A. Solving Linear Equations Using Properties of Equality
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Guide to Solving Linear Equations in One Variable
• Eliminate parentheses using the distributive property, then combine any like terms.
• Use the additive property of equality to write the equation with all variable terms
on one side, and all constants on the other. Simplify each side.
• Use the multiplicative property of equality to obtain an equation of the form
x ϭ constant.
• For applications, answer in a complete sentence and include any units of measure
indicated.
For our first example, we’ll use the equation 31x Ϫ 12 ϩ x ϭ Ϫx ϩ 7 from our
initial discussion.
EXAMPLE 1
ᮣ
Solving a Linear Equation Using Properties of Equality
Solve for x: 31x Ϫ 12 ϩ x ϭ Ϫx ϩ 7.
Solution
ᮣ
31x Ϫ 12 ϩ x ϭ Ϫx ϩ 7
3x Ϫ 3 ϩ x ϭ Ϫx ϩ 7
4x Ϫ 3 ϭ Ϫx ϩ 7
5x Ϫ 3 ϭ 7
5x ϭ 10
xϭ2
original equation
distributive property
combine like terms
add x to both sides (additive property of equality)
add 3 to both sides (additive property of equality)
multiply both sides by 15 or divide both sides by 5
(multiplicative property of equality)
As we noted in Table R.1, the solution is x ϭ 2.
Now try Exercises 7 through 12
ᮣ
To check a solution by substitution means we substitute the solution back into the
original equation (this is sometimes called back-substitution), and verify the lefthand side is equal to the right. For Example 1 we have:
31x Ϫ 12 ϩ x ϭ Ϫx ϩ 7
original equation
312 Ϫ 12 ϩ 2 ϭ Ϫ2 ϩ 7
substitute 2 for x
3112 ϩ 2 ϭ 5
5 ϭ 5✓
simplify
solution checks
If any coefficients in an equation are fractional, multiply both sides by the least
common denominator (LCD) to clear the fractions. Since any decimal number can be
written in fraction form, the same idea can be applied to decimal coefficients.
EXAMPLE 2
ᮣ
Solution
ᮣ
A. You’ve just seen how
we can solve linear equations
using properties of equality
Solving a Linear Equation with Fractional Coefficients
Solve for n: 14 1n ϩ 82 Ϫ 2 ϭ 12 1n Ϫ 62.
1
4 1n ϩ 82
1
4n ϩ 2
Ϫ 2 ϭ 12 1n Ϫ 62
Ϫ 2 ϭ 12n Ϫ 3
1
1
4n ϭ 2n Ϫ 3
1
41 4n2 ϭ 41 12n Ϫ 32
n ϭ 2n Ϫ 12
Ϫn ϭ Ϫ12
n ϭ 12
original equation
distributive property
combine like terms
multiply both sides by LCD ϭ 4
distributive property
subtract 2n
multiply by Ϫ1
Verify the solution is n ϭ 12 using back-substitution.
Now try Exercises 13 through 30
ᮣ
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Section R.3 Solving Linear Equations and Inequalities
B. Identities and Contradictions
Example 1 illustrates what is called a conditional equation, since the equation is true for
x ϭ 2, but false for all other values of x. The equation in Example 2 is also conditional. An
identity is an equation that is always true, no matter what value is substituted for the variable. For instance, 21x ϩ 32 ϭ 2x ϩ 6 is an identity with a solution set of all real numbers, written as 5x|x ʦ ޒ6, or x ʦ 1Ϫq, q 2 in interval notation. Contradictions are
equations that are never true, no matter what real number is substituted for the variable.
The equations x Ϫ 3 ϭ x ϩ 1 and Ϫ3 ϭ 1 are contradictions. To state the solution set
for a contradiction, we use the symbol “л” (the null set) or “{ }” (the empty set). Recognizing these special equations will prevent some surprise and indecision in later chapters.
