Recognizing a proportional relationship in an equation
An equation ties two quantities together. How can you tell, just by reading it, whether one is always the same multiple of the other?
Where you are 4 in Proportional relationships
When you reach for it
You already have two ways to test whether two quantities are proportional. In a table, every output divided by its input gives the same number, the constant of proportionality. On a graph, the points lie on a straight line through the origin (Recognizing a proportional relationship in a graph). Both tests need numbers worked out or points plotted first.
An equation often comes before either. How far a car goes on its fuel, how many pages a printer has printed, what a phone plan charges for data: each can be written as a rule connecting two quantities, such as
for the miles d a car goes on g gallons of fuel. This lesson is about reading such a rule and deciding, without a table or a graph, whether the relationship is proportional, and if it is, what the constant is.
Reach for it in three situations. When an equation is given and the question is whether it describes a proportional relationship. When a situation is described in words and you want its equation, since every proportional situation has an equation of one simple shape. And when two quantities change together at a steady rate, so their graph is a straight line, and you need to know whether that line passes through the origin: a fixed fee, a starting amount or a warm-up time makes a relationship linear without making it proportional.
The lesson takes two days. Part 1 reads and writes equations of the form y = kx. Part 2 tells a proportional equation apart from one that is only linear, or not linear at all.
The procedure
Part 1: the form y = kx
The definition. A quantity y is proportional to a quantity x when
for one fixed number k that is not 0. The number k is the constant of proportionality. “Fixed” means it does not change when x changes: the same k multiplies every input.
Why this is the same test as the table. Suppose y = kx, and pick any input x other than 0. Dividing both sides by x is allowed, because x is not zero, and gives
So every pair in a table built from the equation has the same ratio, k, which is exactly the table test. The argument also runs backwards: if \displaystyle \frac{y}{x} = k for every pair with x \neq 0, multiplying both sides by x gives y = kx. The equation settles the input 0 as well, which the ratio cannot: putting x = 0 into y = kx gives y = k \cdot 0 = 0. That is the point (0, 0), and the reason the graph passes through the origin. Putting x = 1 gives y = k, the point (1, k): the constant is the output for one unit of input, the unit rate.
Reading the constant. In y = kx the constant is the coefficient of x, the number multiplying it. Sometimes the multiplying is hidden.
- \displaystyle y = \frac{x}{8} is proportional with \displaystyle k = \frac{1}{8}, because dividing by 8 is the same as multiplying by \displaystyle \frac{1}{8}.
- y = x is proportional with k = 1, because x = 1 \cdot x.
The constant carries units: output units per input unit. In d = 28g, d is in miles and g in gallons, so k = 28 miles per gallon.
Rewriting into the form. An equation need not arrive with y alone on the left. You may rewrite it with any move that keeps exactly the same pairs (x, y) true: doing the same thing to both sides (Inverse operations and the properties of equality), such as dividing both sides by the same nonzero number, or rewriting one side with the distributive property or by collecting like terms. A move that keeps the same true pairs cannot change whether the relationship is proportional, so rewrite first and then read. For example,
becomes y = 4x after dividing both sides by 3, so it is proportional with k = 4.
Either quantity can come first. If y = kx with k \neq 0, dividing both sides by k gives \displaystyle x = \frac{1}{k}y. So x is proportional to y as well, with the reciprocal constant. In d = 28g the car uses \displaystyle \frac{1}{28} of a gallon for each mile: \displaystyle g = \frac{1}{28}d. Which of the two equations to write depends on which quantity you want to find.
Writing the equation from a situation. If a situation is proportional, its equation is y = kx and only k is unknown. One known pair of values gives it: \displaystyle k = \frac{y}{x} for that pair, as Example 3 shows.
Worked examples 1 to 4 go with Part 1.
Part 2: proportional or only linear
A constant added on the end. Compare
Each time x goes up by 1, both outputs go up by 2, so both graphs are straight lines, equally steep. Only the first is proportional.
For y = 2x + 3 three checks fail, and each says the same thing in a different way.
- Put x = 0: y = 2 \cdot 0 + 3 = 3, not 0. The line crosses the y-axis at (0, 3) and misses the origin.
