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7.1
7.1
al-jabr, algebrista, algebra
The word algebra comes from the
title of the work Hisâb al-jabr w’al
muquâbalah, a ninth-century
treatise by the Arab Mohammed
ibn Mûsâ al-Khowârizmî. The title
translates as “the science of
reunion and reduction,” or more
generally, “the science of
transposition and cancellation.”
In the title of Khowârizmî’s
book, jabr (“restoration”) refers to
transposing negative quantities
across the equals sign in solving
equations. From Latin versions of
Khowârizmî’s text, “al-jabr”
became the broad term covering
the art of equation solving. (The
prefix al means “the.”)
In Spain under Moslem rule,
the word algebrista referred to
the person who restored (reset)
broken bones. Signs outside
barber shops read Algebrista y
Sangrador (bonesetter and
bloodletter). Such services were
part of the barber’s trade. The
traditional red-and-white striped
barber pole symbolizes blood and
bandages.
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Linear Equations
Solving Linear Equations An algebraic expression involves the basic
operations of addition, subtraction, multiplication, or division (except by 0), or raising to powers or taking roots on any collection of variables and numbers. Some
examples of algebraic expressions are
x 3y 8
and
8x 9 ,
.
y 4 ,
z
An equation is a statement that two algebraic expressions are equal. A linear equation in one variable involves only real numbers and one variable. Examples include
x 1 2 ,
y 3 5,
and
2k 5 10 .
Linear Equation in One Variable
An equation in the variable x is linear if it can be written in the form
ax b c ,
where a, b, and c are real numbers, with a 0.
A linear equation in one variable is also called a first-degree equation, since the
greatest power on the variable is one.
If the variable in an equation is replaced by a real number that makes the statement true, then that number is a solution of the equation. For example, 8 is a solution of the equation x 3 5, since replacing x with 8 gives a true statement. An
equation is solved by finding its solution set, the set of all solutions. The solution
set of the equation x 3 5 is 8.
Equivalent equations are equations with the same solution set. Equations generally are solved by starting with a given equation and producing a series of simpler
equivalent equations. For example,
8x 1 17 ,
8x 16 ,
and
x2
are equivalent equations since each has the same solution set, 2. We use the addition and multiplication properties of equality to produce equivalent equations.
Addition and Multiplication Properties of Equality
Addition Property of Equality For all real numbers a, b, and c, the equa-
tions
ab
and
acbc
are equivalent. (The same number may be added to both sides of an
equation without changing the solution set.)
Multiplication Property of Equality For all real numbers a, b, and c, where
c 0, the equations
ab
and
ac bc
are equivalent. (Both sides of an equation may be multiplied by the
same nonzero number without changing the solution set.)
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CHAPTER 7
The Basic Concepts of Algebra
Because subtraction and division are defined in terms of addition and multiplication,
respectively, the addition and multiplication properties can be extended: The same
number may be subtracted from both sides of an equation, and both sides may be divided by the same nonzero number without affecting the solution set.
The distributive property allows us to combine like terms, such as 4y and 2y. For
example, 4y 2y 4 2y 2y.
EXAMPLE 1
Solve 4x 2x 5 4 6x 3.
First, combine like terms separately on both sides of the equation to get
2x 5 7 6x .
Next, use the addition property to get the terms with x on the same side of the equation and the remaining terms (the numbers) on the other side. One way to do this is
to first add 5 to both sides.
2x 5 5 7 6x 5
2x 12 6x
Add 5.
Now subtract 6x from both sides.
2x 6x 12 6x 6x
4x 12
Subtract 6x.
Finally, divide both sides by 4 to obtain the term x on the left.
4x
12
4
4
Graphing calculators can solve
equations of the form
expression in X 0 .
The TI-83 Plus requires that you
input the expression, the variable
for which you are solving, and a
“guess,” separated by commas.
The screen here shows how to
solve the equation in Example 1.
Divide by 4.
x 3
To be sure that 3 is the solution, check by substituting back into the original
equation (not an intermediate one).
4x 2x 5 4 6x 3
43 23 5 4 63 3
12 6 5 4 18 3
11 11
Original equation
?
?
Let x 3.
Multiply.
True
Since a true statement is obtained, 3 is the solution. The solution set is 3.
The steps used to solve a linear equation in one variable are as follows. (Not all
equations require all of these steps.)
Solving a Linear Equation in One Variable
Step 1: Clear fractions. Eliminate any fractions by multiplying
both sides of the equation by a common denominator.
