Chapter 3 · Ratios and proportional relationships

Finding the Constant of Proportionality

Find the constant of proportionality, the unit rate that links two quantities, from a table, a graph, an equation, or a description in words.

20 min read

You already know how to tell whether a table or a graph shows a proportional relationship. Now you will find the one number that drives such a relationship.

Picture an elevator that rises at a steady 2 meters every second. Let xx be the time in seconds and let yy be the height the elevator has risen, in meters. After 1 second, y=2y = 2. After 2 seconds, y=4y = 4. After 4 seconds, y=8y = 8. Each height is 2 times its time.

That multiplier, 2 meters per second, is called the constant of proportionality. It is the number you multiply xx by to get yy, and it is the same number for every pair of values. In this lesson you will find it in three views: a table, a graph, and an equation or a description in words.

Key concept

The constant of proportionality

Two quantities xx and yy are in a proportional relationship when y÷xy \div x, also written yx\frac{y}{x}, equals the same number kk for every pair of values with x≠0x \neq 0. That number kk is the constant of proportionality.

Multiplying both sides of y÷x=ky \div x = k by xx gives y=kxy = kx. The graph of y=kxy = kx is a straight line. It passes through (0,0)(0, 0) because k⋅0=0k \cdot 0 = 0, and through (1,k)(1, k) because k⋅1=kk \cdot 1 = k. So kk is also the unit rate: the amount of yy that goes with exactly 1 unit of xx.

If y÷xy \div x does not give the same number for every such pair, the relationship is not proportional and has no constant of proportionality.

k=yx for every pair with x≠0,y=kxk = \frac{y}{x} \text{ for every pair with } x \neq 0, \qquad y = kx

In a table. Each row gives one pair of values, xx and yy. In a proportional relationship, y=kxy = kx. In any row where x≠0x \neq 0, you can divide both sides by xx and they stay equal, so y÷x=ky \div x = k. That gives you a method: pick a row where xx is not 0 and divide yy by xx.

One row is enough only when you already know the relationship is proportional. If you are not told, divide in every row where x≠0x \neq 0. The results of the divisions, called quotients, must all be the same number.

The table below shows how much flour a bakery uses for its bread. Here xx is the number of loaves and yy is the amount of flour in cups. The last column works out y÷xy \div x for each row, in cups per loaf.

Flour used for loaves of bread
Number of loaves, xFlour in cups, yy divided by x (cups per loaf)
22
55
5÷2=2.55 \div 2 = 2.5
44
1010
10÷4=2.510 \div 4 = 2.5
66
1515
15÷6=2.515 \div 6 = 2.5
1010
2525
25÷10=2.525 \div 10 = 2.5

Worked example

Find k from a table

Use the table above, where xx is the number of loaves and yy is the amount of flour in cups. Find the constant of proportionality and explain what it means.

  1. If the relationship is proportional, then y = kx, so dividing y by x in a row where x is not 0 gives k. Start with the first row, where x = 2 and y = 5.

    yx=52=2.5\frac{y}{x} = \frac{5}{2} = 2.5
  2. Check the other rows. The relationship is proportional only if y divided by x gives the same quotient in every row where x is not 0.

    104=2.5,156=2.5,2510=2.5\frac{10}{4} = 2.5, \quad \frac{15}{6} = 2.5, \quad \frac{25}{10} = 2.5
  3. All four quotients equal 2.5, so the relationship is proportional and k = 2.5. Test this value by multiplying, because k times x must give y. Try the row where x = 6.

    2.5×6=152.5 \times 6 = 15
  4. Say what k means, using the units. Here y is in cups and x is in loaves, so k is in cups per loaf. Each loaf uses 2.5 cups of flour, which is the unit rate.

    y=2.5xy = 2.5x

Answer

The constant of proportionality is k=2.5k = 2.5. The bakery uses 2.52.5 cups of flour for each loaf, so y=2.5xy = 2.5x.

On a graph. The graph of a proportional relationship y=kxy = kx is a straight line through (0,0)(0, 0). Each point on the line is a pair (x,y)(x, y), just like a row of a table. So the method is the same: choose a point with x≠0x \neq 0 and divide its yy-value by its xx-value.

