Download 7.2 Systems of Linear Equations in Three Variables

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748 Chapter 7 Systems of Equations and Inequalities
Section
7.2
Objectives
� Verify the solution of a system
�
�
�
of linear equations in three
variables.
Solve systems of linear
equations in three variables.
Solve problems using
systems in three variables.
Verify the solution of a system of
linear equations in three
variables.
Systems of Linear Equations in Three Variables
A
ll animals sleep, but the length
of time they sleep varies
widely: Cattle sleep for only a
few minutes at a time. We
humans seem to need more
sleep than other animals,
up to eight hours a day.
Without enough sleep, we
have difficulty concentrating, make mistakes in routine tasks, lose energy, and
feel bad-tempered. There is a
relationship between hours of
sleep and death rate per year per
100,000 people. How many hours of sleep will put you in the group with the minimum
death rate? In this section, we will answer this question by solving a system of linear
equations with more than two variables.
Systems of Linear Equations in Three Variables and Their
Solutions
An equation such as x + 2y - 3z = 9 is called a linear equation in three variables.
In general, any equation of the form
Ax + By + Cz = D,
Point of
intersection
P
Figure 7.9
where A, B, C, and D are real numbers such that A, B, and C are not all 0, is a linear
equation in three variables: x, y, and z. The graph of this linear equation in three
variables is a plane in three-dimensional space.
The process of solving a system of three linear equations in three variables
is geometrically equivalent to finding the point of intersection (assuming that
there is one) of three planes in space. (See Figure 7.9.) A solution of a system of
linear equations in three variables is an ordered triple of real numbers that
satisfies all equations of the system. The solution set of the system is the set of all
its solutions.
EXAMPLE 1
Determining Whether an Ordered Triple
Satisfies a System
Show that the ordered triple 1-1, 2, -22 is a solution of the system:
x + 2y - 3z = 9
c 2x - y + 2z = - 8
-x + 3y - 4z = 15.
Solution
Because -1 is the x-coordinate, 2 is the y-coordinate, and - 2 is
the z-coordinate of 1- 1, 2, - 22, we replace x with - 1, y with 2, and z with -2 in
each of the three equations.
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Section 7.2 Systems of Linear Equations in Three Variables
x + 2y - 3z = 9
2x - y + 2z = - 8
-x + 3y - 4z = 15
- 1 + 2122 - 31 - 22 ⱨ 9
21-12 - 2 + 21-22 ⱨ -8
-1 + 4 + 6 ⱨ 9
-2 - 2 - 4 ⱨ - 8
- 1-12 + 3122 - 41 -22 ⱨ 15
9 = 9,
-8 = - 8,
true
749
1 + 6 + 8 ⱨ 15
15 = 15,
true
true
The ordered triple 1-1, 2, -22 satisfies the three equations: It makes each equation
true. Thus, the ordered triple is a solution of the system.
Check Point
system:
1
Show that the ordered triple 1- 1, -4, 52 is a solution of the
x - 2y + 3z = 22
c 2x - 3y - z = 5
3x + y - 5z = - 32.
�
Solve systems of linear
equations in three variables.
Solving Systems of Linear Equations in Three Variables
by Eliminating Variables
The method for solving a system of linear equations in three variables is similar to
that used on systems of linear equations in two variables. We use addition to
eliminate any variable, reducing the system to two equations in two variables. Once
we obtain a system of two equations in two variables, we use addition or substitution
to eliminate a variable. The result is a single equation in one variable. We solve this
equation to get the value of the remaining variable. Other variable values are found
by back-substitution.
Study Tip
It does not matter which variable you
eliminate, as long as you do it in two
different pairs of equations.
Solving Linear Systems in Three Variables by Eliminating Variables
1. Reduce the system to two equations in two variables. This is usually
accomplished by taking two different pairs of equations and using the
addition method to eliminate the same variable from both pairs.
2. Solve the resulting system of two equations in two variables using addition
or substitution. The result is an equation in one variable that gives the value
of that variable.
3. Back-substitute the value of the variable found in step 2 into either of the
equations in two variables to find the value of the second variable.
4. Use the values of the two variables from steps 2 and 3 to find the value of the
third variable by back-substituting into one of the original equations.
5. Check the proposed solution in each of the original equations.
EXAMPLE 2
Solving a System in Three Variables
Solve the system:
5x - 2y - 4z = 3
c 3x + 3y + 2z = - 3
- 2x + 5y + 3z = 3.