EXAMPLE 3
ᮣ
Solving Equations (Special Cases)
Solve each equation and state the solution set.
a. 21x Ϫ 42 ϩ 10x ϭ 8 ϩ 413x ϩ 12
b. 8x Ϫ 16 Ϫ 10x2 ϭ 24 ϩ 613x Ϫ 52
Solution
ᮣ
a. 21x Ϫ 42 ϩ 10x ϭ 8 ϩ 413x ϩ 12
2x Ϫ 8 ϩ 10x ϭ 8 ϩ 12x ϩ 4
12x Ϫ 8 ϭ 12x ϩ 12
Ϫ8 ϭ 12
original equation
distributive property
combine like terms
subtract 12x; contradiction
Since Ϫ8 is never equal to 12, the original equation is a contradiction.
The solution set is empty: { }
b. 8x Ϫ 16 Ϫ 10x2 ϭ 24 ϩ 613x Ϫ 52
8x Ϫ 6 ϩ 10x ϭ 24 ϩ 18x Ϫ 30
18x Ϫ 6 ϭ 18x Ϫ 6
Ϫ6 ϭ Ϫ6
original equation
distributive property
combine like terms
subtract 18x; identity
The result shows that the original equation is an identity, with an infinite
number of solutions: 5x | x ʦ ޒ6 . You may recall this notation is read, “the set
of all numbers x, such that x is a real number.”
Now try Exercises 31 through 36
B. You’ve just seen how
we can recognize equations
that are identities or
contradictions
ᮣ
In Example 3(a), our attempt to solve for x ended with all variables being eliminated,
leaving an equation that is always false —a contradiction (Ϫ8 is never equal to 12).
There is nothing wrong with the solution process, the result is simply telling us the
original equation has no solution. In Example 3(b), all variables were again eliminated
but the end result was always true —an identity (Ϫ6 is always equal to Ϫ6). Once
again we’ve done nothing wrong mathematically, the result is just telling us that the
original equation will be true no matter what value of x we use for an input.
C. Solving Linear Inequalities
A linear inequality resembles a linear equality in many respects:
Linear Inequality
Related Linear Equation
(1)
x 6 3
xϭ3
(2)
3
p Ϫ 2 Ն Ϫ12
8
3
p Ϫ 2 ϭ Ϫ12
8
A linear inequality in one variable is one that can be written in the form ax ϩ b 6 c,
where a, b, and c ʦ ޒand a 0. This definition and the following properties also apply
when other inequality symbols are used. Solutions to simple inequalities are easy to spot.
For instance, x ϭ Ϫ2 is a solution to x 6 3 since Ϫ2 6 3. For more involved inequalities we use the additive property of inequality and the multiplicative property of
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inequality. Similar to solving equations, we solve inequalities by isolating the variable
on one side to obtain a solution form such as variable 6 number.
The Additive Property of Inequality
If A, B, and C represent algebraic expressions and A 6 B,
then A ϩ C 6 B ϩ C
Like quantities (numbers or terms) can be added to both sides of an inequality.
While there is little difference between the additive property of equality and the
additive property of inequality, there is an important difference between the multiplicative property of equality and the multiplicative property of inequality. To illustrate, we
begin with Ϫ2 6 5. Multiplying both sides by positive three yields Ϫ6 6 15, a true
inequality. But notice what happens when we multiply both sides by negative three:
Ϫ2 6 5
original inequality
Ϫ21Ϫ32 6 51Ϫ32
6 6 Ϫ15
multiply by negative three
false
The result is a false inequality, because 6 is to the right of Ϫ15 on the number line.
Multiplying (or dividing) an inequality by a negative quantity reverses the order relationship between two quantities (we say it changes the sense of the inequality). We
must compensate for this by reversing the inequality symbol.
6 7 Ϫ15
change direction of symbol to maintain a true statement
For this reason, the multiplicative property of inequality is stated in two parts.
The Multiplicative Property of Inequality
EXAMPLE 4
ᮣ
If A, B, and C represent algebraic
expressions and A 6 B,
then AC 6 BC
If A, B, and C represent algebraic
expressions and A 6 B,
then AC 7 BC
if C is a positive quantity
(inequality symbol remains the same).
if C is a negative quantity
(inequality symbol must be reversed).
Solving an Inequality
Solve the inequality, then graph the solution set and write it in interval notation:
Ϫ2
1
5
3 x ϩ 2 Յ 6.