- The ratio is not fixed: x = 1 gives y = 5, ratio 5, while x = 3 gives y = 9, ratio 3.
- Doubling the input does not double the output: x = 1 gives 5, but x = 2 gives 7, not 10.
In general, an equation y = kx + b with b \neq 0 is linear but not proportional, because at x = 0 it gives y = b, and a proportional relationship must give 0 there. The number b is where the line meets the y-axis, its intercept. The proportional relationships are exactly the lines with b = 0 and k \neq 0.
Not a line at all. Some equations have no constant added on and still are not proportional, because x is not simply multiplied by a fixed number.
- y = x^2 gives y = 0 at x = 0, so its graph passes through the origin, but the ratio \displaystyle \frac{y}{x} = x changes with x. Doubling x multiplies y by 4.
- \displaystyle y = \frac{12}{x} divides by x instead of multiplying. Doubling x halves y: this is an inversely proportional relationship, a different thing.
So passing through the origin is needed, but it is not enough. What decides is the shape of the equation.
The test, all together. Rewrite the equation, with moves that keep the same pairs true, until y is alone on one side and the other side is simplified: brackets multiplied out, like terms collected. The relationship is proportional exactly when the other side is then a single term, a fixed nonzero number times x, with nothing added. The simplifying matters, because constants can cancel: y = 3(x + 2) - 6 has two constants in it, yet 3(x + 2) - 6 = 3x + 6 - 6 = 3x, so it is proportional.
In a situation. The added b is whatever is already there when the input is 0: a monthly fee charged even when no data is used, fuel already in a tank, a printer’s warm-up time before the first page. A quantity counted from zero that grows at a steady rate is proportional to what drives it. The same quantity with a starting amount added is only linear.
Worked examples 5 to 8 go with Part 2.
Worked examples
Reading the constant
Decide whether each equation describes a proportional relationship. If it does, give the constant of proportionality.
(a) y = 1.2x \qquad (b) \displaystyle y = \frac{x}{8} \qquad (c) y = x \qquad (d) y = 6x - 1
Solution(a) Proportional, with k = 1.2. The right side is one term, the fixed number 1.2 times x, with nothing added.
(b) Proportional, with \displaystyle k = \frac{1}{8}. Dividing by 8 means multiplying by \displaystyle \frac{1}{8}, so the equation is \displaystyle y = \frac{1}{8}x.
(c) Proportional, with k = 1. The equation is y = 1 \cdot x: every output equals its input, and every ratio \displaystyle \frac{y}{x} is 1.
(d) Not proportional. The -1 on the end means x = 0 gives y = -1, not 0. The ratio agrees: x = 1 gives y = 5, a ratio of 5, and x = 2 gives y = 11, a ratio of \displaystyle \frac{11}{2}.
Rewriting before reading
Each equation describes a proportional relationship. Rewrite it in the form y = kx and give k.
(a) 4y = 10x \qquad (b) y = 3(2x) - x \qquad (c) x = 0.25y \qquad (d) \displaystyle y = \frac{2x}{5} + \frac{x}{5}
Solution(a) Divide both sides by 4, which keeps the same pairs true because 4 is not zero: \displaystyle y = \frac{10}{4}x = 2.5x. So k = 2.5.
(b) By the associative property of multiplication, 3(2x) = (3 \cdot 2)x = 6x. Then 6x - x = (6 - 1)x = 5x by the distributive property, so y = 5x and k = 5.
(c) Here y is on the right. Multiply both sides by 4: 4x = y, that is, y = 4x, so k = 4. Read the other way round, the original equation already says x is proportional to y with constant 0.25, and 4 is the reciprocal of 0.25.
(d) The two terms are like terms, so they combine: \displaystyle \frac{2x}{5} + \frac{x}{5} = \frac{3x}{5}. So \displaystyle y = \frac{3}{5}x and \displaystyle k = \frac{3}{5}.
Writing the equation for a car’s fuel
On a long drive at a steady speed, a car goes 168 miles on 6 gallons of fuel, and the distance is proportional to the fuel used.
(a) Write an equation for the distance d in miles in terms of the fuel g in gallons. (b) Write an equation for g in terms of d. (c) How far does the car go on 9.5 gallons? (d) How much fuel does it need for 434 miles?