Equation 1
Equation 2
Equation 3
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750 Chapter 7 Systems of Equations and Inequalities
5x - 2y - 4z = 3 Equation 1
c 3x + 3y + 2z = - 3 Equation 2
-2x + 5y + 3z = 3 Equation 3
The given system (repeated)
Solution There are many ways to proceed. Because our initial goal is to reduce
the system to two equations in two variables, the central idea is to take two different
pairs of equations and eliminate the same variable from both pairs.
Step 1 Reduce the system to two equations in two variables. We choose any two
equations and use the addition method to eliminate a variable. Let’s eliminate z using
Equations 1 and 2. We do so by multiplying Equation 2 by 2. Then we add equations.
(Equation 1)
(Equation 2)
b
No change "
Multiply by 2.
"
5x - 2y - 4z = 3
3x + 3y + 2z = - 3
e
5x - 2y - 4z = 3
6x + 6y + 4z = - 6
11x + 4y
Add:
= -3
Equation 4
Now we must eliminate the same variable using another pair of equations. We can
eliminate z from Equations 2 and 3. First, we multiply Equation 2 by - 3. Next, we
multiply Equation 3 by 2. Finally, we add equations.
(Equation 2)
(Equation 3)
b
3x + 3y + 2z = - 3
-2x + 5y + 3z = 3
Multiply by - 3.
"
Multiply by 2.
"
e
Add:
- 9x - 9y - 6z = 9
- 4x + 10y + 6z = 6
-13x +
y
= 15 Equation 5
Equations 4 and 5 give us a system of two equations in two variables:
b
11x + 4y = - 3
-13x + y = 15.
Equation 4
Equation 5
Step 2 Solve the resulting system of two equations in two variables. We will use
the addition method to solve Equations 4 and 5 for x and y. To do so, we multiply
Equation 5 on both sides by - 4 and add this to Equation 4.
(Equation 4)
(Equation 5)
b
11x + 4y = - 3
-13x + y = 15
No change
"
Multiply by - 4.
"
b
Add:
11x + 4y = - 3
52x - 4y = - 60
63x
= - 63
x
= - 1 Divide both sides by 63.
Step 3 Use back-substitution in one of the equations in two variables to find the
value of the second variable. We back-substitute -1 for x in either Equation 4 or 5
to find the value of y.
-13x + y = 15
-131- 12 + y = 15
13 + y = 15
y = 2
Equation 5
Substitute - 1 for x.
Multiply.
Subtract 13 from both sides.
Step 4 Back-substitute the values found for two variables into one of the original
equations to find the value of the third variable. We can now use any one of the
original equations and back-substitute the values of x and y to find the value for z.
We will use Equation 2.
3x + 3y + 2z = - 3
31-12 + 3122 + 2z = - 3
3 + 2z = - 3
2z = - 6
z = -3
Equation 2
Substitute -1 for x and 2 for y.
Multiply and then add:
31-12 + 3122 = - 3 + 6 = 3.
Subtract 3 from both sides.
Divide both sides by 2.
With x = - 1, y = 2, and z = - 3, the proposed solution is the ordered triple
1- 1, 2, -32.
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Section 7.2 Systems of Linear Equations in Three Variables
751
Step 5 Check. Check the proposed solution, 1- 1, 2, -32, by substituting the
values for x, y, and z into each of the three original equations. These substitutions
yield three true statements. Thus, the solution set is 51-1, 2, - 326.
Check Point
2
Solve the system:
x + 4y - z = 20
c 3x + 2y + z =
8
2x - 3y + 2z = - 16.
In some examples, one of the variables is already eliminated from a given
equation. In this case, the missing variable should be eliminated from the other two
equations, thereby making it possible to omit one of the elimination steps. We
illustrate this idea in Example 3.
EXAMPLE 3
Solving a System of Equations with a Missing Term
Solve the system:
x +
z = 8
c x + y + 2z = 17
x + 2y + z = 16.
Equation 1
Equation 2
Equation 3
Solution
Step 1 Reduce the system to two equations in two variables. Because Equation 1
contains only x and z, we could omit one of the elimination steps by eliminating y
using Equations 2 and 3. This will give us two equations in x and z. To eliminate y
using Equations 2 and 3, we multiply Equation 2 by - 2 and add Equation 3.
