Download Document

582760656
Name _______________________
1.1 Properties of Real Numbers
Different number systems came about because of a need to express different solutions:
Eli has 3 sheep but owes
Joseph 5 sheep.
3  5  2
Allie found $20 and agreed
to share it equally with 2 of
her friends.
20
2
 6  $6.66
3
3
A need for integers,
negative whole numbers.
A need for rational numbers.
The decimal only
approximates the solution.
Find the area of a circle with radius 5 inches.
A   52  25  78.54 in2
OR
Find the diagonal of a square with a side of 6 cm.
h  6 2  8.49 cm
A need for irrational numbers.
The decimals only approximate the solutions.
Number Systems
Real Numbers, R
Combination of all rational and irrational numbers.
Irrational Numbers
Rational Numbers, Q
Any signed number that can be written as a quotient of 2
integers: a/b (where b0). Terminating and repeating decimals.
7
3
1
,  , 9, -1.2, 0, 4 , 0.3 ,
5
2
8
CAN’T be written as a
quotient of 2 integers.
 Non-repeating and
non-terminating
decimals.
 Many square roots.
Integers, I
Any signed number without a fractional part.
I = {…-2, -1, 0, 1, 2, ...}
Whole Numbers, W
W = {0, 1, 2, 3...}
, 2 ,  7 ,
1.0110111…,
2
,
3
Natural Numbers
N = {1, 2, 3...}
Remember that a decimal is a fraction with a power of 10 in the denominator, so many decimals are
5
1
fractions. For example, 0.5 is
or
and that makes 0.5 a rational number.
10
2
S. Stirling
Page 1 of 18
582760656
Name _______________________
Always remember to simply a number before determining the set(s) of numbers in which it belongs.
You may use a calculator for the square roots, but you should be able to the rest with out one.
Example 1 alt.
To which sets of numbers does each number belong?
d. 27 is Integer, Rational and Real
is Natural, Whole, Integer,
Rational and Real
a. 7
b.  0.04
e. 0.3 is Rational and Real
is Rational and Real
0.2 which is 2/10 so rational
c.
2
1
is Irrational and Real
2
5
f.
1.4142... non-repeat &
non-terminating
Do Example 1, page 6.
situations.
is the fraction for 0.3 , repeating
3
is Irrational, Real
0.6324... nonterminating
Determine which type of numbers are used to express which type of values in
Graphing Real Numbers
Find an approximate and put a dot on the axis.
Write the plotted value above the dot.
Do Example 2, page 6.




3
2
1.7



5



Ordering Real Numbers
If a and b are real numbers,
then a = b, a < b, or a > b.
Example 3, page 6. Compare  0.25 and  0.01 .
To prove that a < b…
 the graph of a is to the left of b
or
 b – a is a positive number
Since –0.5 is to the left of –0.1 on the number
line,  0.25   0.01

 0.25  0.5 and  0.01  0.1
OR since 0.1 0.5  0.4 is positive
 0.25   0.01 .
S. Stirling
Page 2 of 18
582760656
Name _______________________
Properties of Real Numbers
The opposite or additive inverse of any number a is a .
The sum of opposites must be 0. a  (a )  0
The reciprocal or multiplicative inverse of any nonzero
1
number a is
or 1  a .
a
1
a b
1
The product of reciprocals is 1. a  1 or
a
b a
Example 4, b.
Opposite:   3.2  3.2
2
1
16
 3  
10
5
5
1
5
Reciprocal:

16
16

5
Rewrite: 3
3
.
5
Opposite:     
5
5
3
3
Reciprocal:
Example 4, page 7. Find the opposite and
reciprocal of
a. 3.2
1 5

3
3
5
Example 5 & QC, page 7.
Which property is illustrated?
Property
Addition
Multiplication
Closure
Commutative
Associative
a  b is a real number
a b  ba
 a  b  c  a  b  c 
ab is a real number
ab  ba
 ab  c  a bc 
Identity
Inverse
a0  a, 0a  a
a   a   0
a. 6  (6)  0
Add. Inverse
a 1 a, 1 a  a
1
a  1, a  0
a
a  b  c   ab  ac
Distributive
Mult. over add.
b.
 4 1  2  4  2
Mult. Identity
a.
 3  0  5  3  5
Add. Identity
Absolute Value
b. 5   2  (3)  (5  2)  (3)
The absolute value of a real number is its distance from zero on
Associative of Add
the number line.
Read x as “the absolute value of x”. Interpret 3 as “all
points which are a distance of 3 units from the origin.” So both 3 and –3 make that sentence true.
If you think of it as a function…you put in a positive or
negative number and you get out an “unsigned” positive
number. But it is best to think of it as the distance from zero.
distance of 3




distance of 3





Do not get absolute value confused with parenthesis! Also, when there is an expression inside the
absolute value, you must simplify first, before you take the absolute value.
Example 6, page 8.
Find the absolute values.
4 = 4
S. Stirling
0 =0
1 (2) = 2 = 2
0  3 = 3 = 3
Page 3 of 18

582760656
Name _______________________
1.2 Algebraic Expressions
Evaluating Algebraic Expressions
Variable is a symbol, usually a letter, which represents one or more numbers.
Algebraic expression is any combination of numbers, variables and/or operators.
To evaluate an expression you
 substitute in values for the variables
 simplify (by following the order of the operations).
Vocabulary:
What is an algebraic expression? Give some examples and non-examples. (Think: How is an
algebraic expression different from an equation?)
Examples:
5, a, 5x, 2x – 8, 8.2x2 + 15,
(a + b) – c
Non-examples:
3x  7  8
3x 2  2  7.23
y  mx  b
A   r2
What is the root word in evaluate? What does it mean to evaluate an expression? Can the value of
an expression change? What would make it change?
“Value” is in evaluate, so it
means to find the value of an
expression.
The value of an expression
changes based on the values
of its variables.
WARNING! a should be read “the opposite of a”.
If a  3 , then a  3 . If a  3 , then a  (3)  3 .
Example 1
Evaluate a  2b  ab for a  3 and b  1.
Example 2 Alt.
Evaluate d 2  4  d  2c  for c  3 and d  5 .
a  2b  ab
(3)  2(1)  (3)(1) substitute
3  2  3 multiplication & division (left to right)
2
addition & subtraction (left to right)
d 2  4  d  2c 
(5)2  4  (5)  2(3)  substitute
(5)2  4 11 inside paranthesis M then A.
25  4 11 exponents
multiplication & division (left to right)
25  44
addition & subtraction (left to right)
69
See Example 2, page 12, for more problems with evaluating expressions. See Example 3, page 13, for a
real-world problem.
S. Stirling
Page 4 of 18
582760656
Name _______________________
Simplifying Algebraic Expressions
Terms are the things being added together. It
could consist of just a number, just a
variable or numbers and variables that are
being multiplied together, or divided.
The numerical factor in a term is called
the coefficient. It tells you how many of
that term.
Like terms have the same variables raised to
the same powers. You should combine
like terms when simplifying!
Read: 5a  3b  2a  b as
“ 5 apples and 3 bananas and 2 apples and
1 banana”
Simplified: “ 7 apples and 4 bananas”
Algebraically: 7a  4b
Properties for Simplifying
Let a, b and c represent real numbers.
Definition of
Subtraction
Definition of
Division
Distributive Property
for Subtraction
Multiplication by
zero
Multiplication by
negative one
Opposite of a sum
Examples:
The expression 2 x  5 y  3 y  7 x
has 4 terms (two pairs of like terms)
–2x, 5y, –3y and 7x
the coefficient of the term 3 y is –3
the coefficient of the term 7x is 7
Simplified is 2 y  5 x
2
2
The expression 3r t  5.2  8r t
has 3 terms (one pair of like terms)
3r 2t ,  5.2, 8r 2t
the coefficient of the term 3r 2t is –3
the coefficient of the term 8r 2t is 8
Simplified is 5r 2t  5.2
Subtraction is “add the opposite of” 3  5  3   5  2
a  b  a  (b)
a
1
a , b0
b
b
a  b  c   ab  ac
a b 
2   6  2   6  4
Division is “multiply by the reciprocal of” 2  1  2 6  4
3 6 3 1
3
3 1 3
5 

