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The Humongous Book of Algebra Problems by W. Michael Kelley

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Chapter Twenty-One — Rational Equations and Inequalities
The Humongous Book of Algebra Problems
475
Direct and Indirect Variation
Turn a word problem into a rational equation
21.21 Assume that the value of x varies directly with the value of y according to
the constant of proportionality k. Identify two equations that describe the
relationship between x and y.
If x and y vary proportionally, then y = kx and .
Note: Problems 21.22–21.23 refer to the direct variation relationship described below.
21.22 A professional sports team notes that the ambient crowd noise at home
games is directly proportional to the attendance at those games. During one
game, the cheering of a maximum capacity crowd of 75,000 fans averages 90
decibels. Identify the constant of proportionality k.
Let a represent attendance and n represent the noise (in decibels) generated by
that population. If n varies directly with a, then (according to Problem 21.21)
n = ka. Substitute n = 90 and a = 75,000 into the equation and solve for k.
Note: Problems 21.22–21.23 refer to the direct variation relationship described in Problem
21.22.
21.23 Approximately how loud is a crowd of 62,000 fans?
Substitute a = 62,000 and k = 0.0012 into the variation equation n = ka to
calculate n.
Varying
proportionally means
the same thing as
varying directly.
Let’s say
k = 2. According
to the equation
y = kx, y is always twice
as big as x. Because
y is always two times
as large as x, y ÷ x
always equals 2.
This value
of k (and the
equation n = ka
that you plug it into)
come from Problem
21.22.
Chapter Twenty-One — Rational Equations and Inequalities
The Humongous Book of Algebra Problems
476
21.24 Assume x and y vary proportionally. If x = 16 when y = –3, calculate the value of
y when x = 6.
If x and y vary proportionally, then x = ky. Substitute x = 16 and y = –3 into the
equation and solve for k.
To determine the value of y when x = 6, substitute x = 6 and into the
proportionality equation.
21.25 Assume that the growth of a vine is directly proportional to the time it is
exposed to light. If the vine is exposed to 72 hours of light and grows 1.5
inches, how long will the vine grow when exposed to 200 hours of light?
Round the answer to the hundredths place.
Let l represent the length the vine grows when exposed to h hours of light. The
values vary proportionally, so l = kh, where k is a constant of proportionality.
Substitute l = 1.5 and h = 72 into the equation to calculate k.
To determine how long the vine grows after 200 hours of light, substitute
h = 200 and into the equation l = kh.
The vine grows approximately 4.17 inches when exposed to 200 hours of light.
To cancel out
the coefcient of
y, multiply both sides
of the equation by
its reciprocal.
Rounding
4.166666… to
the hundredths
place gives you
4.17.

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