(Equation 2)
x + y + 2z = 17
(Equation 3) b x + 2y + z = 16
Multiply by - 2.
"
No change
"
Add:
b
-2x - 2y - 4z = - 34
x + 2y + z = 16
-x
- 3z = - 18 Equation 4
Equation 4 and the given Equation 1 provide us with a system of two equations in
two variables:
b
Equation 1
x + z = 8
-x - 3z = - 18. Equation 4
Step 2 Solve the resulting system of two equations in two variables. We will solve
Equations 1 and 4 for x and z.
b
x + z =
8
-x - 3z = - 18
Add:
-2z = - 10
z =
5
Equation 1
Equation 4
Divide both sides by - 2.
Step 3 Use back-substitution in one of the equations in two variables to find
the value of the second variable. To find x, we back-substitute 5 for z in either
Equation 1 or 4. We will use Equation 1.
x + z = 8
x + 5 = 8
x = 3
Equation 1
Substitute 5 for z.
Subtract 5 from both sides.
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752 Chapter 7 Systems of Equations and Inequalities
x +
z = 8
c x + y + 2z = 17
x + 2y + z = 16
Equation 1
Equation 2
Equation 3
Step 4 Back-substitute the values found for two variables into one of the original
equations to find the value of the third variable. To find y, we back-substitute 3 for
x and 5 for z into Equation 2 or 3. We cannot use Equation 1 because y is missing in
this equation. We will use Equation 2.
The given system (repeated)
x + y + 2z = 17
3 + y + 2152 = 17
y + 13 = 17
y = 4
Equation 2
Substitute 3 for x and 5 for z.
Multiply and add.
Subtract 13 from both sides.
We found that z = 5, x = 3, and y = 4. Thus, the proposed solution is the ordered
triple (3, 4, 5).
Step 5 Check. Substituting 3 for x, 4 for y, and 5 for z into each of the three
original equations yields three true statements. Consequently, the solution set is
513, 4, 526.
Check Point
3
Solve the system:
2y - z = 7
c x + 2y + z = 17
2x - 3y + 2z = - 1.
A system of linear equations in three variables represents three planes. The
three planes may not always intersect at one point. The planes may have no
common point of intersection and represent an inconsistent system with no
solution. By contrast, the planes may coincide or intersect along a line. In these
cases, the planes have infinitely many points in common and represent systems with
infinitely many solutions. Systems of linear equations in three variables that are
inconsistent or that contain dependent equations will be discussed in Chapter 8.
�
Solve problems using systems in
three variables.
Applications
Systems of equations may allow us to find models for data without using a graphing
utility. Three data points that do not lie on or near a line determine the graph of a
quadratic function of the form y = ax2 + bx + c, a Z 0. Quadratic functions often
model situations in which values of y are decreasing and then increasing, suggesting
the bowl-like shape of a parabola.
EXAMPLE 4
Modeling Data Relating Sleep and Death Rate
In a study relating sleep and death rate, the following data were obtained. Use the
function y = ax2 + bx + c to model the data.
x
(Average Number of
Hours of Sleep)
y
(Death Rate per Year
per 100,000 Males)
4
1682
7
626
9
967
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Section 7.2 Systems of Linear Equations in Three Variables
753
Solution We need to find values for a, b, and c in y = ax2 + bx + c. We can do
so by solving a system of three linear equations in a, b, and c. We obtain the three
equations by using the values of x and y from the data as follows:
y ⴝ ax 2 ⴙ bx ⴙ c
When x = 4, y = 1682:
When x = 7, y = 626:
When x = 9, y = 967:
Technology
Use the quadratic function to model the data.
1682 = a # 4 2 + b # 4 + c
16a + 4b + c = 1682
2
#
#
c 626 = a 7 + b 7 + c or c 49a + 7b + c = 626
967 = a # 92 + b # 9 + c
81a + 9b + c = 967.
The graph of
y = 104.5x2 - 1501.5x + 6016
is displayed in a [3, 12, 1] by
[500, 2000, 100] viewing rectangle. The
minimum function feature shows that
the lowest point on the graph, the
vertex, is approximately (7.2, 622.5).
Men who average 7.2 hours of sleep
are in the group with the lowest death
rate, approximately 622.5 deaths per
100,000 males.
