for | function form | standard form | y = kx |
Each method has the case it was designed for and the case it cannot handle. Noticing which form the equation arrives in is what tells you which one to reach for.
Socratic
The graph passes through two points you can name without computing.
Discussion prompt
Explain why every direct variation graph contains both the origin and the point whose x-coordinate is one and y-coordinate is k. Then say what those two points let you conclude about the slope.
Hint: Substitute one into the model.
Answer:
Substituting x equal to one into y equals kx gives y equal to k, so (1, k) is always a solution. Together with the origin, that gives two named points on the graph before any work is done.
The slope between (0, 0) and (1, k) is k minus zero over one minus zero, which is k. So the constant of variation is the slope of the line — which is exactly what the textbook's summary on page 238 states, and it means everything you learned about slope in Lesson 4.5 applies directly to k.
Section
Section 4
Concept
The graph of y equals kx is a line through the origin whose slope is k. A positive k gives a rising line and a negative k a falling one, exactly as in Lesson 4.5.
So the constant of variation can be read off a graph as easily as it can be computed from a pair.
Figure (svg): Two direct variation lines, one with a positive constant and one with a negative constant
McDougal Littell Algebra 1: Concepts and Skills, Ch. 4 Graphing Linear Equations and Functions — Lesson 4.6 Direct Variation §4.6, pp. 238-238 — the Properties of Graphs of Direct Variation Models summary
Picture it
Both through the origin, tilting opposite ways.
Figure (svg): Two direct variation lines, one with a positive constant and one with a negative constant
Everything from Lesson 4.5 carries over. A larger size of k means a steeper line, and the sign decides the direction.
Worked example
The slope and the constant of variation are the same number.
\[ \text{A direct variation graph passes through } (0, 0) \text{ and } (4, 10). \text{ Find } k. \]
Compute the slope
Why: Ten minus zero over four minus zero.
\[ \frac{10}{4} \]
Simplify
Why: Five halves, or two and a half.
\[ \frac{5}{2} \]
Identify it as k
Why: The slope of a direct variation graph is the constant of variation.
\[ k = \frac{5}{2} \]
Write the model
Why: Substitute k into y equals kx.
\[ y = (\frac{5}{2}) x \]
Figure (svg): The solution to Worked example read k off a graph shown as a ladder of expressions, one row per algebraic move
\[ k = \dfrac{10}{4} = \dfrac{5}{2} \;\Longrightarrow\; y = \tfrac{5}{2}x \]
Verify: check the ratio of the given point
Why: Ten divided by four is five halves, the same as the slope. That is not a coincidence: computing the slope from the origin to any point is the same arithmetic as computing the ratio of that point's coordinates, since the subtractions are both from zero.
Sorting
Read the size and the sign of each constant.
Sort into buckets
Sort each direct variation model by the line it graphs.
A slope of one is the dividing line between steep and gentle, since it makes the rise and the run equal. That landmark comes straight from the ramp investigation in Lesson 4.5.
Worked example
The size of k measures steepness and its sign the direction.
\[ \text{Compare the graphs of } \; y = 3x, \quad y = \tfrac{1}{3}x \; \text{ and } \; y = -3x. \]
Take k equal to 3
Why: Positive and larger than one, so a steep rising line.
Take k equal to one third
Why: Positive and less than one, so a gentle rising line.
Take k equal to negative 3
Why: Negative, so a falling line, as steep as the first.
Note what they share
Why: All three pass through the origin.
Figure (svg): Two direct variation lines, one with a positive constant and one with a negative constant
\[ k = 3, \; \tfrac{1}{3}, \; -3 \]
Verify: check the steepness claim against Lesson 4.5
Why: The first and third have slopes of the same size and opposite signs, so they are equally steep and tilt opposite ways — mirror images in the horizontal axis. Comparing steepness means comparing how far the slopes are from zero, which is why three and negative three are equally steep.
Trap
A line passes through (0, 3) and (1, 7). A student computes the slope as 4 and writes y = 4x.
Take the slope as the constant of variation
Why: The rule that k is the slope was applied without checking that the relationship is direct variation.
At x equal to zero the model gives zero, but the line is at three. The equation is y equals 4x plus 3, and there is no constant of variation because this is not direct variation at all.
Check that the line passes through the origin before calling its slope k
Why: The property k equals the slope belongs to direct variation models specifically.