Step 2: Simplify each side separately. Simplify each side of the
equation as much as possible by using the distributive property to clear parentheses and by combining like terms as
needed.
(continued)
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7.1
The problem-solving strategy of
guessing and checking, discussed
in Chapter 1, was actually used by
the early Egyptians in equation
solving. This method, called the
Rule of False Position,
involved making an initial guess at
the solution of an equation, and
then following up with an
adjustment in the likely event that
the guess was incorrect. For
example (using our modern
notation), if the equation
6x 2x 32
was to be solved, an initial guess
might have been x 3.
Substituting 3 for x gives
63 23 32
?
18 6 32
?
24 32 .
False
The guess, 3, gives a value (24)
which is smaller than the desired
value (32). Since 24 is 34 of 32,
the guess, 3, is 34 of the actual
solution. The actual solution,
therefore, must be 4, since 3 is
34 of 4.
Use the methods explained in
this section to verify this result.
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Step 3: Isolate the variable terms on one side. Use the addition
property of equality to get all terms with variables on one
side of the equation and all numbers on the other.
Step 4: Transform so that the coefficient of the variable is 1.
Use the multiplication property of equality to get an equation
with just the variable (with coefficient 1) on one side.
Step 5: Check. Check by substituting back into the original equation.
In Example 1 we did not use Step 1 and the distributive property in Step 2
as given above. Many other equations, however, will require one or both of these
steps.
EXAMPLE 2
Solve 2k 5 3k k 6.
Since there are no fractions in this equation, Step 1 does not apply. Begin by
using the distributive property to simplify and combine terms on the left side of the
equation (Step 2).
2k 5 3k k 6
2k 10 3k k 6
Distributive property
5k 10 k 6
Combine like terms.
Next, add 10 to both sides (Step 3).
5k 10 10 k 6 10
5k k 16
Add 10.
Now subtract k from both sides.
5k k k 16 k
4k 16
Subtract k.
Combine like terms.
Divide both sides by 4 (Step 4).
4k 16
4
4
k4
Divide by 4.
Check that the solution set is 4 by substituting 4 for k in the original equation
(Step 5).
In the remaining examples in this section, we will not identify the steps by
number.
When fractions or decimals appear as coefficients in equations, our work can be
made easier if we multiply both sides of the equation by the least common denominator of all the fractions. This is an application of the multiplication property of
equality, and it produces an equivalent equation with integer coefficients.
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CHAPTER 7
The Basic Concepts of Algebra
EXAMPLE
3
Solve
x 7 2x 8
4.
6
2
Start by eliminating the fractions. Multiply both sides by 6.
6
6
x 7 2x 8
6 4
6
2
x7
2x 8
6
64
6
2
Distributive property
x 7 32x 8 24
x 7 6x 24 24
François Viète (1540 – 1603)
was a lawyer at the court of Henry
IV of France and studied
equations. Viète simplified the
notation of algebra and was among
the first to use letters to represent
numbers. For centuries, algebra
and arithmetic were expressed in a
cumbersome way with words and
occasional symbols. Since the
time of Viète, algebra has gone
beyond equation solving; the
abstract nature of higher algebra
depends on its symbolic language.
7x 17 24
7x 17 17 24 17
Distributive property
Combine terms.
Add 17.
7x 7
7x 7
7
7
Divide by 7.
x 1
Check to see that 1 is the solution set.
FOR FURTHER THOUGHT
The Axioms of Equality
When we solve an equation, we must make
sure that it remains “balanced”— that is, any
operation that is performed on one side of an
equation must also be performed on the other
side in order to assure that the set of solutions
remains the same.
Underlying the rules for solving equations are
four axioms of equality, listed below. For all
real numbers a, b, and c,
aa
If a b , then b a.
If a b and b c ,
then a c .
4. Substitution axiom If a b, then a may
replace b in any
statement without
affecting the truth or
falsity of the statement.
1. Reflexive axiom
2. Symmetric axiom
3. Transitive axiom
A relation, such as equality, which satisfies
the first three of these axioms (reflexive,
symmetric, and transitive), is called an
equivalence relation.
For Group Discussion
1. Give an example of an everyday relation
that does not satisfy the symmetric axiom.
2. Does the transitive axiom hold in sports
competition, with the relation “defeats”?
3. Give an example of a relation that does not
satisfy the transitive axiom.
