5.1 Linear Equations
• A linear equation in one variable can be written in
the form: Ax + B = 0
• Linear equations are solved by getting “x” by
itself on one side of the equation
• Examples of non-linear equations:
x 2  3x  5
1
x2
x
5.1 Linear Equations
• Example: Solve by getting x by itself on one side of
the equation.
Subtract 7 from both sides:
Divide both sides by 3:
3x  7  52
3x  45
x  15
5.1 Linear Equations
• Example: Renting a car for one day costs $20 plus $.25
per mile. How much would it cost to rent the car for
one day if 68 miles are driven?
$20 = fixed cost,
$.25  68 = variable cost
$20  68  $.25 
$20  $17 
$37
5.1 Linear Equations
• A linear equation in two variables can be put in
the form (called standard form):
Ax  By  C
where A, B, and C are real numbers and
A and B are not zero
5.1 Linear Equations
• Example (substitution):
2x  y  7
3 x  y  13
From the first equation we get y=2x-7, so
substituting into the second equation:
3 x  2 x  7  13
5 x  7  13  5 x  20  x  4
24  y  7  8 y  7
 y  1  y  1
5.1 Linear Equations
• Example (elimination): 2 x  3 y  5
4 x  y  17
• Multiply the second equation by 3 to get:
2x  3y  5
12 x  3 y  51
• Adding equations you get:
14 x  56  x  4
4  4  y  17  y  17  16  1
5.2 Graphs of Linear Functions
• Graph by
plotting
points:
3 x  2 y  12
x
0
2
4
y
6
3
0
5.2 Graphs of Linear Functions
• The slope of a line through points (x1,y1)
and (x2,y2) is given by the formula:
y2  y1
rise
m