The graph below shows the elevator from the start of the lesson. The horizontal axis shows the time xx in seconds, and the vertical axis shows the height yy in meters. Neither quantity is negative here. Line A is straight and passes through the labeled points (0,0)(0, 0), (1,2)(1, 2), (2,4)(2, 4) and (4,8)(4, 8). Each point is one pair of values: (4,8)(4, 8) means that after 4 seconds the elevator has risen 8 meters.

Divide at each labeled point with x≠0x \neq 0: 2÷1=22 \div 1 = 2, 4÷2=24 \div 2 = 2 and 8÷4=28 \div 4 = 2. The quotient is always 2, so k=2k = 2 meters per second.

Coordinate grid from x 0 to 6 and y 0 to 10, with a grid step of 1, showing 5 plotted elements. Details: line 2 x plus -1 y equals 0, named Line A; point at (0, 0), named (0, 0); point at (1, 2), named (1, 2); point at (2, 4), named (2, 4); point at (4, 8), named (4, 8).
Line A passes through (0,0)(0, 0), (1,2)(1, 2), (2,4)(2, 4) and (4,8)(4, 8). Time in seconds runs across, and height in meters runs up.

Often the point with x=1x = 1 is not labeled, or kk is not a whole number. You do not need that point. On a straight line through (0,0)(0, 0), every point with x≠0x \neq 0 gives the same quotient y÷xy \div x, so any labeled point with x≠0x \neq 0 will do. Use the exact labeled coordinates, not a guess from the grid.

The next graph shows an empty tank filling with water. The horizontal axis shows the time xx in minutes, and the vertical axis shows the water in the tank, yy, in liters. Line B is straight and passes through the labeled points (0,0)(0, 0), (2,3)(2, 3), (4,6)(4, 6) and (6,9)(6, 9).

Coordinate grid from x 0 to 8 and y 0 to 10, with a grid step of 1, showing 5 plotted elements. Details: line 3 x plus -2 y equals 0, named Line B; point at (0, 0), named (0, 0); point at (2, 3), named (2, 3); point at (4, 6), named (4, 6); point at (6, 9), named (6, 9).
Line B passes through (0,0)(0, 0), (2,3)(2, 3), (4,6)(4, 6) and (6,9)(6, 9). Time in minutes runs across, and water in liters runs up.

Worked example

Find k from a graph

Use the graph above. The horizontal axis shows the time xx in minutes, and the vertical axis shows the water yy in liters. Line B is straight and passes through the labeled points (0,0)(0, 0), (2,3)(2, 3), (4,6)(4, 6) and (6,9)(6, 9). Find the constant of proportionality.

  1. Check that there is a constant to find. Line B is a straight line and it passes through the labeled point (0, 0), so the graph shows a proportional relationship, y = kx.

  2. Choose a labeled point where x is not 0. Use (2, 3), which means that after 2 minutes the tank holds 3 liters. The point fits y = kx, so dividing its y-value by its x-value gives k.

    k=32=1.5k = \frac{3}{2} = 1.5
  3. Confirm with the other labeled points where x is not 0. On a straight line through (0, 0), every such point gives the same quotient.

    64=1.5,96=1.5\frac{6}{4} = 1.5, \quad \frac{9}{6} = 1.5
  4. Say what k means. Here y is in liters and x is in minutes, so the tank gains 1.5 liters each minute. To see this unit rate in the equation y = 1.5x, put in x = 1.

    y=1.5×1=1.5y = 1.5 \times 1 = 1.5

Answer

The constant of proportionality is k=1.5k = 1.5. The tank fills at 1.51.5 liters per minute, so y=1.5xy = 1.5x.

In an equation. A proportional relationship can be written as y=kxy = kx, so kk is the number that multiplies xx. The elevator's equation is y=2xy = 2x, and you can read k=2k = 2 straight from it. This works only when the equation has the form y=kxy = kx: one number times xx, with nothing added or subtracted.

The letters can change. Let tt be a time in hours and let dd be the distance in miles that a car travels in that time. In the equation d=55td = 55t, the number that multiplies tt is 55, so k=55k = 55 miles per hour.