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7.1
Algebra dates back to the
Babylonians of 2000 B.C. The
Egyptians also worked problems in
algebra, but the problems were not
as complex as those of the
Babylonians. In about the sixth
century, the Hindus developed
methods for solving problems
involving interest, discounts, and
partnerships.
Many Hindu and Greek works
on mathematics were preserved
only because Moslem scholars
from about 750 to 1250 made
translations of them. The Arabs
took the work of the Greeks and
Hindus and greatly expanded it.
For example, Mohammed ibn
Musa al-Khowârizmî wrote books
on algebra and on the Hindu
numeration system (the one we
use) that had tremendous
influence in Western Europe; his
name is remembered today in the
word algorithm.
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Problem Solving
In the next section we solve problems involving interest rates and concentrations of solutions. These problems involve percents that are converted to decimal
numbers. The equations that are used to solve such problems involve decimal coefficients. We can clear these decimals by multiplying by the largest power
of 10 necessary to obtain integer coefficients. The next example shows how this
is done.
EXAMPLE 4
Solve .06x .0915 x .0715.
Since each decimal number is given in hundredths, multiply both sides of
the equation by 100. (This is done by moving the decimal points two places to the
right.)
.06x .0915 x .0715
.06x .0915 x .0715
Multiply by 100.
6x 915 x 715
6x 915 9x 105
Distributive property
3x 135 105
Combine like terms.
3x 135 135 105 135
Subtract 135.
3x 30
3x 30
3
3
Divide by 3.
x 10
Check to verify that the solution set is 10.
When multiplying the term .0915 x by 100 in Example 4, do not multiply
both .09 and 15 x by 100. This step is not an application of the distributive property, but of the associative property. The correct procedure is
100
.0915 x 100.09 15 x
915 x .
Associative property
Multiply.
Special Kinds of Linear Equations Each of the equations above had a
solution set containing one element; for example, 2x 1 13 has solution set 6,
containing only the single number 6. An equation that has a finite (but nonzero)
number of elements in its solution set is a conditional equation. Sometimes an
equation has no solutions. Such an equation is a contradiction and its solution set
is 0. It is also possible for an equation to have an infinite number of solutions. An
equation that is satisfied by every number for which both sides are defined is called
an identity. The next example shows how to recognize these types of equations.
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CHAPTER 7
The Basic Concepts of Algebra
EXAMPLE 5
Solve each equation. Decide whether it is a conditional equation, an identity, or a contradiction.
(a) 5x 9 4x 3
5x 9 4x 12
5x 9 4x 4x 12 4x
x 9 12
x 9 9 12 9
x 3
Distributive property
Subtract 4x.
Combine like terms.
Add 9.
The solution set, 3, has one element, so 5x 9 4x 3 is a conditional
equation.
Sofia Kovalevskaya
(1850 – 1891) was the most widely
known Russian mathematician in
the late nineteenth century. She
did most of her work in the theory
of differential equations —
equations invaluable for
expressing rates of change. For
example, in biology, the rate of
growth of a population, say of
microbes, can be precisely stated
by differential equations.
Kovalevskaya studied privately
because public lectures were not
open to women. She eventually
received a degree (1874) from the
University of Göttingen, Germany.
In 1884 she became a lecturer at
the University of Stockholm and
later was appointed professor of
higher mathematics.
Kovalevskaya was well known
as a writer. Besides novels about
Russian life, notably The Sisters
Rajevski and Vera Vorontzoff, she
wrote her Recollections of
Childhood, which has been
translated into English.
(b) 5x 15 5x 3
5x 15 5x 15
Distributive property
Both sides of the equation are exactly the same, so any real number would make
the equation true. For this reason, the solution set is the set of all real numbers,
and the equation 5x 15 5x 3 is an identity.
(c) 5x 15 5x 4
5x 15 5x 20
5x 15 5x 5x 20 5x
15 20
Distributive property
Subtract 5x.
False
Since the result, 15 20, is false, the equation has no solution. The solution set is 0. The equation 5x 15 5x 4 is a contradiction.
Formulas
The solution of a problem in algebra often depends on the use of a
mathematical statement or formula in which more than one letter is used to express
a relationship. Examples of formulas are
d rt ,
I prt ,
and
P 2L 2W .