The easiest way to solve this system is to eliminate c from two pairs of equations,
obtaining two equations in a and b. Solving this system gives a = 104.5, b = - 1501.5,
and c = 6016. We now substitute the values for a, b, and c into y = ax2 + bx + c.
The function that models the given data is
y = 104.5x2 - 1501.5x + 6016.
We can use the model that we obtained in Example 4 to find the death rate of
males who average, say, 6 hours of sleep. First, write the model in function notation:
f1x2 = 104.5x2 - 1501.5x + 6016.
Substitute 6 for x:
f162 = 104.51622 - 1501.5162 + 6016 = 769.
According to the model, the death rate for males who average 6 hours of sleep is 769
deaths per 100,000 males.
4
Find the quadratic function y = ax2 + bx + c whose graph
passes through the points (1, 4), (2, 1), and (3, 4).
Check Point
Exercise Set 7.2
Practice Exercises
In Exercises 1–4, determine if the given ordered triple is a solution
of the system.
1. 12, - 1, 32
x + y + z = 4
c x - 2y - z = 1
2x - y - 2z = - 1
3. (4, 1, 2)
x - 2y
= 2
c 2x + 3y
= 11
y - 4z = - 7
2. 15, - 3, - 22
x + y + z = 0
c x + 2y - 3z = 5
3x + 4y + 2z = - 1
4. 1 - 1, 3, 22
x - 2z = - 5
c y - 3z = - 3
2x - z = - 4
Solve each system in Exercises 5–18.
3x + 2y - 3z = - 2
9. c 2x - 5y + 2z = - 2
4x - 3y + 4z = 10
2x + 3y + 7z = 13
10. c 3x + 2y - 5z = - 22
5x + 7y - 3z = - 28
2x - 4y + 3z = 17
11. c x + 2y - z = 0
4x - y - z = 6
x +
z = 3
12. c x + 2y - z = 1
2x - y + z = 3
2x + y
= 2
13. c x + y - z = 4
3x + 2y + z = 0
x + 3y + 5z = 20
14. c
y - 4z = -16
3x - 2y + 9z = 36
15. c
x + y
= -4
y - z =
1
2x + y + 3z = - 21
x + y = 4
16. c x + z = 4
y + z = 4
x + y + 2z = 11
5. c x + y + 3z = 14
x + 2y - z = 5
2x + y - 2z = - 1
6. c 3x - 3y - z = 5
x - 2y + 3z = 6
312x + y2 + 5z = - 1
17. c 21x - 3y + 4z2 = - 9
411 + x2 = - 31z - 3y2
4x - y + 2z = 11
7. c x + 2y - z = - 1
2x + 2y - 3z = - 1
x - y + 3z = 8
8. c 3x + y - 2z = - 2
2x + 4y + z = 0
7z - 3 = 21x - 3y2
18. c 5y + 3z - 7 = 4x
4 + 5z = 312x - y2
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754 Chapter 7 Systems of Equations and Inequalities
In Exercises 19–22, find the quadratic function y = ax2 + bx + c
whose graph passes through the given points.
In Exercises 29–30, solve each system for 1x, y, z2 in terms of the
nonzero constants a, b, and c.
21. 1 -1, - 42, 11, -22, (2, 5)
ax - by - 2cz = 21
29. c ax + by + cz = 0
2ax - by + cz = 14
19. 1 -1, 62, (1, 4), (2, 9)
20. 1 -2, 72, 11, - 22, (2, 3)
22. (1, 3), 13, -12, (4, 0)
In Exercises 23–24, let x represent the first number, y the second
number, and z the third number. Use the given conditions
to write a system of equations. Solve the system and find the
numbers.
23. The sum of three numbers is 16. The sum of twice the first
number, 3 times the second number, and 4 times the third
number is 46. The difference between 5 times the first number
and the second number is 31. Find the three numbers.
24. The following is known about three numbers: Three times
the first number plus the second number plus twice the third
number is 5. If 3 times the second number is subtracted from
the sum of the first number and 3 times the third number, the
result is 2. If the third number is subtracted from 2 times the
first number and 3 times the second number, the result is 1.
Find the numbers.