Solution
WORTHY OF NOTE
As an alternative to multiplying or
dividing by a negative value, the
additive property of inequality can
be used to ensure the variable term
will be positive. From Example 4,
the inequality Ϫ4x Յ 2 can be
written as Ϫ2 Յ 4x by adding 4x to
both sides and subtracting 2 from
both sides. This gives the solution
Ϫ12 Յ x, which is equivalent to
xՆ
Ϫ12 .
ᮣ
1
5
Ϫ2
xϩ Յ
3
2
6
Ϫ2
1
5
6 a x ϩ b Յ 162
3
2
6
Ϫ4x ϩ 3 Յ 5
Ϫ4x Յ 2
Ϫ4x
2
Ն
Ϫ4
Ϫ4
1
xՆϪ
2
original inequality
clear fractions (multiply by LCD)
simplify
subtract 3
divide by Ϫ4, reverse inequality sign
result
Ϫ1
2
• Graph:
Ϫ3 Ϫ2 Ϫ1
[
0
1
2
3
• Interval notation: x ʦ 3Ϫ12, q 2
4
Now try Exercises 37 through 46
ᮣ
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Section R.3 Solving Linear Equations and Inequalities
29
To check a linear inequality, you often have an infinite number of choices—any
number from the solution set/interval. If a test value from the solution interval results
in a true inequality, all numbers in the interval are solutions. For Example 4, using
x ϭ 0 results in the true statement 12 Յ 56 ✓.
Some inequalities have all real numbers as the solution set: 5x| x ʦ ޒ6, while other
inequalities have no solutions, with the answer given as the empty set: { }.
EXAMPLE 5
ᮣ
Solving Inequalities
Solve the inequality and write the solution in set notation:
a. 7 Ϫ 13x ϩ 52 Ն 21x Ϫ 42 Ϫ 5x
b. 31x ϩ 42 Ϫ 5 6 21x Ϫ 32 ϩ x
Solution
ᮣ
a. 7 Ϫ 13x ϩ 52 Ն 21x Ϫ 42 Ϫ 5x
7 Ϫ 3x Ϫ 5 Ն 2x Ϫ 8 Ϫ 5x
Ϫ3x ϩ 2 Ն Ϫ3x Ϫ 8
2 Ն Ϫ8
original inequality
distributive property
combine like terms
add 3x
Since the resulting statement is always true, the original inequality is true for
all real numbers. The solution is all real numbers ޒ.
b. 31x ϩ 42 Ϫ 5
3x ϩ 12 Ϫ 5
3x ϩ 7
7
6
6
6
6
21x Ϫ 32 ϩ x
2x Ϫ 6 ϩ x
3x Ϫ 6
Ϫ6
original inequality
distribute
combine like terms
subtract 3x
Since the resulting statement is always false, the original inequality is false for
all real numbers. The solution is { }.
C. You’ve just seen how
we can solve linear inequalities
Now try Exercises 47 through 52
ᮣ
D. Solving Compound Inequalities
In some applications of inequalities, we must consider more than one solution interval.
These are called compound inequalities, and these require us to take a close look at
the operations of union and intersection ă. The intersection of two sets A and
B, written A ă B, is the set of all elements common to both sets. The union of two sets
A and B, written A ´ B, is the set of all elements that are in either set. When stating the
union of two sets, repetitions are unnecessary.
EXAMPLE 6
ᮣ
Finding the Union and Intersection of Two Sets
Solution
A ă B is the set of all elements in both A and B:
A ʝ B ϭ 51, 2, 36.
A ´ B is the set of all elements in either A or B:
A ´ B ϭ 5Ϫ2, Ϫ1, 0, 1, 2, 3, 4, 56.
WORTHY OF NOTE
For the long term, it may help to
rephrase the distinction as follows.
The intersection is a selection of
elements that are common to two
sets, while the union is a collection
of the elements from two sets (with
no repetitions).
For set A ϭ 5Ϫ2, Ϫ1, 0, 1, 2, 36 and set B ϭ 51, 2, 3, 4, 56,
determine A ă B and A B.