In some applications, the necessary formula is solved for one of its variables, which
may not be the unknown number that must be found. The following examples show
how to solve a formula for any one of its variables. This process is called solving for
a specified variable. Notice how the steps used in these examples are very similar
to those used in solving a linear equation. Keep in mind that when you are solving
for a specified variable, treat that variable as if it were the only one, and treat all
other variables as if they were numbers.
Solving for a Specified Variable
Step 1: Use the addition or multiplication property as necessary to
get all terms containing the specified variable on one side of
the equation.
Step 2: All terms not containing the specified variable should be on
the other side of the equation.
(continued)
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7.1
Linear Equations
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Step 3: If necessary, use the distributive property to write the side
with the specified variable as the product of that variable and
a sum of terms.
In general, follow the steps given earlier for solving linear equations.
L
W
W
L
EXAMPLE 6
Solve the formula P 2L 2W for W.
This formula gives the relationship between the perimeter of a rectangle, P, the
length of the rectangle, L, and the width of the rectangle, W. See Figure 1.
Solve the formula for W by getting W alone on one side of the equals sign. To
begin, subtract 2L from both sides.
P 2L 2W
Perimeter, P, the sum of the
lengths of the sides of a
rectangle, is given by
P 2L 2L 2W 2L
P 2L 2W.
P 2L 2W
FIGURE 1
P 2L 2W
2
2
P 2L
W
2
Subtract 2L.
Divide by 2.
or
W
P 2L
2
Models
Sometimes equations can be used to predict future data based on past
data. Equations such as these are called models. Of course, models must be used
with care, for often there is no guarantee that past performances will continue in
the future.
EXAMPLE 7
By studying the winning times in the 500-meter speed-skating
event at Olympic games back to 1900, it was found that the winning times for men
were closely approximated by the equation
M 46.338 .097x ,
where M is the time in seconds needed to win for men, with x the Olympic year
and x 0 corresponding to 1900. (For example, 1994 winning times would be
estimated by replacing x with 1994 1900 94.) The corresponding equation
for women is
W 57.484 .196x .
(a) Find the year for which M 42.458.
M 46.338 .097x
42.458 46.338 .097x
3.88 .097x
x 40
Given equation
Let M 42.458.
Subtract 46.338.
Divide by .097.
The year 1940 corresponds to x 40.
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CHAPTER 7
The Basic Concepts of Algebra
(b) Predict the winning time for women in 2006.
The year 2006 corresponds to x 106.
W 57.484 .196x
Given equation
W 57.484 .196106
Let x 106.
W 36.708
This model predicts the winning time to be just under 37 seconds.
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7.1
EXERCISES
1. Which of the following equations are linear equations in x?
A. 3x x 1 0
B. 8 x 2
1
1
C. 6x 2 9
D.
x
0
2
x
2. Which of the equations in Exercise 1 are not linear
equations in x? Explain why.
3. Decide whether 6 is a solution of 3x 4 5x by
substituting 6 for x. If it is not a solution, explain why.
4. Use substitution to decide whether 2 is a solution
of 5x 4 3x 6 9x 1. If it is not a
solution, explain why.
5. If two equations are equivalent, they have the same
.
6. The equation 4
x 2 3x 24 4x is an
identity. Let x represent the number of letters in your
last name. Is this number a solution of this equation?
Check your answer.
7. The expression .0610 x 100 is equivalent to
A. .06 .06x
B. 60 6x
C. 6 6x
D. 6 .06x
8. Describe in your own words the steps used to solve a
linear equation.
Solve each equation.
9. 7k 8 1
10. 5m 4 21
11. 8 8x 16
12. 9 2r 15
13. 7x 5x 15 x 8
14. 2x 4 x 4x 5
15. 12w 15w 9 5 3w 5 9
16. 4t 5t 8 4 6t 4
17. 2x 3 4x 1
18. 4x 9 8x 3
19. 32w 1 2w 2 5
20. 4x 2 2x 3 6
21. 2x 3x 4 2x 3
22. 6x 35x 2 41 x
23. 6p 43 2p 5p 4 10
24. 2k 34 2k 2k 3 2
25. 2z 5z 2 2 2z 7
26. 6x 4x 8 9 6x 3
27. 3m 6 5m 1 2m 4 5m 5
28. 4k 2 8k 5 3k 9 2k 6
29. 3x 2x 5 4 32x 4 3x
30. 2
x 1 4 5 6x 7 9x
31. 9 3a 4 2a 4 2 5a a
32. 2 4x 3 4x 4 3 6x x
33. 2m 6 3m 4 4 m 4m 6