Solution(a) Because the relationship is proportional, d = kg for some fixed k, and the one known pair gives it: \displaystyle k = \frac{168}{6} = 28. So d = 28g, and the constant is 28 miles per gallon.
(b) Divide both sides of d = 28g by 28: \displaystyle g = \frac{d}{28}, that is, \displaystyle g = \frac{1}{28}d. The car uses \displaystyle \frac{1}{28} of a gallon for each mile.
(c) Put g = 9.5 into the equation from (a): d = 28 \times 9.5 = 266 miles.
(d) Put d = 434 into the equation from (b): \displaystyle g = \frac{434}{28} = 15.5 gallons.
Check (d) with (a): 28 \times 15.5 = 434. Each equation answers its own question with one multiplication or one division, which is the reason for writing both.
What the numbers in a printer’s equation mean
An office printer prints at a steady rate. The number of pages p it has printed t minutes after it starts printing is p = 22t.
(a) Explain why p is proportional to t, and say what the constant means. (b) What do the points (0, 0) and (1, 22) on its graph say about the printer? (c) How long does it take to print 330 pages? (d) A second printer prints 120 pages in 4 minutes, also at a steady rate. Write its equation and say which printer is faster.
Solution(a) The equation has the form y = kx, with p in place of y, t in place of x, and k = 22. The constant has units of pages per minute: the printer prints 22 pages every minute.
(b) The point (0, 0) says that at the moment printing starts, no pages have been printed. The point (1, 22) says that after one minute there are 22 pages: it is the unit rate, k, read off the graph.
(c) Set p = 330: 330 = 22t. Dividing both sides by 22 gives t = 15 minutes.
(d) The second printer’s constant is \displaystyle \frac{120}{4} = 30 pages per minute, so its equation is p = 30t. Its constant is larger, so it is faster: in the 15 minutes the first printer needs for 330 pages, the second prints 30 \times 15 = 450.
Fuel used and fuel left
A car starts with a full tank of 15 gallons and goes 30 miles on each gallon. After d miles it has used u gallons and has f gallons left, where
Which of u and f is proportional to d? Support the answer with values at d = 60, 150 and 300, and with the graph.
SolutionThe equation \displaystyle u = \frac{d}{30} = \frac{1}{30}d has a single term, a fixed number times d, so u is proportional to d with \displaystyle k = \frac{1}{30} gallon per mile. The equation for f has 15 in it as well as the d term, so f is not proportional: at d = 0 it gives f = 15, a full tank, not 0.
The values agree.
| d (miles) | 0 | 60 | 150 | 300 |
|---|---|---|---|---|
| u (gallons used) | 0 | 2 | 5 | 10 |
| f (gallons left) | 15 | 13 | 10 | 5 |
Every ratio \displaystyle \frac{u}{d} is the same: \displaystyle \frac{2}{60} = \frac{5}{150} = \frac{10}{300} = \frac{1}{30}. The ratios \displaystyle \frac{f}{d} are \displaystyle \frac{13}{60}, \displaystyle \frac{10}{150} = \frac{1}{15} and \displaystyle \frac{5}{300} = \frac{1}{60}, all different.
On the graph both are straight lines, because each changes by the same \displaystyle \frac{1}{30} of a gallon for every mile, one going up and the other going down. Only the line for u passes through the origin. The two lines cross at d = 225, where u = f = 7.5: half the tank used and half left.
Two phone plans
Two plans charge for mobile data. With plan A the monthly cost in dollars for g gigabytes of data is C = 6g. With plan B it is C = 20 + 2g.
(a) Which plan’s cost is proportional to the data used? (b) Find each plan’s cost for 3 gigabytes and for 6 gigabytes, and say what doubling the data does to each cost. (c) Which plan is cheaper at 3 gigabytes, and which at 6?
Solution(a) Plan A: C = 6g is a fixed number times g, so the cost is proportional to the data, with constant 6 dollars per gigabyte. Plan B: the 20 is a fixed monthly charge, paid even when g = 0, so the cost is not proportional to the data, although it still rises steadily, by 2 dollars a gigabyte.