5
5 5 25
a 00
“Anything times zero is 0.”
1 a  a
“Anything times a negative one is the opposite of it”
  a  b   a  (b)
Opposite of a
difference
Opposite of a product
  a  b   a  b  b  a
Opposite of an
opposite
  a   a
  ab   a b  a (b)
“Taking the opposite of a sum or difference is
like distributing a negative one.”
One of the factors is negative and one is positive.
or Opposite of a negative is a positive.
WARNING:
Helpful hints for simplifying:
You DO NOT distribute
Write all subtraction as “add the opposite of”
over multiplication!
Write all division as “multiply by the reciprocal of”
Rewrite variables without a coefficient as 1 times…  x  1x or  x  1x
  ab   a b
Always follow the order of operations.
Rearrange terms to get like terms together (commutative property).
Replace implied operations with real symbols: 3x  3•x or – (x+5)  –1 • (x+5)
Do not remove parenthesis from an expression until you have completed the indicated operation.
S. Stirling
Page 5 of 18
582760656
Example 4 Simplify.
Name _______________________
Example 5 Find the perimeter.
a. 3k  k
 3k 1k  2k
b. 5 z 2  10 z  8 z 2  z
= 5 z 2   10 z    8 z 2   1z write as addition
= 5 z 2   8 z 2    10 z   1z commutative
= 3z 2  9 z combine like terms
c.
  m  n   2  m  3n 
= 1 m  n   2  m  3n  now distribute
= 1m   1n   2m  6n
= 1m  2m   1n   6n commutative
= 2m  7n
combine like terms
Since you can add in any order you want, commutative:
2a  b 2a  b

 b  b  b  a  a  2a
2
2
4a  2b
 3b  4a combine like things
2
2a  b  3b  4a division (distributive)
6a  2b
Choose C.
Extra Example Simplify.
1
4 6 4
x   x
3
3 5 3
1
4 5 4
= x    x rewrite division as multiply by the reciprocal
3
3 6 3
1
10  4 
= x     x  order of operations!
3
9  3 
10
= x 
combine like terms
9
S. Stirling
Page 6 of 18
582760656
Name _______________________
1.3 Solving Equations
Solving Equations
A solution of an equation is number that makes the equation (or sentence) true.
Properties of Equality (aka Properties for Solving Equations)
Let a, b and c represent real numbers.
Property:
Reflexive
Symmetric
aa
If a  b , then b  a .
Transitive
If a  b and b  c , then a  c.
Addition
If a  b , then a  c  b  c .
Subtraction
If a  b , then a  c  b  c .
Multiplication If a  b , then ac  bc .
Division
Substitution
a b
 .
c c
If a  b , then b may be substituted
for a in any expression.
If a  b and c  0 , then
In English…
Anything equals itself.
You may change expressions around an equal
sign.
If you’re as tall as Rob and Rob is as tall as
Andy, then you’re as tall as Andy.
You may add the same number to both sides
of an equation.
You may subtract the same number from both
sides of an equation.
You may multiply both sides of an equation
by the same number.
You may divide both sides of an equation by
the same number, as long as it’s not zero.
You may take something out and replace it
with something equal to it.
Helpful hints for solving:
“Simplify”
Order of Operations
MUST follow
Draw a line down from the equal
“PEMDAS”
sign so you can see the sides of the the order of
operations.
equation.
Parenthesis
You may simplify each side of an
“Solve”
Exponents and Roots
equation separately.
Best to work in
Multiplication and Division
the reverse order
You want to “undo” what is
Addition and Subtraction
of operations.
being done to the variable by using
inverse operations. (i.e. addition
inverse of subtraction, 5x undoes 5x ).
What ever you do to one side of the equation you must do to the other.
It is best to solve in the reverse order of operations.
Try to isolate the variable, get a sentence like x = # or # = x.
The goal: Find the number(s) that makes the original sentence true. (Check your answers!)
Example 1a Solve the equation.
Example 1b Solve the equation.
8z 12  5z  21 starting equation
5z 12  5z 12 Subtraction property
3z  33
Simplify
z  11 division prop. (both sides by 3)
2t  3  9  4t starting equation
4t  3 4t  3 Addition property
6t  12
Simplify
t  2 division prop. (both sides by 6)
8z 12  5z  21
S. Stirling
2t  3  9  4t
Page 7 of 18
582760656
Name _______________________
Could you solve the previous examples in a different way? Check with your group members. As long as
you follow the properties for solving equations, and do not make any silly mistakes like 3  5  15 , you’ll
be fine! To check your solutions, substitute in the value for the variable and check (by simplifying) to see
that the sides are really equal.
Example 2b Solve the equation.
6(t  2)  2(9  2t )
Note: you can’t add to or subtract from both sides
until you simplify each side of the equation!
6(t  2)  2(9  2t ) copy the original
6t 12  18  4t distributive property
4t 12  4t  12 add & subtract prop
10t  30
Simplify
10t 30

