Transcript Chapter 2: Equations, Inequalities and Problem Solving

Chapter 9
Equations,
Inequalities and
Problem Solving
Chapter Sections
9.1 – The Addition Property of Equality
9.2 – The Multiplication Property of Equality
9.3 – Further Solving Linear Equations
9.4 – Introduction to Problem Solving
9.5 – Formulas and Problem Solving
9.6 – Percent and Mixture Problem Solving
9.7 – Solving Linear Inequalities
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§ 9.1
The Addition Property of
Equality
Linear Equations
Linear equation in one variable
can be written in the form ax + b = c, a  0
Equivalent equations
are equations with the same solutions in the form of
variable = number, or
number = variable
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Addition Property of Equality
Addition Property of Equality
a = b and a + c = b + c are equivalent equations
Example
a.) 8 + z = – 8
8 + (– 8) + z = – 8 + – 8
z = – 16
(Add –8 to each side)
(Simplify both sides)
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Solving Equations
Example
4p – 11 – p = 2 + 2p – 20
3p – 11 = 2p – 18 (Simplify both sides)
3p + (– 2p) – 11 = 2p + (– 2p) – 18
p – 11 = – 18
(Simplify both sides)
p – 11 + 11 = – 18 + 11
p=–7
(Add –2p to both sides)
(Add 11 to both sides)
(Simplify both sides)
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Solving Equations
Example
5(3 + z) – (8z + 9) = – 4z
15 + 5z – 8z – 9 = – 4z
6 – 3z = – 4z
(Use distributive property)
(Simplify left side)
6 – 3z + 4z = – 4z + 4z (Add 4z to both sides)
6+z=0
6 + (– 6) + z = 0 +( – 6)
z=–6
(Simplify both sides)
(Add –6 to both sides)
(Simplify both sides)
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§ 9.2
The Multiplication
Property of Equality
Multiplication Property of Equality
Multiplication property of equality
a = b and ac = bc are equivalent equations
Example
–y=8
(– 1)(– y) = 8(– 1)
y=–8
(Multiply both sides by –1)
(Simplify both sides)
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Solving Equations
Example
1
5
x
7
9
1  5
7 x    7
7  9
35
x
9
(Multiply both sides by 7)
(Simplify both sides)
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Solving Equations
Example
8
x6
3
8 3
8
  x  6 
38
 3
x  16
(Multiply both sides by fraction)
(Simplify both sides)
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Solving Equations
Recall that multiplying by a number is equivalent to
dividing by its reciprocal
Example
3z – 1 = 26
3z – 1 + 1 = 26 + 1
(Add 1 to both sides)
3z = 27
3 z 27

3
3
(Simplify both sides)
z=9
(Simplify both sides)
(Divide both sides by 3)
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Solving Equations
Example
12x + 30 + 8x – 6 = 10
20x + 24 = 10
20x + 24 + (– 24) = 10 + (– 24)
(Simplify left side)
(Add –24 to both sides)
20x = – 14
(Simplify both sides)
20 x  14

20
20
(Divide both sides by 20)
7
x
10
(Simplify both sides)
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§ 9.3
Further Solving Linear
Equations
Solving Linear Equations
Solving linear equations in one variable
1)
2)
3)
4)
Multiply to clear fractions
Use distributive property
Simplify each side of equation
Get all variable terms on one side and number terms
on the other side of equation (addition property of
equality)
5) Get variable alone (multiplication property of
equality)
6) Check solution by substituting into original problem
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Solving Linear Equations
Example
3( y  3)
 2y  6
5
5  3( y  3)
 52 y  6 
5
3 y  9  10 y  30
(Multiply both sides by 5)
(Simplify)
3 y  (3 y )  9  10 y  (3 y )  30 (Add –3y to both sides)
9  (30)  7 y  30  (30) (Simplify; add –30 to both sides)
 21 7 y