Practice Plus
Solve each system in Exercises 25–26.
y + 4
z
x + 2
+
= 0
6
3
2
y - 1
z
9
x + 1
+
=
25. f
2
2
4
2
y + 1
z - 2
19
x - 5
+
+
=
4
3
2
4
y - 1
x + 3
+
2
2
y + 1
x - 5
+
26. f
2
3
y + 1
x - 3
+
4
2
z + 2
3
=
4
2
z
25
= 4
6
5
z - 3
= 2
2
In Exercises 27–28, find the equation of the quadratic function
y = ax2 + bx + c whose graph is shown. Select three points
whose coordinates appear to be integers.
27.
y
1
−1
−2
2
4
6
8
x
Application Exercises
31. You throw a ball straight up from a rooftop. The ball misses
the rooftop on its way down and eventually strikes the
ground. A mathematical model can be used to describe the
relationship for the ball’s height above the ground, y, after x
seconds. Consider the following data:
x, seconds after the ball is
thrown
y, ball’s height, in feet, above
the ground
1
224
3
176
4
104
a. Find the quadratic function y = ax2 + bx + c whose
graph passes through the given points.
b. Use the function in part (a) to find the value for y when
x = 5. Describe what this means.
32. A mathematical model can be used to describe the relationship between the number of feet a car travels once the brakes
are applied, y, and the number of seconds the car is in motion
after the brakes are applied, x. A research firm collects the
following data:
x, seconds in motion
after brakes are applied
46
2
84
3
114
a. Find the quadratic function y = ax2 + bx + c whose
graph passes through the given points.
b. Use the function in part (a) to find the value for y when
x = 6. Describe what this means.
Use a system of linear equations in three variables to solve
Exercises 33–39.
The bar graph compares the body composition of a 160-pound
adult male and a 135-pound adult female. Exercises 33–34 are
based on the data displayed by the graph.
Body Composition of an
Adult Male and Female
−6
70%
−8
135-pound female
160-pound male
Percentage of Total
Body Weight
60%
y
8
6
4
50%
40%
30%
20%
10%
2
−1
y, feet car travels
once the brakes are applied
1
−4
28.
ax - by + 2cz = - 4
30. c ax + 3by - cz = 1
2ax + by + 3cz = 2
2
4
6
8
x
Water
Fat
Protein
Minerals
Source: Thompson et al., The Science of Nutrition, Benjamin Cummings, 2008
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Section 7.2 Systems of Linear Equations in Three Variables
34. This exercise refers to the body composition of a 160pound adult male, shown by the graph on the previous
page. Ninety-four percent of his total body weight consists
of water, fat, and protein. The difference between the percentage of body weight consisting of water and fat is 47%.
The difference between the percentage of body weight
consisting of protein and fat is 2%. Find the percentage of
total body weight consisting of water, of fat, and of protein.
35. At a college production of Streetcar Named Desire, 400 tickets
were sold. The ticket prices were $8, $10, and $12, and the total
income from ticket sales was $3700. How many tickets of each
type were sold if the combined number of $8 and $10 tickets
sold was 7 times the number of $12 tickets sold?
36. A certain brand of razor blades comes in packages of 6, 12,
and 24 blades, costing $2, $3, and $4 per package, respectively.
A store sold 12 packages containing a total of 162 razor
blades and took in $35. How many packages of each type
were sold?
37. A person invested $6700 for one year, part at 8%, part at
10%, and the remainder at 12%. The total annual income
from these investments was $716. The amount of money
invested at 12% was $300 more than the amount invested at
8% and 10% combined. Find the amount invested at
each rate.
38. A person invested $17,000 for one year, part at 10%, part at
12%, and the remainder at 15%. The total annual income
from these investments was $2110. The amount of money
invested at 12% was $1000 less than the amount invested at
10% and 15% combined. Find the amount invested at
each rate.
39. In the following triangle, the degree measures of the three interior angles and two of the exterior angles are represented
with variables. Find the measure of each interior angle.
African Life Span
70
Projections without AIDS
60
50
Projections with AIDS
1985 1990 1995 2000 2005 2010
Year
Source: United Nations
Technology Exercises
44. Does your graphing utility have a feature that allows you
to solve linear systems by entering coefficients and
constant terms? If so, use this feature to verify the
solutions to any five exercises that you worked by hand
from Exercises 5–16.
45. Verify your results in Exercises 19–22 by using a graphing
utility to graph the resulting parabola. Trace along the curve
and convince yourself that the three points given in the
exercise lie on the parabola.
Critical Thinking Exercises
Make Sense? In Exercises 46–49, determine whether each
statement makes sense or does not make sense, and explain
your reasoning.