division prop. (both sides by 10)
10 10
t 3
Simplify
Example 2 Solve the equation.
3x  7(2 x  13)  3(2 x  9)
3x 14x  91  6x  27 distributive prop
11x  91  6x  27 simplify
11x  27 11x  27 add & subtract prop
64  5x
Simplify
64 x
 division prop. (both sides by 5)
5 5
x  64  12 4  12.8
5
5
Solving Literal Equations
When you solve an equation that has more than one variable in it, the process for solving is the same as
it was in Example 1 and Example 2. The key here is to focus on the variable you are solving for and
isolate it. The “solution” will be an expression with variables in it rather than a number.
Problem Solving Example: You have to build a 4 sided rectangular fence and have 200 feet of fencing.
You want to try different lengths and need to know the corresponding widths. Find the dimensions of the
rectangular fence for the lengths of 5 feet, 10 feet, 15.5 feet and any number of feet.
Analyze:
Need to find widths of rectangles for different lengths.
The total feet of fencing = perimeter of the rectangle
Perimeter is given by P  2l  2w
How to solve:
Could do each length separately?
For l = 5, 200  2(5)  2w .
Now solve:
200  10  2w
subtract 10
10 10
get result
190  2w
divide by 2
w  95
S. Stirling
Since you need to do this many times it
would be better to write a formula for w =.
P  2l  2w
subtract whatever 2l is
2l  2l
get result
P  2l  2w
 P  2l   2w divide by 2
2
2
 P  2l 
w
2
Now for P = 200 and l = 5
 200  2(5)   190  95 feet.
w
2
2
Page 8 of 18
582760656
Name _______________________
Example 3alt Solve the formula for b1.
Example 3alt Solve for h.
V   r 2h
1
A  h  b1  b2 
2
h
A   b1  b2  rewrite (by multiplying)
2
2 h 2
A      b1  b2  multiply by the reciprocal
h 2 h
2A
 b1  b2 simplify
h
2A
 b2  b1 subtraction property, b2 both sides
h
V    r 2  h undo multiplication with division
V
  r2  h