The origin test is quick and it is the difference between a valid model and one that is wrong at every point.
Faded example
The graph passes through the origin and one other point.
Fill in the blanks
\text0 (0, 0) \text2 (4, 10): \quad k = \dfrac______}} = \dfrac______ = \dfrac______}
Why: Both subtractions are from zero, so the slope computation reduces to the ratio of the second point's coordinates. That is why the slope of a direct variation graph and its constant of variation are the same number rather than two related ones.
Prediction
Two models, y equals 2x and y equals 4x.
Predict first
How does the graph of y equals 4x compare with that of y equals 2x?
Correct: Twice as steep, through the same origin.
\[ y = 2x: \; (1, 2) \qquad y = 4x: \; (1, 4) \]
Why: Doubling k doubles the slope, so the line climbs twice as fast, and both models still contain the origin since k times zero is zero regardless. Nothing shifts vertically, because a direct variation model has no constant term to shift it — that is the difference between changing k here and changing the constant in y equals mx plus b.
Socratic
It is defined as a ratio and behaves as a slope.
Discussion prompt
Explain why the constant of variation, defined as the ratio y over x, turns out to be the slope of the graph. Then say why this fails for a line that does not pass through the origin.
Hint: Compute the slope between the origin and any point on the line.
Answer:
Slope between the origin and a point (x, y) is y minus zero over x minus zero, which is just y over x — the ratio itself. So for a line through the origin the ratio of any point's coordinates and the slope are literally the same computation, and the two definitions of k coincide.
For a line through (0, 3), the slope between (0, 3) and (1, 7) is four, while the ratio of the second point's coordinates is seven. The subtractions are no longer from zero, so the two computations come apart. This is why the property is stated for direct variation models specifically, and why checking the origin first matters.
Section
Section 5
Concept
Real data rarely gives identical ratios. When the ratios cluster around one value, a direct variation model with that constant is a reasonable approximation, even though no data point fits it exactly.
The textbook's alligator ratios range from 0.85 to 0.94 and 0.90 is chosen.
Figure (svg): Alligator tail and body lengths with their ratios all near the same value
McDougal Littell Algebra 1: Concepts and Skills, Ch. 4 Graphing Linear Equations and Functions — Lesson 4.6 Direct Variation §4.6, pp. 238-238 — Example 4, Use a Direct Variation Model, with the alligator data
Picture it
Eight ratios, none of them equal, all of them close.
Figure (svg): Alligator tail and body lengths with their ratios all near the same value
The data come from the St. Augustine Alligator Farm and cover animals from two to over fifty years old. That the ratio barely moves across that range is a real biological fact, and it is what makes the model worth having.
Worked example
This is Example 4 from the textbook.
\[ \text{Tail and body lengths give ratios } 0.94, \; 0.85, \; 0.90, \; 0.86, \; 0.93, \; 0.93, \; 0.93, \; 0.89. \text{ Write a model.} \]
Compute every ratio
Why: Divide each tail length by its body length.
Look at the spread
Why: They run from 0.85 to 0.94, a narrow band.
Choose a representative value
Why: 0.90 sits near the middle of the band.
\[ k = 0.90 \]
State the model
Why: Tail length is about nine tenths of body length.
\[ T = 0.90 B \]
Figure (svg): Alligator tail and body lengths with their ratios all near the same value
\[ T \approx 0.90B \]
Verify: test the model against one of the original animals
Why: For a body length of 4.28 the model predicts a tail of 3.85, and the measured value was 3.99 — close but not exact. That gap is expected and is the price of a model: it describes the group well while fitting no individual perfectly.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 4 Graphing Linear Equations and Functions — Lesson 4.6 Direct Variation §4.6, pp. 238-238
Elimination
The ratios are 0.94, 0.85, 0.90, 0.86, 0.93, 0.93, 0.93 and 0.89.
Eliminate the wrong options
Which constant best represents them?
Survives elimination: A
Why: A representative constant should sit among the ratios rather than at an extreme, and 0.90 is near the middle of the band. Choosing an end value is a common instinct because those numbers are the ones that stand out, and it makes the model systematically wrong in one direction.
