Download Chapter 1 Section 1 - Canton Local Schools

Chapter 1
Equations and
Inequalities
© 2010 Pearson Education, Inc.
All rights reserved
© 2010 Pearson Education, Inc. All rights reserved
1
SECTION 1.1
Linear Equations in One Variable
OBJECTIVES
1
Learn the vocabulary and concepts used in
studying equations.
2
Solve linear equations in one variable.
3
Solve rational equations with variables in the
denominators.
4
Solve formulas for a specific variable.
5
Solve applied problems by using linear equations.
© 2010 Pearson Education, Inc. All rights reserved
2
Definitions
An equation is a statement that two
mathematical expressions are equal. For
example, 7  5  2.
An equation in one variable, is a statement
that two expressions, with at least one
containing the variable, are equal. For example,
2 x  3  7.
The domain of the variable in an equation is
the set of all real numbers for which both sides
of the equation are defined.
© 2010 Pearson Education, Inc. All rights reserved
3
EXAMPLE 1a Finding the Domain of the Variable
Find the domain of the variable x in each of the
following equations.
5
a. 2 
x 1
Solution
5
a. 2 
x 1
Simply write
Left side is defined for all
values of x.
Right side is not defined if x = 1.
5
2
, x  1.
x 1
© 2010 Pearson Education, Inc. All rights reserved
4
EXAMPLE 1b Finding the Domain of the Variable
Find the domain of the variable x in each of the
following equations.
b. x  2  x
Solution
b. x  2  x
Right side is defined for x ≥ 0.
Left side is is defined for all
values of x.
The domain is  x x  0.
Interval notation  0, 
© 2010 Pearson Education, Inc. All rights reserved
5
EXAMPLE 1c Finding the Domain of the Variable
Find the domain of the variable x in each of the
following equations.
c. 2 x  3  7
Solution
c. 2 x  3  7 Both sides are defined for all
values of x.
The domain is  x x is a real number.
Interval notation
 ,  
© 2010 Pearson Education, Inc. All rights reserved
6
LINEAR EQUATIONS
A conditional linear equation in one
variable, such as x, is an equation that can
be written in the standard form
ax  b  0,
where a and b are real numbers with a ≠ 0.
© 2010 Pearson Education, Inc. All rights reserved
7
PROCEDURE FOR SOLVING LINEAR
EQUATIONS IN ONE VARIABLE
Step 1 Eliminate Fractions. Multiply both sides
of the equation by the least common
denominator (LCD) of all the fractions.
Step 2 Simplify. Simplify both sides of the
equation by removing parentheses and
other grouping symbols (if any) and
combining like terms.
© 2010 Pearson Education, Inc. All rights reserved
8
PROCEDURE FOR SOLVING LINEAR
EQUATIONS IN ONE VARIABLE
Step 3 Isolate the Variable Term. Add
appropriate expressions to both sides, so
that when both sides are simplified, the
terms containing the variable are on one
side and all constant terms are on the other
side.
Step 4 Combine Terms. Combine terms
containing the variable to obtain one term
that contains the variable as a factor.
© 2010 Pearson Education, Inc. All rights reserved
9
PROCEDURE FOR SOLVING LINEAR
EQUATIONS IN ONE VARIABLE
Step 5 Isolate the Variable. Divide both sides by
the coefficient of the variable to obtain the
solution.
Step 6 Check the Solution. Substitute the
solution into the original equation.
© 2010 Pearson Education, Inc. All rights reserved
10
EXAMPLE 3
Solving a Linear Equation
Solve: 6 x  3 x  2  x  2    11.
Solution
Step 2
6 x  3 x  2  x  2    11
6 x  3 x  2 x  4  11
6 x  3 x  2 x  4  11
5 x  4  11
Step 3
Step 4
5x  4  4  11  4
5 x  15
© 2010 Pearson Education, Inc. All rights reserved
11
EXAMPLE 3
Solving a Linear Equation
Solve: 6 x  3 x  2  x  2    11.
Solution continued
5
x
15
Step 5

5
5
x3
Step 6 Check: 6  3  3  3  2   3  2    11
18  9  2  11
18  7  11
The solution set is {3}.
© 2010 Pearson Education, Inc. All rights reserved
12
EXAMPLE 4
Solving a Linear Equation
Solve: 1  5 y  2  y  7   2 y  5  3  y  .
Solution
Step 2
1  5 y  2  y  7   2 y  5 3  y 
Step 3
1  5 y  2 y  14  2 y  15  5 y
15  3 y  15  3 y
15  3 y  3 y  15  3 y  3 y
15  15
00
© 2010 Pearson Education, Inc. All rights reserved
13
EXAMPLE 4
Solving a Linear Equation
Solve: 1  5 y  2  y  7   2 y  5  3  y .
Solution continued
0 = 0 is equivalent to the original equation.
The equation 0 = 0 is always true and its
solution set is the set of real numbers. So the
solution set to the original equation is the set
of real numbers. The original equation is an
identity.
© 2010 Pearson Education, Inc. All rights reserved
14
TYPES OF LINEAR EQUATIONS
There are three types of linear equations.
1. A linear equation that is satisfied by all
values in the domain of the variable is an
identity. For example, 2(x –1) = 2x – 2.
2. A linear equation that is not an identity,
but is satisfied by at least one number is
a conditional equation. For example,
2x = 6.
© 2010 Pearson Education, Inc. All rights reserved
15
TYPES OF LINEAR EQUATIONS
3. A linear equation that is not satisfied for
any value of the variable is an inconsistent
equation. For example, x = x + 2 is an
inconsistent equation. Since no number is
2 more than itself, the solution set of the
equation x = x + 2 is . When you try to
solve an inconsistent equation, you will
obtain a false statement, such as 0 = 2.
© 2010 Pearson Education, Inc. All rights reserved
16
RATIONAL EQUATIONS
If at least one rational expression appears in
an equation, then the equation is called a
rational equation.
© 2010 Pearson Education, Inc. All rights reserved
17
RATIONAL EQUATIONS
A solution that satisfies the new equation but
does not satisfy the original equation is called
an extraneous solution or extraneous root.
So whenever we multiply an equation by an
expression containing the variable, we must
check all solutions obtained to reject
extraneous solutions (if any).
© 2010 Pearson Education, Inc. All rights reserved
18
EXAMPLE 5
Solving a Conditional Rational Equation
1 1 3 1
Solve
 