then make ones, simplify
  r2
  r2
V
h
 r2
Hint: If you can’t figure out how to solve the literal equation, replace the variables (the ones you are not solving
for) with numbers. Solve the equation. The process with the variables will be the same as it was with the numbers.
Hint Example 4b Solve 7 
2x
 5 then do Ex
3
Example 4b Solve for x. Find any restrictions.
d
2x
 b isolate x-term
a
d b 
2x
a
subtract b from both sides
a
a 2x
 d  b  
2
2 a
x
multiply by the reciprocal
a  d  b
a
d  b 
2
2
either answer
a  0 because it would make the
denominator zero.
Note:
Hint Example 4a Solve 2x  3x 15  0 then
do Ex 4a.
Example 4a Solve for x. Find any restrictions.
ax  bx 15  0
ax  bx 15  0 add 15 to both sides, isolate x-terms
ax  bx  15 need to get x isolated, so…
x  a  b   15
un-distribute (factor out) the x
x a  b
15

divide both sides by  a  b 
a  b a  b
x
15
 a  b
Note:
a  b because it would make the
denominator zero.
Think: Compare the processes of simplifying and solving. How are they the same? different?
S. Stirling
Page 9 of 18
582760656
Name _______________________
Writing Equations to Solve Problems
Example 5
Example 6
A dog kennel owner has 100 feet of fencing to enclose
a rectangular dog run. She wants it to be five times as
long as it is wide. Find the dimensions of the dog run.
The lengths of the sides of a triangle are in a ratio 3:4:5
The perimeter of the triangle is 18 in. Find the lengths
of the sides.
width = w length = 5w
perim  2width  2length now substitute to
get the equation in terms of w.
100  2( w)  2(5w)
w
100  12w simplify
5w
100 25
1
w

 8 ft.
12
3
3
sides have a common factor = x
sides: 3x : 4 x : 5x
length  5
25 125
2

 41 ft.
3
3
3
dimensions: 8
1
2
ft by 41 ft
3
3
no decimals here because you would need to
round, not exact! That’s bad!
5x
4x
perim  3x  4 x  5 x
18  12x simplify
18 3
x 
12 2
3x
1
3 9
3x  3     4 in
2
2 2
 3  12
4 x  4     6 in
2 2
1
 3  15
5 x  5     7 in
2
2 2
Check: 3: 4: 5  9: 12: 15 
9 12 15
: :
2 2 2
Example 7
Radar detected an unidentified plane 5000 miles away, approaching at 700 mi/h. Fifteen minutes later and interceptor
plane was dispatched, traveling at 800 mi/h. How long did the interceptor take to reach the approaching plane?
distance = rate * time = miles/hour * hour
5000 mi
Plane 1: leaves at t = 0, d1  700t
Plane 2: leaves at t – 0.25, d2  800(t  0.25)
They will meet each other when the total distance is 5000 mi.
P2
P1
800 mi/hr
700 mi/hr
.25 hr later
start t=0
700t  800(t  0.25)  5000
700t  800t  200  5000
1500t  5200
52
7
7 1
t
 3 hours so t  0.25  3   3.2167 hours OR 3 h 13 min.
15
15
15 4
S. Stirling
Page 10 of 18
582760656
Name _______________________
1.4 Solving Inequalities
Solving and Graphing Inequalities
The solution of an inequality are the number(s) that makes the sentence true.
The properties and procedures are almost the same as those for equations, except when you multiply or
divide both sides by a negative number. With equations there is typically one solution, but with
inequalities there are an infinite number of solutions.
Investigation:
Use substitution to find
solutions to the sentence:
x Is 3x  6 ?
–4 3(4)  6 yes
0
3(0)  6 yes
2
3(2)  6 yes
3
3(3)  6 no
3x  6
Graph the solution set
below:









Use substitution to find
solutions to the sentence:
2x  8
Graph the solution set
below:


Use solving equation properties to find the
solution set…
x Is 2x  8 ?
–6 2(6)  8 no
–4 2(4)  8 yes
0
2(0)  8 yes
2
2(2)  8 yes