x2  x1
run
5.2 Graphs of Linear Functions
• A positive slope rises
from left to right.
• A negative slope falls
from left to right.
5.2 Graphs of Linear Functions
•
Finding the slope of a line from its equation
1. Solve the equation for y.
2. The slope is given by the coefficient of x
• Example: Find the slope of the equation.
3x  2 y  5
2 y  3 x  5
y   32 x  52  m   32
5.2 Graphs of Linear Functions
• Standard form:
Ax  By  C
• Slope-intercept form: y  mx  b
(where m = slope and b = y-intercept)
5.2 Graphs of Linear Functions
• Example: Put the equation 2x + 3y = 6 in slopeintercept form, determine the slope and intercept,
then graph.
2 x  3 y  6  3 y  2 x  6
y  32 x  2  m  32 , b  2
Since b = 2, (0,2) is a point on the line.
Since m   23 , go down 2 and across 3 to point
(3,0) a second point on the line, then connect the
two points to draw the line.
5.2 Graphs of Linear Functions
• Example: Graph the equation.
2x  3y  6
x
0
3
y
2
0
6.1 Special Products
• Special product:
ax  y   ax  ay
• Example:
9(2 x  3 y )  18 x  27 y
6.1 Special Products
• Difference of 2 squares:
2
2
x  y   x  y   x  y
• Example:
w  5w  5  w
2
 5  w  25
2
2
6.1 Special Products
• Squaring binomials:
x  y   x  2 xy  y
2
x  y   x 2  2 xy  y 2
2
2
2
• Examples:
2
2
2
2
m  3  m  23m   3  m  6m  9
5 z  1
2
 5 z   25 z   12  25 z 2  10 z  1
2
6.1 Special Products
•
Multiplying binomials using FOIL (First –
Inner – Outer - Last):
1.
2.
3.
4.
5.
F – multiply the first 2 terms
O – multiply the outer 2 terms
I – multiply the inner 2 terms
L – multiply the last 2 terms
Combine like terms
6.1 Special Products
• Multiplying binomials:
x  a  x  b  x 2  (a  b) x  ab
• Example:
x  3 x  7  x 2  (3  7) x  3(7)  x 2  4 x  21
6.1 Special Products
• Multiplying binomials:
ax  b cx  d   acx 2  (ad  bc) x  bd
• Example:
3x  2  2 x  5
 (3 x)( 2 x)  [3(5)  2(2)] x  5(2)
 6 x 2  11x  10
6.1 Special Products
• Multiplying two polynomials (note: the book does
this by grouping and using special products):
x2  x  2
x 3
 3 x 2  3x  6
x3  x 2  2 x
x3  4 x 2  5x  6
6.2 Factoring: Common Factor and
Difference of Squares
•
Finding the Greatest Common Factor:
1. Factor – write each number in factored form.
2. List common factors
3. Choose the smallest exponents – for variables
and prime factors
4. Multiply the primes and variables from step 3
•
Always factor out the GCF first when
factoring an expression
6.2 Factoring: Common Factor and
Difference of Squares
•
Example: factor 5x2y + 25xy2z
5x y  5 x y z
2
1
2 1 0
25 xy z  5 x y z
2
2 1 2 1
GCF  5 x y z  5 xy
1 1 1 0
5 x y  25 xy z  5 xy( x  5 yz )
2
2
6.2 Factoring: Common Factor and
Difference of Squares
• Difference of 2 squares:
2
2
x  y  x  y   x  y 
• Example:
9  w  3  w  3  w 3  w
2
2
2
• Note: the sum of 2 squares (x2 + y2) cannot
be factored.
6.2 Factoring: Common Factor and
Difference of Squares
• Factor by Grouping – Introductory Example:
x 2  5 xy  4 x  20 y
 x( x  5 y )  4( x  5 y )
 ( x  4)( x  5 y )
• Note: this will be covered in more detail in the
next section.
6.2 Factoring: Common Factor and
Difference of Squares
•
Factoring by grouping
1. Group Terms – collect the terms in 2 groups
that have a common factor
2. Factor within groups
3. Factor the entire polynomial – factor out a
common binomial factor from step 2
4. If necessary rearrange terms – if step 3 didn’t
work, repeat steps 2 & 3 until you get 2
binomial factors
6.2 Factoring: Common Factor and
Difference of Squares
• Example:
10 x 2  12 y 2  15 xy  8 xy
 2(5 x 2  6 y 2 )  xy(15  8)
This arrangement doesn’t work.
• Rearrange and try again
10 x 2  15 xy  12 y 2  8 xy
 5 x(2 x  3 y )  4 y (3 y  2 x)
 (5 x  4 y )( 2 x  3 y )
6.3 Factoring Trinomials
•
Factoring x2 + bx + c (no “ax2” term yet)
Find 2 integers: product is c and sum is b
Sign hints:
1. Both integers are “+” if b and c are “+”
2. Both integers are “-” if c is “+” and b is “-”
3. One integer is “+” and one is “-” if c is “-”
6.3 Factoring Trinomials
• Example: x 2  5 x  4
(4)  (1)  5; (-4)  (1)  4
( x  4)( x  1)
• Example: x 2  4 x  21
7  3  4; 7  (3)  21
( x  7)( x  3)
6.3 Factoring Trinomials
•
Factoring ax2 + bx + c by using FOIL (in
reverse)
1. The first terms must give a product of ax2
(pick two)
2. The last terms must have a product of c (pick
two)
3. Check to see if the sum of the outer and inner
products equals bx
4. Repeat steps 1-3 until step 3 gives a sum = bx
6.3 Factoring Trinomials
• Example:
2 x  7 x  6  (2 x  ?)( x  ?)
2
try (2 x  1)( x  6)  2 x  13 x  6 incorrect
2
try (2 x  6)( x  1)  2 x 2  8 x  6 incorrect
try (2 x  3)( x  2)  2 x 2  7 x  6 correct
6.3 Factoring Trinomials
• Box Method (not in book):
2 x 2  7 x  6  (2 x  ?)( x  ?)
2x
2
x 2x
?
?
6
6.3 Factoring Trinomials
• Box Method – keep guessing until crossproduct terms add up to the middle value
2x
2
x 2x
2 4x
3
3x
6
so 2 x 2  7 x  6  (2 x  3)( x  2)
6.3 Factoring Trinomials
• Perfect square trinomials:
x  2 xy  y  x  y 
2
2
2
x  2 xy  y  x  y 
2
2
2
• Examples:
2
2
2
2
m  6m  9  m  23m   3  m  3
25 z  10 z  1  5 z   25 z   1  5 z  1
2
2
2
2
6.3 Factoring Trinomials
•
Factoring ax2 + bx + c by grouping
1. Multiply a times c
2. Find a factorization of the number from step 1
that also adds up to b
3. Split bx into these two factors multiplied by x
4. Factor by grouping (always works)
6.3 Factoring Trinomials
• Example:
8 x 2  14 x  15
ac  120  60  ( 2)  30  ( 4)
 15  ( 8)  20  (6)
b  14  20  6
• Split up and factor by grouping
8 x 2  14 x  15  8 x 2  20 x  6 x  15
 4 x(2 x  5)  3(2 x  5)
 (4 x  3)( 2 x  5)
6.4 Sum and Difference of Cubes
• Difference of 2 cubes:

x  y  x  y  x  xy  y
3
3
• Example:
2

2

w  27  w  3  w  3 w  3w  9)
3
3
3
2

6.4 Sum and Difference of Cubes
• Sum of 2 cubes:

x  y  x  y   x  xy  y
3
3
• Example:
2

2

w  27  w  3  w  3 w  3w  9)
3
3
3
2

Summary of Factoring
•
Summary of Factoring
1. Factor out the greatest common factor
2. Count the terms:
– 4 terms: try to factor by grouping
– 3 terms: check for perfect square trinomial. If not
a perfect square, use general factoring methods
– 2 terms: check for difference of 2 squares,
difference of 2 cubes, or sum of 2 cubes
3. Can any factors be factored further?
6.5 Equivalent Fractions
•
P
Polynomial Fraction– has the form:
Q
where P and Q are polynomials with Q not
equal to zero.
6.5 Equivalent Fractions
• Lowest terms – A fraction P/Q is in lowest terms
if the greatest common factor of the numerator and
the denominator is 1.
• Fundamental property of fractions – If P/Q is a
polynomial fraction and if K represents any
polynomial where K  0, then:
PK P

QK Q
6.5 Equivalent Fractions
•
Example: Write the fraction
in lowest terms:
3x  12
5 x  20
3x  12 3( x  4)

5 x  20 5( x  4)
3
(
x

4
)
3
2. By the fundamental property:

1. Factor:
5( x  4)
3. The fraction is undefined for:
x4
5
6.6 Multiplication and Division of
Fractions
• Multiplying Fractions– product of two
fractions is given by:
P R
PR


Q S
QS
• Dividing Fractions– quotient of two
fractions is given by:
P R
P S
PS




Q S
Q R
QR
6.6 Multiplication and Division of
Fractions
•
Multiplying or Dividing Fractions:
1. Factor completely
2. Multiply (multiply by reciprocal for division)
3. Write in lowest terms using the fundamental
property
6.6 Multiplication and Division of
Fractions
• Example - multiply:
• Factor:
x 2  3x x 2  5 x  4
 2
2
x  3x  4 x  2 x  3
x( x  3)
( x  4)( x  1)

( x  4)( x  1) ( x  1)( x  3)
• Cancel to get in lowest terms:
x
x 1
6.6 Multiplication and Division of
Fractions
x2  4
( x  2)( x  3)
• Example - divide:

( x  3)( x  2)
2x
x2  4
2x


( x  3)( x  2) ( x  2)( x  3)
• Factor:
( x  2)( x  2)
2x


( x  3)( x  2) ( x  2)( x  3)
• Cancel to get in lowest terms:
2x
( x  3) 2
6.7 Addition and Subtraction of Fractions
•
Finding the least common denominator for
rational expressions:
1. Factor each denominator
2. List the factors using the maximum number
of times each one occurs
3. Multiply the factors from step 2 to get the
least common denominator
6.7 Addition and Subtraction of Fractions
•
5
3
and 3
Find the LCD for:
2
6r
4r
1. Factor both denominators
6r 2  2  3  r 2
4r  2  r
3
2
2. The LCD is the product of the
largest power of each factor: 22  3  r 3
3
 12r
3
6.7 Addition and Subtraction of Fractions
• Adding Fractions:
P
If
and R are fractions, then
Q
Q
P
Q
 
R
Q
P R
Q
• Subtracting Fractions:
If P and R are fractions, then
Q
Q
P
Q
 
R
Q
PR
Q
6.7 Addition and Subtraction of Fractions
•
Adding/Subtracting when the denominators are
different fractions:
1. Find the LCD
2. Rewrite fractions – multiply top and bottom
of each to get the LCD in the denominator
3. Add the numerators (the LCD is the
denominator
4. Write in lowest terms
6.4 Adding/Subtracting Rational Expressions
•
Add:
2x
1