46. Solving a system in three variables, I found that x = 3 and
y = - 1. Because z represents a third variable, z cannot
equal 3 or - 1.
47. A system of linear equations in three variables, x, y, and z,
cannot contain an equation in the form y = mx + b.
2x + 5
48. I’m solving a three-variable system in which one of the given
equations has a missing term, so it will not be necessary to
use any of the original equations twice when I reduce the
system to two equations in two variables.
z
49. Because the percentage of the U.S. population that was
foreign-born decreased from 1910 through 1970 and then
increased after that, a quadratic function of the form
f1x2 = ax2 + bx + c, rather than a linear function of the
form f1x2 = mx + b, should be used to model the data.
y
2x − 5
x
43. AIDS is taking a deadly toll on southern Africa. Describe
how to use the techniques that you learned in this section to
obtain a model for African life span using projections with
AIDS, shown by the red graph in the figure. Let x represent
the number of years after 1985 and let y represent African
life span in that year.
Life Span (years)
33. This exercise refers to the body composition of a 135-pound
adult female, shown by the graph on the previous page.
Ninety-five percent of her total body weight consists of
water, fat, and protein. The difference between the percentage of body weight consisting of water and fat is 35%. The
difference between the percentage of body weight
consisting of fat and protein is 9%. Find the percentage of
total body weight consisting of water, of fat, and of protein.
755
50. Describe how the system
Writing in Mathematics
40. What is a system of linear equations in three variables?
41. How do you determine whether a given ordered triple is a
solution of a system in three variables?
42. Describe in general terms how to solve a system in three
variables.
x
x
d
2x
2x
+
+
+
y
2y
2y
y
+
+
-
z
3z
2z
z
+
+
2w
w
2w
w
= -8
= 18
= 10
= 3
could be solved. Is it likely that in the near future a graphing
utility will be available to provide a geometric solution (using
intersecting graphs) to this system? Explain.
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Preview Exercises
51. A modernistic painting consists of triangles, rectangles, and
pentagons, all drawn so as to not overlap or share sides.
Within each rectangle are drawn 2 red roses and each
pentagon contains 5 carnations. How many triangles,
rectangles, and pentagons appear in the painting if the painting contains a total of 40 geometric figures, 153 sides of
geometric figures, and 72 flowers?
Exercises 53–55 will help you prepare for the material covered in
the next section.
2
3
53. Subtract:
.
x - 4
x + 2
2x
5x - 3
54. Add:
+
.
2
x2 + 1
1x2 + 12
Group Exercise
55. Solve:
52. Group members should develop appropriate functions that
model each of the projections shown in Exercise 43.
Section
7.3
A + B = 3
c 2A - 2B + C = 17
4A - 2C = 14.
Partial Fractions
Objectives
� Decompose P , where Q has
Q
only distinct linear factors.
� Decompose P , where Q has
Q
repeated linear factors.
� Decompose P , where Q has
Q
a nonrepeated prime quadratic
factor.
� Decompose P , where Q has
Q
a prime, repeated quadratic
factor.
T
he rising and setting of the sun
suggest the obvious: Things
change over time. Calculus is
the study of rates of change,
allowing the motion of the
rising sun to be measured
by “freezing the frame” at
one instant in time. If you
are given a function, calculus
reveals its rate of change at any
“frozen” instant. In this section,
you will learn an algebraic technique
used in calculus to find a function if its rate of change is known. The technique
involves expressing a given function in terms of simpler functions.
The Idea behind Partial Fraction Decomposition
We know how to use common denominators to write a sum or difference of rational
expressions as a single rational expression. For example,
31x + 22 - 21x - 42
3
2
=
x - 4
x + 2
1x - 421x + 22
=
3x + 6 - 2x + 8
x + 14
=
.
1x - 421x + 22
1x - 421x + 22
For solving the kind of calculus problem described in the section opener, we must
reverse this process:
Partial fraction
x + 14
is
(x − 4)(x + 2)
expressed as the
sum of two
simpler fractions.
Partial fraction
x+14
3
–2
=
+
.
(x-4)(x+2)
x-4
x+2
This is the partial fraction
x + 14
decomposition of
.
(x − 4)(x + 2)
Each of the two fractions on the right is called a partial fraction. The sum of these
fractions is called the partial fraction decomposition of the rational expression on
the left-hand side.