(b) Plan A costs 6 \times 3 = 18 dollars and 6 \times 6 = 36 dollars. Doubling the data doubles the cost, as it must when C = 6g: 6(2g) = 2(6g). Plan B costs 20 + 2 \times 3 = 26 dollars and 20 + 2 \times 6 = 32 dollars. Doubling the data does not double the cost, because the 20 is paid once, not once per gigabyte.
(c) At 3 gigabytes plan A is cheaper, since 18 < 26. At 6 gigabytes plan B is cheaper, since 32 < 36. The proportional plan suits light use; the fixed fee with a lower rate pays off for heavier use.
Simplify before deciding
Decide whether each equation describes a proportional relationship. Give k where it does.
(a) y = 4(x + 5) - 20 \qquad (b) y = 3(x - 2) + 5 \qquad (c) y = x(x + 1) - x \qquad (d) \displaystyle y = \frac{8}{x}
Solution(a) Distribute: 4(x + 5) = 4x + 20, so y = 4x + 20 - 20 = 4x. Proportional, with k = 4. The two constants cancel, and only simplifying shows it.
(b) Distribute: 3(x - 2) = 3x - 6, so y = 3x - 6 + 5 = 3x - 1. Not proportional: at x = 0, y = -1.
(c) Distribute: x(x + 1) = x^2 + x, so y = x^2 + x - x = x^2. The graph passes through the origin, since x = 0 gives y = 0, but the relationship is not proportional: the ratio \displaystyle \frac{y}{x} = x changes with x. At x = 1, y = 1; at x = 2, y = 4.
(d) Not proportional. Here 8 is divided by x; x is not multiplied by a fixed number. At x = 1, y = 8, and at x = 2, y = 4: doubling the input halves the output. The equation does not even give a value at x = 0.
A printer’s warm-up time
A printer needs 30 seconds to warm up and then prints 24 pages a minute. The total time T in minutes to print a job of n pages, warm-up included, is
(a) Is T proportional to n? (b) Someone reasons that 48 pages take twice as long as 24 pages. Find both times and test this. (c) Which quantity in this situation is proportional to n? (d) How long does a 120-page job take?
Solution(a) No. The 0.5 is the warm-up, the time spent before the first page, and it is added on, not multiplied by n. By the equation, a job of 0 pages would take 0.5 minutes, not 0.
(b) For 24 pages, \displaystyle T = 0.5 + \frac{24}{24} = 1.5 minutes. For 48 pages, \displaystyle T = 0.5 + \frac{48}{24} = 2.5 minutes. That is not twice 1.5, which would be 3. Doubling works only for a proportional relationship, and the warm-up is paid once per job, not once per page.
(c) The printing time after the warm-up, \displaystyle P = \frac{n}{24} minutes, is proportional to n, with constant \displaystyle \frac{1}{24} of a minute per page. Doubling the pages doubles it: 1 minute for 24 pages, 2 minutes for 48.
(d) \displaystyle T = 0.5 + \frac{120}{24} = 0.5 + 5 = 5.5 minutes.
Boundaries
- The constant must be fixed. y = x \cdot x is “a number times x”, but the number is x itself and changes with every input. The k in y = kx is one number for every input.
- The constant must not be zero. The equation y = 0x, that is y = 0 for every x, is excluded. Every output is 0, so the output says nothing about the input, and the relationship cannot be turned round: there is no \displaystyle x = \frac{1}{k}y when k = 0.
- A negative constant. As an equation, y = -2x is proportional: every ratio \displaystyle \frac{y}{x} is -2, and doubling x doubles y. Amounts such as miles, gallons and pages are never negative, so in situations like the ones in this lesson the constant is positive.
- The ratio form leaves out 0. The equation \displaystyle \frac{y}{x} = 3 says y = 3x for every x except 0, where the left side has no value. Its graph is the line y = 3x with the origin missing. The relationship it describes is proportional, but to include the starting point (0, 0), write it as y = 3x.
- A model holds only where the situation does. p = 22t describes the printer while it has paper and keeps its rate, and d = 28g describes the car only at a steady speed and only up to the fuel its tank holds. Inputs are never negative in either. Outside that range the equation still produces numbers, but they no longer describe anything.