Use solving equation properties to find the
solution set…
3x 6
 divide both sides by 3.
3 3
2 x 8

or x  4
2 2
Does x  2 yield the same solution set? YES
If no, what does?
Does
x  4 yield the same solution set?
NO
If no, what does?
x  4 When dividing both sides by a
negative, change the order.
Properties for Solving Inequalities
Let a, b and c represent real numbers.
Property:
Transitive
If a  b and b  c , then a  c.
Addition
If a  b , then a  c  b  c .
Subtraction
If a  b , then a  c  b  c .
Multiplication
If a  b and c  0 , then
ac  bc .
If a  b and c  0 , then
ac  bc .
Division
S. Stirling
a b
 .
c c
a b
If a  b and c  0 , then  .
c c
If a  b and c  0 , then
In English…
If Al is shorter than Bob and Bob is shorter than Cal,
then Al is shorter than Cal.
You may add the same number to both sides of an
inequality.
You may subtract the same number from both sides
of an inequality.
You may multiply both sides of an inequality by the
same POSITIVE number.
If you multiply both sides of an inequality by the
same NEGATIVE number, you must change the
ORDER (reverse the inequality symbol).
You may divide both sides of an inequality by the
same POSITIVE number.
If you divide both sides of an inequality by the
same NEGATIVE number, you must change the
ORDER (reverse the inequality symbol).
Page 11 of 18
582760656
Name _______________________
Note: When graphing solutions use an open dot, , for < or > and a closed dot,  , for  and  .
Also when checking solutions for an inequality, you should check the boundary, numbers above and
numbers below the boundary.
Examples Solve the inequality. Check your answer.
5x 12  8
5 x  12  8
5 x  20
5 x 20

5
5
x4
Example QC 1b Solve the inequality.
12  2  3n  1  22
12  2  3n  1  22
starting equation
Subtraction property –12
division prop. (both sides
by –5 ) CHANGE
ORDER!
solution
Check boundary & check another.
x = 4? 5(4)  12  8 , 20 12  8 yes
x = 0? 5(0)  12  8 , 12  8 yes
x = 6? 5(6)  12  8 , 18  8 no
12  6n  2  22
12  6n  24
12  6n
2  n
or
n  2
starting equation
Distribute
Simplify
Add prop –24
division prop. 6
solution (note
arrow pointing
same way)
Example 1b Solve the inequality. Graph the solution..
Check your answer.
Example 1a Solve the inequality. Graph the solution.
Check your answer.
3x 12  3
3 x  12  3
3 x 15

3
3
x5
starting equation
Add. prop 12
division prop. 3
solution
Check boundary & check another.
x = 5? 3(5)  12  3 , 15 12  3 yes
x = 0? 3(0)  12  3 , 12  3 yes
x = 6? 3(6)  12  3 , 6  3 no
S. Stirling
6  5(2  x)  41
6  10  5 x  41
16  5 x  41
5 x  25
5 x 25