Worked example
Guided Practice 7. The model can be run in either direction.
\[ \text{Using } T = 0.90B, \text{ estimate the body length of an alligator whose tail is } 4.5 \text{ ft.} \]
Substitute the known tail length
Why: Four and a half for T.
\[ 4.5 = 0.90 B \]
Solve for B
Why: Divide both sides by nine tenths.
\[ B = 5 \]
State the estimate
Why: The body is about five feet long.
\[ \text{about } 5 \text{ft} \]
Say how confident to be
Why: The model is approximate, so the answer is an estimate rather than a measurement.
Figure (svg): The solution to Worked example use the fitted model backwards shown as a ladder of expressions, one row per algebraic move
\[ B = \dfrac{4.5}{0.90} = 5 \text{ ft} \]
Verify: check the estimate against the original data
Why: The measured animals with tails near 4.67 had bodies near 5.04, so a five-foot body for a 4.5-foot tail sits comfortably inside the observed range. Checking a prediction against the data it came from is what separates a usable estimate from a number produced by arithmetic.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 4 Graphing Linear Equations and Functions — Lesson 4.6 Direct Variation §4.6, pp. 238-238
Trap
\[ T = 0.90B \text{ at } T = 4.5: \; B = 5 \]
Report that the alligator's body is 5 feet long
Why: The arithmetic is exact, so the answer feels exact too.
The model was fitted to ratios ranging from 0.85 to 0.94. Using either end instead gives a body between 4.8 and 5.3 feet, so five feet is the middle of a range rather than a measurement.
The body is about 5 feet long, based on an approximate model.
Carry the approximation through to the answer
Why: An estimate built on an approximate constant is itself an estimate, however exact the division was.
Saying about, and knowing roughly how wide the uncertainty is, is part of using a fitted model honestly.
Faded example
A tail of 4.5 feet, and T equals 0.90B.
Fill in the blanks
4.5 = 0.90B \;\Longrightarrow\; B = \dfrac0.905} = ___
Why: Dividing by the constant undoes the multiplication, giving a body length of about five feet. Because the constant was chosen rather than measured, the answer is an estimate, and it should be reported with an about in front of it.
Hypothesis
Predict before you decide.
Predict first
Which of these would most weaken your confidence in a fitted direct variation model?
Correct: The ratios drift steadily upward as x increases.
A drifting ratio suggests a curve rather than a line, which is what a relationship like y equals kx squared would produce.
Why: A steady drift means the ratio is not constant but is itself changing with x, so the relationship is systematically something other than direct variation and no single k will do. Random scatter within a narrow band is exactly what measurement noise looks like and is expected. A small sample weakens confidence mildly, and no point matching exactly is normal for any fitted model. Direction in the residuals is the warning sign; size alone is not.
Socratic
The model is wrong about every alligator measured.
Discussion prompt
Explain what a fitted direct variation model gives you that the raw table of eight measurements does not. Then say what you would do if the ratios ranged from 0.4 to 1.6 instead.
Hint: Ask what question the table cannot answer.
Answer:
The table only answers questions about the eight animals measured. The model answers questions about any alligator: given a tail of 4.5 feet, no row of the table applies, but the model gives an estimate. Trading exactness on the measured cases for coverage of the unmeasured ones is what a model is for.
Ratios spread from 0.4 to 1.6 would not cluster at all, and any chosen constant would be badly wrong for much of the data. The honest response is to say the relationship is not direct variation and look for a different kind of model — or to conclude that the two quantities are not closely related. A model should be reported with the evidence for it, not just its equation.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| y = kx | y = mx + b, with b not zero | |
|---|---|---|
| Passes through the origin | always | never |
| The ratio y over x | is constant | changes with x |
| Points needed to determine it | one | two |
Direct variation is the special case with the constant term removed, and every difference in this table comes from that one missing number.
Pattern
Whether you are testing for direct variation, fitting a model or graphing one, the same five moves cover it.
Step four is one substitution and it catches the whole class of errors where the division was done the wrong way round.
OpenStax Elementary Algebra 2e, §8.9 Use Direct and Inverse Variation §8.9
Check
Substitute the pair and solve for the constant.
Check your understanding
y varies directly with x, and y is 54 when x is 6. What is the constant of variation?
Answer: A
Why: Fifty-four equals k times six, so dividing both sides by six gives k equal to nine and the model y equals 9x. Substituting six back in gives fifty-four, which confirms it.