Now try Exercises 53 through 58
ᮣ
Notice the intersection of two sets is described using the word “and,” while the
union of two sets is described using the word “or.” When compound inequalities are
formed using these words, the solution is modeled after the ideas from Example 6. If
“and” is used, the solutions must satisfy both inequalities. If “or” is used, the solutions
can satisfy either inequality.
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EXAMPLE 7
ᮣ
Solving a Compound Inequality
Solve the compound inequality, then write the solution in interval notation:
Ϫ3x Ϫ 1 6 Ϫ4 or 4x ϩ 3 6 Ϫ6.
Solution
ᮣ
Begin with the statement as given:
Ϫ3x Ϫ 1 6 Ϫ4
Ϫ3x 6 Ϫ3
x 7 1
4x ϩ 3 6 Ϫ6
4x 6 Ϫ9
9
x 6 Ϫ
4
or
or
or
original statement
isolate variable term
solve for x, reverse first inequality symbol
The solution x 7 1 or x 6 Ϫ94 is better understood by graphing each
interval separately, then selecting both intervals (the union).
WORTHY OF NOTE
x Ͼ 1:
The graphs from Example 7 clearly
show the solution consists of two
disjoint (disconnected) intervals.
This is reflected in the “or”
statement: x 6 Ϫ94 or x 7 1, and in
the interval notation. Also, note the
solution x 6 Ϫ94 or x 7 1 is not
equivalent to Ϫ94 7 x 7 1, as there
is no single number that is both
greater than 1 and less than Ϫ94 at
the same time.
x Ͻ Ϫ9 :
4
)
Ϫ8 Ϫ7 Ϫ6 Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1
0
1
2
3
4
5
6
Ϫ9
4
)
Ϫ8 Ϫ7 Ϫ6 Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1
x Ͻ Ϫ 9 or x Ͼ 1:
4
0
1
2
3
4
5
6
Ϫ9
4
)
)
Ϫ8 Ϫ7 Ϫ6 Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1
0
1
2
3
4
5
6
9
Interval notation: x ʦ aϪq, Ϫ b ´ 11, q 2.
4
Now try Exercises 59 and 60
EXAMPLE 8
ᮣ
ᮣ
Solving a Compound Inequality
Solve the compound inequality, then write the solution in interval notation:
3x ϩ 5 7 Ϫ13 and 3x ϩ 5 6 Ϫ1.
Solution
ᮣ
Begin with the statement as given:
3x ϩ 5 7 Ϫ13
3x 7 Ϫ18
x 7 Ϫ6
3x ϩ 5 6 Ϫ1
3x 6 Ϫ6
x 6 Ϫ2
and
and
and
original statement
subtract five
divide by 3
The solution x 7 Ϫ6 and x 6 Ϫ2 can best be understood by graphing each interval
separately, then noting where they intersect.
WORTHY OF NOTE
x Ͼ Ϫ6:
x Ͻ Ϫ2:
x Ͼ Ϫ6 and x Ͻ Ϫ2:
)
Ϫ8 Ϫ7 Ϫ6 Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1
)
Ϫ8 Ϫ7 Ϫ6 Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1
)
The inequality a 6 b (a is less than
b) can equivalently be written as
b 7 a (b is greater than a). In
Example 8, the solution is read,
“x 7 Ϫ6 and x 6 Ϫ2,” but if we
rewrite the first inequality as
Ϫ6 6 x (with the “arrowhead” still
pointing at Ϫ62, we have Ϫ6 6 x
and x 6 Ϫ2 and can clearly see
that x must be in the single interval
between Ϫ6 and Ϫ2.
0
1
2
3
4
5
6
0
1
2
3
4
5
6
0
1
2
3
4
5
6
)
Ϫ8 Ϫ7 Ϫ6 Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1
Interval notation: x ʦ 1Ϫ6, Ϫ22.
Now try Exercises 61 through 72
ᮣ
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Section R.3 Solving Linear Equations and Inequalities
The solution from Example 8 consists of the single interval 1Ϫ6, Ϫ22, indicating the original inequality could actually be joined and written as Ϫ6 6 x 6 Ϫ2,
called a joint inequality. We solve joint inequalities in much the same way as linear inequalities, but must remember they have three parts (left, middle, and right).