5 5
x  5
starting equation
Distribute
Simplify
Subtr prop 16
division prop. –5
CHANGE
ORDER!
solution
Check boundary & check another.
x = –5? 6  5(2  ( 5))  41 ,
6  5(7)  41 , 41  41 yes
x = 0? 6  5(2  0)  41 , 6  16  41 yes
x = –6? 6  5(2  ( 6))  41 ,
6  5(8)  41 , 46  41 no
Page 12 of 18
582760656
Name _______________________
Sometimes an inequality will always be true (infinite solutions) or will never be true. For example,
5  2 is always true and 5  2 is never true.
Example 2a Solve the inequality. Graph the solution.
Example 2b Solve the inequality. Graph the solution.
7 x  6  7( x  4)
7 x  6  7 x  28
0  34 is never true.
2 x  3  2( x  5)
2x  3  2x 10
0  7 is always true.
Try any value for x and it will make the
sentence false. Solution: “No solution.”
Try any value for x and it will make the
sentence true. Solution: “All real numbers.”
See additional Examples, page 27, for more problems if needed.
Example 3 (Application)
A band agrees to play for $200 plus 25% of the ticket sales. Find the ticket sales needed for the
band to receive at least $500.
So ticket sales ≥ 500
Strategy: Try it with number(s).
What if you sold 100 tickets?
25% of 100 = .25 * 100 = 25
then 200 + 25 = 225
compare to 500
So 225 not at least 500 not a
solution.
Build the inequality:
25% of 100 0.25t
then +200 0.25t  200
compare: at least 500
200  0.25t  500
Solve the
inequality:
200  0.25t  500
1
t  300
4
t  1200
Interpret solution: Ticket sales must be greater than or equal to $1200.
See additional problem, top of page 28, for another problem, if needed.
S. Stirling
Page 13 of 18
582760656
Name _______________________
Compound Inequalities
A compound inequality is a pair of inequalities joined by and or or.
How to read & write:
x  1 or x  3
x  1 and x  3
Find all the values that make both of the
inequalities true.
Rewrite: 1  x  3
Read: “x is between –1 and 3, inclusive”









Example 4 AND Graph the solution.
AND means in both so find the points that are in
BOTH of the graphs, the OVERLAP!
3x 1  28 and 2x  7  19
3x  27 and
2x  12
x  9 and
x6
or
9  x  6
Find all the values that make either of the
inequalities true.
Read: “x is less than or equal to –1 or greater
than or equal to 3”









Example 5 OR Graph the solution.
OR = in either or both so find the
ACCUMULATION of the points on the graphs!
4 y  2  14 or 3 y  4  13
4 y  16 or 3 y  9
y  4 or y  3
rewritten in between form
 









 









 









 









Example 4alt Graph the solution.
2x  x  6 and x  7  2
x6
x9
and
or
6 x9
S. Stirling
 




















 









x is between 6 and 9.
Page 14 of 18
582760656
Name _______________________
Example 4alt Find the solution.
Example 4alt Find the solution.
7  2x  5  11
3 y  5  1 or 20  3y  y
3 y  6 or 20  4 y
y  2 or 5  y
On “between” inequalities, you can use solving
equation techniques to all 3 parts.
 









 









 