2
x 1 x 1
1. Factor denominators
to get the LCD:
2. Multiply to get a
common denominator:
2x
1

( x  1)( x  1) x  1
LCD is ( x  1)( x  1)
2x
1 x 1


( x  1)( x  1) x  1 x  1
2x
 x 1


( x  1)( x  1) ( x  1)( x  1)

3. Add and
2x  x  1
x 1
1
simplify:  ( x  1)( x  1)  ( x  1)( x  1)  ( x  1)
6.7 Addition and Subtraction of Fractions
•
Complex Fraction – a fraction with fractions in
the numerator, denominator or both
• To simplify a complex fraction (method 1):
1. Write both the numerator and denominator as
a single fraction
2. Change the complex fraction to a division
problem
3. Perform the division by multiplying by the
reciprocal
6.7 Addition and Subtraction of Fractions
3
6

x
• Example:
x
1

4
8
1. Write top and 6  x  3
x
x
bottom as a

single fraction 4x  22  18
2. Change to
division problem
6 x 3

x
3. Multiply by the
reciprocal and  6 x  3  8
x
2 x 1
simplify
6x
x
2x
8
 3x

 18

2 x 1
8

3( 2 x 1)
x
6 x 3
x
2 x 1
8
 2 x81  24x
6.7 Addition and Subtraction of Fractions
•
To simplify a complex fraction
(method 2):
1. Find the LCD of all fractions within the
complex fraction
2. Multiply both the numerator and the
denominator of the complex fraction by
this LCD. Write your answer in lowest
terms
6.7 Addition and Subtraction of Fractions
•
3
6

x
Example:
x
1

4
8
1. Find the LCD: the denominators are 4, 8, and x
so the LCD is 8x.
3
3


8
x
6

8
x
(
6
)

8
x
(
)
2. Multiply top
x
x


x
1
and bottom
8 x 4  8  8 x( 4x )  8 x( 18 )
by this LCD.
3. Simplify:  48 x  24  24(2 x  1)  24
2
2x  x
x(2 x  1)
x
6.8 Equations Involving Fractions
1. Multiply both sides of the equation by the
LCD
2. Solve the resulting equation
3. Check each solution you get – reject any
answer that causes a denominator to equal
zero.
6.8 Equations Involving Fractions
•
Solve:
2
1
 2
2
x  x x 1
1. Factor to get LCD
LCD = x(x - 1)(x + 1)
2
1

x( x  1) ( x  1)( x  1)
2. Multiply both 2 x( x  1)( x  1) x( x  1)( x  1)

sides by LCD
x( x  1)
( x  1)( x  1)
2( x  1)  x
6.8 Equations Involving Fractions
2
1
• Example (continued):
 2
2
x  x x 1
3. Solve the equation 2( x  1)  x
2x  2  x
x  2
4. Check solution
2
1

2
- 2  - 2 - 22  1
2 1

6 3
6.8 Equations Involving Fractions
• Distance, Rate, and time:
d  rt , r  dt , and t  dr
• Rate of Work - If one job can be completed in t units of
time, then the rate of work is:
1
r  job per unit time
t
6.8 Equations Involving Fractions
•
Example: If the same number is added to the
numerator and the denominator of the fraction
2/5, the result is 2/3. What is the number?
1. Equation 2  x  2
5 x 3
2 x
2
3(5  x)
 3(5  x)
2. Multiply by
5 x
3
LCD: 3(5+x)
3(2  x)  2(5  x)
6  3x  10  2 x
3. Subtract 2x and 6
x4
6.8 Equations Involving Fractions
•
1.
2.
3.
Example: It takes a mail carrier 6 hr to cover her route. It
takes a substitute 8 hr. How long does it take if they work
together?
Table:
Rate Time
Part of Job Done
Regular
1/6
x
x/6
Substitute
1/8
x
x/8
x x
 1
6 8
Multiply by LCD: 24 4 x  3x  24
Equation:
7 x  24
4.
Solve:
x
24
7
hours