This means operations must be applied to all three parts in each step of the solution
process, to obtain a solution form such as smaller number 6 x 6 larger number.
The same ideas apply when other inequality symbols are used.
EXAMPLE 9
ᮣ
Solving a Joint Inequality
Solve the joint inequality, then graph the solution set and write it in interval
2x ϩ 5
Ն Ϫ6.
notation: 1 7
Ϫ3
Solution
ᮣ
2x ϩ 5
Ն Ϫ6
Ϫ3
Ϫ3 6 2x ϩ 5 Յ 18
Ϫ8 6 2x Յ 13
13
Ϫ4 6 x Յ
2
1 7
original inequality
multiply all parts by Ϫ3; reverse the inequality symbols
subtract 5 from all parts
divide all parts by 2
13
2
• Graph:
)
Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1
[
0
• Interval notation: x ʦ 1Ϫ4,
D. You’ve just seen how
we can solve compound
inequalities
1
13
2 4
2
3
4
5
6
7
8
Now try Exercises 73 through 78
ᮣ
E. Solving Basic Applications of Linear
Equations and Inequalities
Applications of linear equations and inequalities come in many forms. In most cases,
you are asked to translate written relationships or information given verbally into an
equation using words or phrases that indicate mathematical operations or relationships.
Here, we’ll practice this skill using ideas that were introduced in Section R.1, where
we translated English phrases into mathematical expressions. Very soon these skills
will be applied in much more significant ways.
EXAMPLE 10
ᮣ
Translating Written Information into an Equation
Translate the following relationships into equations, then solve:
In an effort to lower the outstanding balance on her credit card, Laura paid $10 less
than triple her normal payment. If she sent the credit card company $350.75, how
much was her normal payment? (See Section R.1, Examples 2 and 3.)
Solution
ᮣ
Let p represent her normal payment. Then “triple her normal payment” would be
3p, and “ten less than triple” would be 3p Ϫ 10. Since “she sent the company
$350.75,” we have
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3p Ϫ 10 ϭ 350.75
3p ϭ 360.75
p ϭ 120.25
equation form
add 10
divide by 3
Laura’s normal payment is $120.25 per month.
A calculator check is shown in the figure.
Now try Exercises 83 through 90
ᮣ
Inequalities are widely used to help gather information, and to make comparisons
that will lead to informed decisions.
EXAMPLE 11
ᮣ
Using an Inequality to Compute Desired Test Scores
Justin earned scores of 78, 72, and 86 on the first three out of four exams. What
score must he earn on the fourth exam to have an average of at least 80?
Solution
ᮣ
The current scores are 78, 72, and 86. An average of at least 80 means A Ն 80.
In organized form:
Test 1
Test 2
Test 3
Test 4
Computed Average
Minimum
78
72
86
x
78 ϩ 72 ϩ 86 ϩ x
4
80
Let x represent Justin’s score on the fourth exam, then
represents his average score.
78 ϩ 72 ϩ 86 ϩ x
Ն 80
4
78 ϩ 72 ϩ 86 ϩ x Ն 320
236 ϩ x Ն 320
x Ն 84
E. You’ve just seen how
we can solve applications
of linear equations and
inequalities
78 ϩ 72 ϩ 86 ϩ x
4
average must be greater than or equal to 80
multiply by 4
simplify
solve for x (subtract 236)
Justin must score at least an 84 on the last exam to earn an 80 average.
Now try Exercises 91 through 100
ᮣ
F. Solving Applications of Basic Geometry
As your translation skills grow, your ability to solve a wider range of more significant
applications will grow as well. In many cases, the applications will involve some basic
geometry and the most often used figures and formulas appear here. For a more complete review of geometry, see Appendix II Geometry Review with Unit Conversions,
which is posted online at www.mhhe.com/coburn.
Perimeter and Area
Perimeter is a measure of the distance around a two dimensional figure. As this is a linear measure, results are stated in linear units as in centimeters (cm), feet (ft), kilometers (km), miles (mi), and so on. If no unit is specified, simply write the result as x
units. Area is a measure of the surface of a two dimensional figure, with results stated
in square units as in x units2. Some of the most common formulas involving perimeter
and area are given in Table R.2A.