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Barnett/Ziegler/Byleen
Precalculus: Functions & Graphs, 5th Edition
Chapter One
Equations & Inequalities
Copyright © 2001 by the McGraw-Hill Companies, Inc.
Properties of Equality
1. If a = b, then a + c = b + c.
Addition Property
2. If a = b, then a – c = b – c.
Subtraction Property
3. If a = b, then ca = cb, c  0.
Multiplication Property
a b
4. If a = b, then c = c , c  0.
Division Property
5. If a = b, then either may replace the other in
any statement without changing the truth
or falsity of the statement.
Substitution Property
1-1-1
Strategy for Solving Word Problems
1. Read the problem carefully—several times if necessary; that is, until you
understand the problem, know what is to be found, and know what is given.
2. Let one of the unknown quantities be represented by a variable, say x, and try
to represent all other unknown quantities in terms of x. This is an important
step and must be done carefully.
3. If appropriate, draw figures or diagrams and label known and unknown parts.
4. Look for formulas connecting the known quantities with the unknown quantities.
5. Form an equation relating the unknown quantities to the known quantities.
6. Solve the equation and write answers to all questions asked in the problem.
7. Check and interpret all solutions in terms of the original problem—not just
the equation found in step 5—since a mistake may have been made in
setting up the equation in step 5.
1-1-2
Quantity-Rate-Time Formulas
Q
R= T
Quantity
Rate = Time
Q = RT
Quantity = (Rate)(Time)
Q
T=R
Time =
Quantity
Rate
If Q is distance D, then
D
R=T
D = RT
D
T=R
[Note: R is an average or uniform rate.]
1-1-3
Systems of Linear Equations
ax + by = h System of two linear
cx + dy = k equations in two variables
Note that x and y are variables and a, b, c, d, h, and k are real constants.
A pair of numbers x = x0 and y = y0 is a solution of this system if each
equation is satisfied by the pair. The set of all such pairs of numbers is
called the solution set for the system. To solve a system is to find its
solution set.
To solve a system by substitution, choose one of the two equations in a
system and solve for one variable in terms of the other.
Then substitute the result in the other equation and solve the resulting
linear equation in one variable.
Finally, substitute this result back into the expression obtained in the
first step to find the second variable.
1-2-4
Interval Notation
Interval
Notation
Inequality
Notation
[a, b]
a x b
[a, b)
a x<b
(a, b]
a<x b
(a, b)
a<x<b
Line Graph
Type
[
]
x
Closed
[
)
x
Half-open
(
]
x
Half-open
(
)
x
Open
a
a
a
a
b
b
b
b
1-3-5-1
Interval Notation
Interval
Notation
[b ,  )
( b,  )
( –, a]
( –, a)
Inequality
Notation
Line Graph
Type
x b
[
b
x
x
x> b
(
b
x a
x< a
Closed
Open
]
a
x
)
a
x
Closed
Open
1-3-5-2
Inequality Properties
For a, b, and c any real numbers:
1. If a < b and b < c, then a < c.
Transitive Property
2. If a < b, then a + c < b + c.
Addition Property
3. If a < b, then a – c < b – c.
Subtraction Property


5. If a < b and c is negative, then ca > cb. 
Multiplication Property
(Note difference between
4 and 5.)



Division Property
(Note difference between
6 and 7.)
4. If a < b and c is positive, then ca < cb.
a b
6. If a < b and c is positive, then c < .
c
a b
7. If a < b and c is negative, then c > c .
1-3-6
Absolute Value Equations
and Inequalities
d
| x – c | =d
{c – d, c + d}
| x – c | <d
(c – d, c + d)
c–d
c–d
d
x
c
c+d
x
c
c+d
( c – d , c )  c,
( c +d)
0 < |x – c | <d
x
c–d
c
c+d
(  , c – d )  c( + d , )
| x – c | >
d
x
c–d
c
c+d
1-4-7
Particular Kinds of Complex Numbers
Imaginary Unit:
i
Complex Number:
a + bi
a and b real numbers
Imaginary Number:
a + bi
b0
Pure Imaginary Number:
0 + bi = bi
b0
Real Number:
a + 0i = a
Zero:
0 + 0i = 0
Conjugate of a + bi :
a – bi
1-5-8
Subsets of the Set of Complex Numbers
Natural
numbers (N)
Zero
Integers (Z)
Rational
numbers (Q)
Negative
Integers
Noninteger
rational
numbers
Real
numbers (R)
Irrational
numbers (I)
Imaginary
numbers
Complex
numbers (C)
N  Z  Q  R  C
1-5-9
Quadratic Formula
If ax2 + bx + c = 0, a  0, then
x =
–b ±
b2 – 4ac
2a
Discriminant and Roots
Roots of ax2 + bx + c = 0
Discrimant
b 2 – 4ac
a , b , and c real numbers, a  0
Positive
Two distinct real roots
0
One real root (a double root)
Negative
Two imaginary roots, one the conjugate of the other
1-6-10
Squaring Operation on Equations
If both sides of an equation are squared, then the solution
set of the original equation is a subset of the solution set of
the new equation.
Equation
Solution Set
x= 3
x2= 9
{3}
{–3, 3}
Every solution of the new equation must be checked in the
original equation to eliminate extraneous solutions.
1-7-11
Key Steps in Solving Polynomial Inequalities
Step 1.
Write the polynomial inequality in standard form (a form where
the right-hand side is 0.)
Step 2.
Find all real zeros of the polynomial (the left side of the standard form.)
Step 3.
Plot the real zeros on a number line, dividing the number line
into intervals.
Step 4.
Choose a test number (that is easy to compute with) in each interval,
and evaluate the polynomial for each number (a small table is useful.)
Step 5.
Use the results of step 4 to construct a sign chart, showing the sign
of the polynomial in each interval.
Step 6.
From the sign chart, write down the solution of the original polynomial
inequality (and draw the graph, if required.)
1-8-12