Sometimes kk is hidden because the equation shows a division. Then rewrite the equation so that one number multiplies xx. The next example shows how.

Worked example

Find k when an equation hides it

Let xx be an amount of liquid in cups and let yy be the same amount in quarts. The two are related by the equation y=x4y = \frac{x}{4}. Find the constant of proportionality.

  1. The constant k is the number that multiplies x in y = kx. This equation shows x divided by 4, so rewrite the division as a multiplication. Dividing by 4, which is not 0, gives the same result as multiplying by one fourth, so the same pairs of values still fit the equation.

    y=x4=14⋅xy = \frac{x}{4} = \frac{1}{4} \cdot x
  2. Now the equation has the form y = kx. Read the number that multiplies x.

    k=14=0.25k = \frac{1}{4} = 0.25
  3. Check with a pair of values. Choose x = 8 cups. The equation gives y = 8 divided by 4, which is 2 quarts. Because 8 is not 0, you can divide y by x, and the quotient should equal k.

    yx=28=14\frac{y}{x} = \frac{2}{8} = \frac{1}{4}
  4. Say what k means. Here y is in quarts and x is in cups, so each cup is one fourth of a quart. That is the unit rate.

Answer

The constant of proportionality is k=14k = \frac{1}{4}, or 0.250.25. Each cup is 14\frac{1}{4} of a quart, so y=14xy = \frac{1}{4}x.

In a description in words. Some descriptions give you kk directly. Read this one: A sprinkler uses 9 liters of water each minute. It tells you that 1 minute goes with 9 liters. Let xx be the time in minutes and let yy be the water used in liters. Then k=9k = 9 liters per minute. Words such as each and per often signal the unit rate.

Other descriptions give one pair of values that is not the unit rate. Then make three moves:

  1. Name the variables and their units. Let xx be the quantity you start from and let yy be the quantity you want to find from it.
  2. Write the pair of values from the description.
  3. Divide yy by xx. This gives kk because the relationship is proportional and x≠0x \neq 0.

Worked example

Find k from a description in words

A machine fills 54 bottles in 4 minutes. The number of bottles it fills is proportional to the time it runs. Find the constant of proportionality that gives the number of bottles from the time, and write the equation.

  1. Name the variables and their units. You want the number of bottles from the time, so let x be the time in minutes and let y be the number of bottles filled.

  2. Write the pair from the description: 4 minutes goes with 54 bottles. The relationship is proportional, so this pair must fit y = kx.

    x=4,y=54,54=k⋅4x = 4, \quad y = 54, \quad 54 = k \cdot 4
  3. Divide both sides by 4 to get k alone. Dividing both sides of an equation by the same number keeps the two sides equal, and it is allowed here because 4 is not 0.

    k=544=272=13.5k = \frac{54}{4} = \frac{27}{2} = 13.5
  4. Check by multiplying, because k times x must give back y. Then say what k means: the machine fills 13.5 bottles per minute, which is the unit rate.

    13.5×4=5413.5 \times 4 = 54
  5. Write the equation by putting this value of k into y = kx.

    y=13.5xy = 13.5x

Answer

The constant of proportionality is k=13.5k = 13.5 bottles per minute, and the equation is y=13.5xy = 13.5x.

The big idea. In a proportional relationship, one number kk links the two quantities: y=kxy = kx, so k=y÷xk = y \div x for every pair with x≠0x \neq 0. It is the unit rate, the amount of yy for 1 unit of xx. Here is where to find it:

  • Table: divide yy by xx in a row where x≠0x \neq 0, and check that the other rows agree.
  • Graph: on a straight line through (0,0)(0, 0), divide yy by xx at a labeled point with x≠0x \neq 0, or read the yy-value of the labeled point (1,k)(1, k).
  • Equation or words: write the equation in the form y=kxy = kx and read the number that multiplies xx. For a proportional relationship described in words, name xx and yy, take one pair with x≠0x \neq 0, and divide yy by xx.

In every view, check that kk times xx gives yy. Next, you will use kk to complete a table and graph a proportional relationship.