Where this goes next
Reading y = kx is the start of reading every straight-line equation. In Linear relationships and slope-intercept form, y = mx + b is studied as a whole family: m is the steepness, which for a proportional relationship is the constant k, and b is where the line meets the y-axis. The proportional relationships are exactly the members with b = 0, and Part 2 of this lesson is the first place that difference matters.
In algebra the same idea returns as Direct variation, “y varies directly with x”, where the constant may be any nonzero number, negative ones included. Its partner, inverse proportion, has the equation \displaystyle y = \frac{k}{x}, the shape met in Example 7(d). And once an equation y = kx is known, any missing value comes straight from it, which is the equation’s way of doing what solving a proportion does with two equal ratios.
Exercises
Part 1: reading the constant
Decide whether each equation describes a proportional relationship. If it does, give the constant of proportionality.
- y = 7x
- \displaystyle y = \frac{x}{5}
- y = 0.4x
- y = x + 7
- \displaystyle y = -\frac{3}{2}x
Part 1: rewriting first
Each equation describes a proportional relationship. Rewrite it in the form y = kx and give k.
- 5y = 2x
- y = 2x + 6x
- x = 0.5y
- \displaystyle y = \frac{1}{2}(10x)
- \displaystyle \frac{y}{3} = 4x
Part 1: from a situation to an equation
- A printer prints 135 pages in 5 minutes at a steady rate. (a) Write an equation for the pages p printed in t minutes. (b) How many pages does it print in 12 minutes? (c) How long does it take to print 486 pages?
- A phone uses data at a steady rate, 4.2 gigabytes in 7 days. (a) Write an equation for the data D, in gigabytes, used in n days. (b) How much data does it use in 30 days? (c) After how many days has it used 9 gigabytes?
- A car goes 372 miles on 12 gallons of fuel, and the distance is proportional to the fuel. (a) Write an equation for the distance d in terms of the fuel g, and another for g in terms of d. (b) How much fuel does it use for 155 miles?
- The cost in dollars of running an electric heater for h hours is c = 0.18h. (a) What does the 0.18 mean? (b) What does running it for 5 hours cost? (c) For how many hours does 2.70 dollars run it?
Part 2: proportional or not
Decide whether each equation describes a proportional relationship. If it does, give k; if it does not, give a reason.
- y = 4x + 1
- y = 5(x - 2) + 10
- y = x^2 + x
- \displaystyle y = \frac{20}{x}
- y - 4 = 3x - 4
- y = 2(3x + 1) - 2(x + 1)
- Find the number c that makes y = 4(x - 3) + c a proportional relationship, and give its constant.
Part 2: situations
- A mobile plan costs 10 dollars a month plus 3 dollars for each gigabyte of data. (a) Write the monthly cost C for g gigabytes. (b) Is C proportional to g? Find the cost for 2 and for 4 gigabytes to support your answer. (c) Which part of the cost is proportional to g?
- A phone plan gives 20 gigabytes of data each month, and a phone uses 0.8 gigabytes a day. After n days of the month, the data used is U = 0.8n and the data left is L = 20 - 0.8n. (a) Which of U and L is proportional to n? (b) Find U and L after 10 days.
- Car A goes d = 26g miles on g gallons. Car B goes 64, 160 and 256 miles on 2, 5 and 8 gallons. (a) Show that car B’s distance is proportional to its fuel, and write its equation. (b) Which car goes farther on a gallon? (c) How much farther does that car go on 10 gallons?
Reasoning
- Explain, using the input 0, why an equation y = kx + b with b \neq 0 can never describe a proportional relationship.
- Show that if y = kx, then multiplying the input by any number m multiplies the output by m too. Then show that this fails for y = 2x + 3.
- The graphs of y = 3x and y = 3x + 2 are parallel straight lines. Explain, using both the graph and the equation, why only one of them describes a proportional relationship.
- Show that if y is proportional to x with constant k, and z is proportional to y with constant m, then z is proportional to x, and find its constant. Then use it: the printer in Example 4 prints p = 22t pages in t minutes, and each page uses 0.04 grams of toner.
Answers
- Proportional, k = 7. One term, 7 times x, with nothing added.
- Proportional, \displaystyle k = \frac{1}{5}. Dividing by 5 is multiplying by \displaystyle \frac{1}{5}.
- Proportional, k = 0.4.
- Not proportional. The 7 is added on, so x = 0 gives y = 7.
