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2. Equations and
Inequalities
2.1
Equations
Copyright © Cengage Learning. All rights reserved.
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Equations
An equation (or equality) is a statement that two quantities
or expressions are equal. Equations are employed in every
field that uses real numbers. As an illustration, the equation
d = rt,
or
distance = (rate)(time),
is used in solving problems involving an object moving at a
constant rate of speed.
If the rate r is 45 mihr (miles per hour), then the distance d
(in miles) traveled after time t (in hours) is given by
d = 45t.
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The following chart applies to a variable x, but any other
variable may be considered. The abbreviations LS and RS
in the second illustration stand for the equation’s left side
and right side, respectively.
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An algebraic equation in x contains only algebraic
expressions such as polynomials, rational expressions,
radicals, and so on.
An equation of this type is called a conditional equation if
there are numbers in the domains of the expressions that
are not solutions.
For example, the equation x2 = 9 is conditional, since the
number x = 4 (and others) is not a solution. If every number
in the domains of the expressions in an algebraic equation
is a solution, the equation is called an identity.
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The most basic equation in algebra is the linear equation,
defined in the next chart, where a and b denote real
numbers.
The illustration in the preceding chart indicates a typical
method of solving a linear equation.
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Following the same procedure, we see that
if
ax + b = 0,
then
x=
,
provided a ≠ 0. Thus, a linear equation has exactly one
solution.
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Example 1 – Solving a linear equation
Solve the equation 6x – 7 = 2x + 5.
Solution:
The equations in the following list are equivalent:
6x – 7 = 2x + 5
given
(6x – 7) + 7 = (2x + 5) + 7
add 7
6x = 2x + 12
simplify
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Example 1 – Solution
6x – 2x = (2x + 12) – 2x
4x = 12
cont’d
subtract 2x
simplify
divide by 4
x=3
simplify
Check:
x = 3 LS: 6(3) – 7 = 18 – 7 = 11
RS: 2(3) + 5 = 6 + 5 = 11
Since 11 = 11 is a true statement, x = 3 checks as a
solution.
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The next example illustrates that a seemingly complicated
equation may simplify to a linear equation.
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Example 2 – Solving an equation
Solve the equation (8x – 2)(3x + 4) = (4x + 3)(6x – 1).
Solution:
The equations in the following list are equivalent:
(8x – 2)(3x + 4) = (4x + 3)(6x – 1)
given
24x2 + 26x – 8 = 24x2 + 14x – 3
multiply factors
26x – 8 = 14x – 3
subtract 24x2
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Example 2 – Solution
12x – 8 = –3
12x = 5
x=
cont’d
subtract 14x
add 8
divide by 12
Hence, the solution of the given equation is
.
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If an equation contains rational expressions, we often
eliminate denominators by multiplying both sides by the lcd
of these expressions.
If we multiply both sides by an expression that equals zero
for some value of x, then the resulting equation may not be
equivalent to the original equation, as illustrated in the next
example.
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Example 3 – An equation with no solutions
Solve the equation
.
Solution:
given
(x – 2) = (1)(x – 2) +
(x – 2)
multiply by x – 2
3x = (x – 2) + 6
simplify
3x = x + 4
simplify
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Example 3 – Solution
cont’d
2x = 4
subtract x
x=2
divide by 2
Check:
x=2
LS:
Since division by 0 is not permissible, x = 2 is not a
solution. Hence, the given equation has no solutions.
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In the process of solving an equation, we may obtain, as a
possible solution, a number that is not a solution of the
given equation.
Such a number is called an extraneous solution or
extraneous root of the given equation.
In Example 3, x = 2 is an extraneous solution (root) of the
given equation.
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The following guidelines may also be used to solve the
equation in Example 3. In this case, observing guideline 2
would make it unnecessary to check the extraneous
solution x = 2.
Guidelines for Solving an Equation Containing Rational
Expressions
1. Determine the lcd of the rational expressions.
2. Find the values of the variable that make the lcd zero.
These are not solutions, because they yield at least one
zero denominator when substituted into the given
equation.
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3. Multiply each term of the equation by the lcd and
simplify, thereby eliminating all of the denominators.
4. Solve the equation obtained in guideline 3.
5. The solutions of the given equation are the solutions
found in guideline 4, with the exclusion of the values
found in guideline 2.
We shall follow these guidelines in the next example.
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Example 4 – An equation containing rational expressions
Solve the equation
Solution:
Guideline 1:
Rewriting the denominator 2x – 4 as 2(x – 2), we see that
the lcd of the three rational expressions is 2(x – 2)(x + 3).
Guideline 2:
The values of x that make the lcd 2(x – 2)(x + 3) zero are 2
and –3, so these numbers cannot be solutions of the
equation.
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Example 4 – Solution
cont’d
Guideline 3:
Multiplying each term of the equation by the lcd and
simplifying gives us the following:
2(x – 2)(x + 3) –
2(x – 2)(x + 3)
=
2(x – 2)(x + 3)
3(x + 3) – 10(x – 2) = 4(x + 3)
3x + 9 – 10x + 20 = 4x + 12
cancel like factors
multiply factors
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Example 4 – Solution
cont’d
Guideline 4:
We solve the last equation obtained in guideline 3.
3x – 10x – 4x = 12 – 9 – 20
–11x = –17
x=
subtract 4x, 9, and 20
combine like terms
divide by –11
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Example 4 – Solution
cont’d
Guideline 5:
Since is not included among the values (2 and –3) that
make the lcd zero (guideline 2), we see that x = is a
solution of the given equation.
We shall not check the solution x = by substitution,
because the arithmetic involved is complicated.
It is simpler to carefully check the algebraic manipulations
used in each step.
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Formulas involving several variables occur in many
applications of mathematics.
Sometimes it is necessary to solve for a specific variable in
terms of the remaining variables that appear in the formula,
as the next example illustrates.
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Example 5 – Relationship between temperature scales
The Celsius and Fahrenheit temperature scales are shown
on the thermometer in Figure 2.
Figure 2
The relationship between the temperature readings C and
F is given by C = (F – 32). Solve for F.
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Example 5 – Solution
To solve for F we must obtain a formula that has F by itself
on one side of the equals sign and does not have F on the
other side. We may do this as follows:
C=
(F – 32)
C = F – 32
C + 32 = F
F=
given
multiply by
add 32
C + 32
equivalent equation
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We can make a simple check of our result in Example 5 as
follows.
Start with C = (F – 32) and substitute 212 (an arbitrary
choice) for F to obtain 100 for C.
Now let C = 100 in F =
C + 32 to get F = 212.
Again, this check does not prove we are correct, but
certainly lends credibility to our result.
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