 .
x 3 4x 2
Solution
Step 1
Step 2
 1 1
 3 1
12 x     12 x   
 x 3
 4x 2 
12  4 x  9  6 x
© 2010 Pearson Education, Inc. All rights reserved
19
EXAMPLE 5
Solving a Conditional Rational Equation
Solution continued
Step 3
12  4 x  12  6 x  9  6 x  12  6 x
Step 4
2 x  3
3 3
x

Step 5
2 2
3
The apparent solution is x  .
2
© 2010 Pearson Education, Inc. All rights reserved
20
EXAMPLE 5
Solving a Conditional Rational Equation
Solution continued
Step 6
Check:
1 1? 3
1
 

3 3 3 2
4 
2
2
2 1?1 1
  
3 3 2 2
?
1 1
3
The solution set is   .
2
© 2010 Pearson Education, Inc. All rights reserved
21
EXAMPLE 6
Solve for x:
Solving a Rational Equation
1
1
2

 2
x 1 x 1 x 1
Solution
Step 1
1 
 1
 2 

  x  1 x  1  2 
 x  1 x  1 

 x  1 x  1
 x  1
Step 2
 2 
1
1
2
  x  1  x  1
  x  1  2
 x  1  x  1

x 1
x 1
 x 1 
 x  1   x  1  2
© 2010 Pearson Education, Inc. All rights reserved
22
EXAMPLE 6
Solving a Rational Equation
Solution continued
Step 3
x 1 x 1  2
Step 4
22
The equation 2 = 2 is always true for all values
of x in its domain.
The domain of x is all real numbers except −1
and 1. Therefore, the solution set is
 ,  1   1, 1  1,  .
© 2010 Pearson Education, Inc. All rights reserved
23
EXAMPLE 7
Solving an Inconsistent Rational
Equation
5
6
2m
Solve for m: 

3 m3 m3
Solution
Step 1
6 
5
 2m 
3  m  3  
 3  m  3 


3
m

3
m

3




Step 2
5
6
2m
3  m  3  3  m  3 
 3  m  3 
3
m3
m3
5m  15  18  6m
5m  3  6m
© 2010 Pearson Education, Inc. All rights reserved
24
EXAMPLE 6
Solving an Inconsistent Rational
Equation
Solution continued
Steps 3-5
Step 6
5m  3  5 m  6 m  5 m
3m
5
6 ? 2  3
Check:


3 33 33
5 6 ? 6
 
3 0 0
Because dividing by zero is undefined, we
reject m = 3 as a solution. The solution set is .
© 2010 Pearson Education, Inc. All rights reserved
25
EXAMPLE 8
Converting temperatures
The formula for converting the temperature in
degrees Celsius (C) to degrees Fahrenheit (F) is
9
F  C  32.
5
If the temperature shows 86° Fahrenheit, what
is the temperature in degrees Celsius?
Solution
9
86  C  32
Substitute 86 for F.
5
9

5  86   5  C  32 
Solve for C.
5

© 2010 Pearson Education, Inc. All rights reserved
26
EXAMPLE 8
Converting temperatures
Solution continued
Solve for C.
9

5  86   5  C  32 
5

430  9C  160
270  9C
270
C
9
30  C
Thus, 86º F converts to 30º C.
© 2010 Pearson Education, Inc. All rights reserved
27
EXAMPLE 9
Solve
Solving for a Specified Variable
9
F  C  32
5
for C.
9


Solution
5 F  5  C  32 
5

5 F  9C  5  32
5 F  5  32  9C  5  32  5  32
5 F  5  32 9C

9
9
5
C   F  32 
9
© 2010 Pearson Education, Inc. All rights reserved
28
EXAMPLE 10 Bungee-Jumping TV Contestants
Suppose that a television adventure program has
its contestants bungee jump from a bridge 120
feet above the water. The bungee cord that is
used has a 140-150% elongation, meaning that
its extended length will be the original length
plus a maximum of an additional 150% of its
original length. The show’s producer wants to be
sure that the jumper doesn’t get closer than 10
feet to the water, and no contestant will be more
than 7 feet tall. How long can the bungee cord
be?
© 2010 Pearson Education, Inc. All rights reserved
29
EXAMPLE 6
Bungee-Jumping TV Contestants
Solution
Let x = length, in feet, of the cord to be used
x + 1.5x = 2.5x = maximum extended length
Height of
Extended
Body
10-foot
bridge above
+
+
=
cord length
length
buffer
the water
2.5 x  7  10  120
2.5 x  103
x  41.2 feet
A cord length of 41.2 feet will meet all conditions.
© 2010 Pearson Education, Inc. All rights reserved
30