Check
Direct variation requires more than both quantities rising.
Check your understanding
Which of these is NOT direct variation?
Answer: A
Why: The boarding charge means the fare is not zero when the distance is zero, so the graph misses the origin and the ratio of fare to miles keeps changing. One mile costs five dollars and two cost seven, giving ratios of five and three and a half.
Check
The constant is the slope.
Check your understanding
A direct variation graph passes through (0, 0) and (2, -6). What is the model?
Answer: A
Why: The slope is negative six minus zero over two minus zero, which is negative three, and that slope is the constant of variation. The line falls from left to right, which is what a negative constant produces.
Real world
This is Example 3 from the textbook. Five gold bars stored at Fort Knox weigh 137.5 pounds.
Discussion prompt
Write a direct variation model relating weight W to the number of bars n, then use it to find the weight of 36 bars. Say why direct variation is the right kind of model here.
Hint: Find the constant first, then substitute.
Answer:
\[ 137.5 = k(5) \;\Longrightarrow\; k = 27.5 \;\Longrightarrow\; W = 27.5n \]
\[ W = 27.5(36) = 990 \text{ pounds} \]
Direct variation is right because the bars are standard mint bars of almost pure gold, so each one weighs the same 27.5 pounds and no bars means no weight. Both conditions matter: identical units give the constant ratio, and the absence of any packaging or container weight is what puts the graph through the origin.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Does the equation y equals 5x plus 1 represent direct variation?
Correct: No, since its graph does not pass through the origin.
\[ x = 1: \; \dfrac{6}{1} = 6 \qquad x = 2: \; \dfrac{11}{2} = 5.5 \]
Why: At x equal to zero the equation gives y equal to one rather than zero, so the origin is not on the graph and the ratio of y to x is not constant — it is six at x equal to one and five and a half at x equal to two. The first and third options each name a true property of this equation that is not the definition, which is exactly how non-examples slip through. Every direct variation graph is a line, and most lines are not direct variation.
Explain it
They can scale a recipe and have never seen the phrase direct variation.
Discussion prompt
In no more than four sentences, explain what direct variation means using something they already do, and give them the quickest test for whether a relationship has it.
Hint: Doubling a recipe is the idea.
Answer:
A usable answer: two quantities vary directly when doubling one doubles the other, tripling one triples the other, and so on — like doubling every ingredient in a recipe. That happens exactly when dividing one by the other always gives the same number, which is the constant of variation.
The quickest test is to ask what happens when one quantity is zero. If the other is zero too, direct variation is possible; if there is a starting fee, a boarding charge or a fixed amount left over, it is not. That single question rules out most of the non-examples without any arithmetic.
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 recognition question is fixed by asking what happens when one quantity is zero. The algebra is fixed by substituting the original pair back into the finished model, which catches a multiplication in one line. Graphing is fixed by remembering that (1, k) is always on the line, so the second point needs no computation. Fitting is fixed by computing every ratio, checking they cluster, and choosing a value near the middle rather than at an end. 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 one pair of values and use it to find a constant of variation, showing the substitution and the division, then write the completed model. Underneath, make a table of four more pairs from that model and write the ratio beside each row, checking that all four match your constant. To the right, draw a coordinate plane and graph the model, marking the origin and the point whose x-coordinate is one, and writing beside the second point that its height is k itself. In the lower half, write down one relationship that is not direct variation, and beside it show the two different ratios that prove it. Finally, in the margin, write the sentence you would use to test any relationship in one question.
Every ratio in your table should equal the constant you found at the top. If one does not, the substitution that produced it went wrong, since the model cannot produce a pair with a different ratio.
Recap
Five things, and the first is the one that separates direct variation from every other straight line.
| If the question says | Your first move is |
|---|---|
| y varies directly with x | Write y = kx |
| Find the constant of variation | Substitute the pair and divide |
| Graph the model | Plot the origin and the point (1, k) |
| Is this direct variation | Ask whether zero maps to zero |
| Fit a model to this data | Compute every ratio and see if they cluster |
Lesson 4.7 puts the slope together with the y-intercept into one form of the equation. Direct variation turns out to be the special case where that intercept is zero, and the general form graphs just as quickly.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 4 Graphing Linear Equations and Functions — Lesson 4.6 Direct Variation §4.6, pp. 236-242 — everything on these slides traces back here
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