- Proportional, \displaystyle k = -\frac{3}{2}. It is a fixed number times x, and every ratio \displaystyle \frac{y}{x} is \displaystyle -\frac{3}{2}.
- \displaystyle y = \frac{2}{5}x, so \displaystyle k = \frac{2}{5}. Divide both sides by 5.
- y = 8x, so k = 8. Collect like terms: 2x + 6x = (2 + 6)x.
- y = 2x, so k = 2. Multiply both sides by 2.
- y = 5x, so k = 5. By the associative property, \displaystyle \frac{1}{2}(10x) = \left(\frac{1}{2} \cdot 10\right)x.
- y = 12x, so k = 12. Multiply both sides by 3.
- (a) p = 27t, since \displaystyle k = \frac{135}{5} = 27 pages per minute. (b) 27 \times 12 = 324 pages. (c) \displaystyle t = \frac{486}{27} = 18 minutes.
- (a) D = 0.6n, since \displaystyle k = \frac{4.2}{7} = 0.6 gigabytes per day. (b) 0.6 \times 30 = 18 gigabytes. (c) \displaystyle n = \frac{9}{0.6} = 15 days.
- (a) d = 31g and \displaystyle g = \frac{d}{31}, since \displaystyle k = \frac{372}{12} = 31 miles per gallon. (b) \displaystyle g = \frac{155}{31} = 5 gallons.
- (a) Each hour of running costs 0.18 dollars, that is, 18 cents. (b) 0.18 \times 5 = 0.90 dollars. (c) \displaystyle h = \frac{2.70}{0.18} = 15 hours.
- Not proportional: at x = 0, y = 1.
- Proportional, k = 5: 5(x - 2) + 10 = 5x - 10 + 10 = 5x.
- Not proportional. The ratio \displaystyle \frac{y}{x} = x + 1 changes with x: x = 1 gives y = 2, and x = 2 gives y = 6.
- Not proportional. Here 20 is divided by x rather than x being multiplied by a fixed number; doubling x halves y, so the relationship is inversely proportional.
- Proportional, k = 3. Add 4 to both sides: y = 3x.
- Proportional, k = 4: 6x + 2 - 2x - 2 = 4x.
- c = 12, which gives y = 4x with k = 4. Since 4(x - 3) + c = 4x - 12 + c, the constant part -12 + c must be 0.
- (a) C = 10 + 3g. (b) No: 2 gigabytes cost 16 dollars and 4 gigabytes cost 22 dollars, not 32; and 0 gigabytes would still cost 10 dollars. (c) The data charge, 3g, is proportional to g, with constant 3 dollars per gigabyte.
- (a) U is proportional to n, with k = 0.8 gigabytes per day. L is not: at n = 0, L = 20. (b) U = 8 gigabytes and L = 12 gigabytes.
- (a) \displaystyle \frac{64}{2} = \frac{160}{5} = \frac{256}{8} = 32, the same ratio each time, so d = 32g. (b) Car B, at 32 miles per gallon against 26. (c) 320 - 260 = 60 miles.
- Putting x = 0 into y = kx + b gives y = b, which is not 0. Any proportional relationship y = kx gives y = k \cdot 0 = 0 at x = 0, so it cannot contain the pair (0, b).
- With input mx, the output of y = kx is k(mx) = m(kx) = my, by the associative and commutative properties of multiplication. For y = 2x + 3, x = 1 gives 5 but x = 2 gives 7, not 10. In general 2(mx) + 3 = m(2x + 3) - 3m + 3, which is m times the output only when m = 1.
- On the graph, both lines rise 3 for each 1 across, but y = 3x passes through (0, 0) and y = 3x + 2 crosses the y-axis at (0, 2). In the equation, y = 3x is a fixed number times x, while y = 3x + 2 has 2 added on, so its ratio \displaystyle \frac{y}{x} = 3 + \frac{2}{x} changes with x.
- Substitute y = kx into z = my: z = m(kx) = (mk)x by the associative property. The number mk is fixed, and it is not 0 because neither m nor k is, so z is proportional to x with constant mk. For the printer the toner used is 0.04p = 0.04(22t) = 0.88t: 0.88 grams of toner per minute of printing.