7  2x  5  11
5  5  5
12  2x  16
12 2 x 16


2
2
2
6 x 8
so summarize the solution, simply y  2
 









since it is the accumulation of both graphs.
Example 6 Application
The ideal length of a bolt is 13.48 cm. The length can vary from the ideal by at most 0.03 cm. A machinist
finds one bolt that is 13.67 cm long. By how much should the machinist decrease the length so the bolt can
be used?
Relate:
Minimum length = 13.48 – 0.03 = 13.45
Maxium length = 13.48 + 0.03 = 13.51
So 13.45  bolt  13.51
Question: how much to remove?
So x = number of cm to remove
bolt = 13.67 – x
Solve:
13.45  13.67  x  13.51
0.22  x  0.16
0.22  x  0.16
or 0.16  x  0.22
Interpret solution: The machinist must remove at least 0.16 cm and no more than
0.22 cm.
See additional Example, page 29, for another problem.
S. Stirling
Page 15 of 18
582760656
Name _______________________
1.5 Absolute Value Equations and Inequalities
Absolute Value Equations
The absolute value of a number is its
distance from zero on the number line and
distance is nonnegative.
Definition Absolute Value
If x  0 , then x  x .
If x positive, the output is x.
If x  0 , then x   x .
If x negative, the output is
the opposite of x.
There are different ways to approach this.
Absolute Value as a Distance.
Absolute Value as a Function.
Definition of Absolute Value
Solve u  5
Read: “The distance from 0
is 5.”
Think: “What points are 5
units from 0?”
Solutions: u = 5 or u = –5
(“or” not “and”)
Solve u  5
Read: “The absolute value
of what is 5?”
Think: “What could I
substitute in for u to
get 5?”
Solutions: u = 5 or u = –5
Solve u  5
Use the definition.
Think: “The expression inside
the absolute value is 5
or the opposite of 5.”
Set up: u = 5 or u = –5
Already solved.
Check:
5 5
yes
5  5
yes
So both 5 and –5 are solutions.
No Solutions?
Solve u  3
Think: “What could I substitute in for u to get –3?”
Answer? Nothing
Solutions: No Solutions
Use the concepts above to help you solve more complex problems.
Example 1 Solve.
2 y  4  12
Let u  2 y  4
Example 2 Solve.
Distance is 12 units, so…
u  12 , then u  12 or u  12
Since u  2 y  4 , substitute…
2 y  4  12 or 2 y  4  12
2 y  16 or 2 y  8
y  8 or y  4
If
Check:
2(8)  4  16  4  12  12
2(4)  4  8  4  12  12
S. Stirling
3 4w 1  5  10
Let u  4w  1
3 u  5  10
3 u  15
u 5
Can’t distribute!
division prop.
must isolate abs. first!!
Distance is 5 units, so…
If u
So
 5 , then u  5 or u  5
4w1  5 or 4w1  5
4w  6 or 4w  4
6 3
w   or w  1
4 2
Page 16 of 18
582760656
Name _______________________
As stated before, sometimes you will have no solution. For example,
of any number can’t be negative.
Definition Extraneous
Solution
An extraneous solution is a
solution of an equation
derived from an original
equation that is not a
solution of the original
equation.
The moral of this story is that you
should always check your solutions
in the original equation to make sure
the solutions you found work. This
will also come up when solving other
types of equations later in Algebra 2.
Since it is really tedious to check
everything, I will put “check your
solutions” in the instructions for the
problems.
Example 3 Solve.
2 x  7  2 .
The absolute value
Hard to use distance here,
so use the definition.
2 x  5  3x  4
u  #,
then u  #
If
2 x  5   3x  4 
2 x  5  3x  4
2 x  4  2 x  4
or
1 x
Check:
2(1)  5  3(1)  4
7 7
works!
or
u  #
2 x  5    3x  4 
2 x  5  3x  4
3x  5 3x  5
5x  9
9
x
5
Check:
 9
 9
2     5  3    4
 5
 5
18 25
27 20
 
 
5 5
5
5
7
7
 
Extraneous!
5
5
The only solution is 1.
–9/5 is an extraneous solution.
S. Stirling
Page 17 of 18
582760656
Name _______________________
Absolute Value Inequalities
The main idea is distance from 0. Use it! There are three main cases:
u 2
Distance from 0 is 2.
u  2 or u  2








u 2
u 2
Distance from 0 is less than 2.
u  2 and u  2
Distance from 0 is greater than 2.
u  2 or u  2
2  u  2





















So to solve absolute value inequalities
1. isolate the absolute value
2. rewrite as a compound equation or inequality
3. solve each equation or inequality
Example 4 Solve. Graph the solution.
3x  6  12 simplify the problem
3x  6  12 or 3x  6  12
3x  18 or 3x  6
x  6 or x  2
Example 5 Solve. Graph the solution.
3 2 x  6  9  15
isolate the abs. value
3 u  9  15
3 u  24
u 8
2x  6  8 and 2x  6  8
2x  14 and 2x  2
x  7 and x  1
or write 7  x  1
S. Stirling
To get the set up
let u  3x  6
u  12
distance greater than 12.
So distance from 0 is greater than 12.
u  12 or u  12
To get the set up
let
u  2x  6
then solve for
u
.
u  8 distance less than 8.
So u  8 and u  8
Example 5 alternate process
after 2 x  6
Use the set up
8
8  u  8 using “between”
8  2x  6  8
14  2x  2
7  x  1
Page 18 of 18