7
7
3  y
(Simplify; divide both sides by 7)
(Simplify both sides)
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Solving Linear Equations
Example
5x – 5 = 2(x + 1) + 3x – 7
5x – 5 = 2x + 2 + 3x – 7 (Use distributive property)
5x – 5 = 5x – 5
(Simplify the right side)
Both sides of the equation are identical. Since this
equation will be true for every x that is substituted into
the equation, the solution is “all real numbers.”
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Solving Linear Equations
Example
3x – 7 = 3(x + 1)
3x – 7 = 3x + 3
(Use distributive property)
3x + (– 3x) – 7 = 3x + (– 3x) + 3 (Add –3x to both sides)
–7=3
(Simplify both sides)
Since no value for the variable x can be substituted
into this equation that will make this a true statement,
there is “no solution.”
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§ 9.4
An Introduction to
Problem Solving
Strategy for Problem Solving
General Strategy for Problem Solving
1) Understand the problem
• Read and reread the problem
• Choose a variable to represent the unknown
• Construct a drawing, whenever possible
• Propose a solution and check
2) Translate the problem into an equation
3) Solve the equation
4) Interpret the result
• Check proposed solution in problem
• State your conclusion
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Finding an Unknown Number
Example
The product of twice a number and three is the same as the
difference of five times the number and ¾. Find the number.
1.) Understand
Read and reread the problem. If we let
x = the unknown number, then “twice a number” translates to 2x,
“the product of twice a number and three” translates to 2x · 3,
“five times the number” translates to 5x, and
“the difference of five times the number and ¾” translates to 5x – ¾.
Continued
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Finding an Unknown Number
Example continued
2.) Translate
The product of
is the same as
twice a
number
2x
the difference of
and 3
·
3
=
5 times the
number
5x
and ¾
–
¾
Continued
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Finding an Unknown Number
Example continued
3.) Solve
2x · 3 = 5x – ¾
6x = 5x – ¾
(Simplify left side)
6x + (– 5x) = 5x + (– 5x) – ¾
x=–¾
(Add –5x to both sides)
(Simplify both sides)
4.) Interpret
Check: Replace “number” in the original statement of the problem
with – ¾. The product of twice – ¾ and 3 is 2(– ¾)(3) = – 4.5. The
difference of five times – ¾ and ¾ is 5(– ¾) – ¾ = – 4.5. We get the
same results for both portions.
State: The number is – ¾.
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Solving a Problem
Example
A car rental agency advertised renting a Buick Century for $24.95
per day and $0.29 per mile. If you rent this car for 2 days, how
many whole miles can you drive on a $100 budget?
1.) Understand
Read and reread the problem. Let’s propose that we drive a total of
100 miles over the 2 days. Then we need to take twice the daily rate
and add the fee for mileage to get 2(24.95) + 0.29(100) = 49.90 + 29
= 78.90. This gives us an idea of how the cost is calculated, and also
know that the number of miles will be greater than 100. If we let
x = the number of whole miles driven, then
0.29x = the cost for mileage driven
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Solving a Problem
Example continued
2.) Translate
Daily costs
mileage costs
plus
2(24.95)
+
maximum budget
is equal to
0.29x
=
100
Continued
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Solving a Problem
Example continued
3.) Solve
2(24.95) + 0.29x = 100
49.90 + 0.29x = 100
(Simplify left side)
49.90 – 49.90 + 0.29x = 100 – 49.90 (Subtract 49.90 from both sides)
0.29x = 50.10
0.29 x 50.10

0.29
0.29
x  172.75
(Simplify both sides)
(Divide both sides by 0.29)
(Simplify both sides)
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Solving a Problem
Example continued
4.) Interpret
Check: Recall that the original statement of the problem asked
for a “whole number” of miles. If we replace “number of miles”
in the problem with 173, then 49.90 + 0.29(173) = 100.07,
which is over our budget. However, 49.90 + 0.29(172) = 99.78,
which is within the budget.
State: The maximum number of whole number miles is 172.
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§ 9.5
Formulas and
Problem Solving
Formulas
A formula is an equation that states a known
relationship among multiple quantities (has more than
one variable in it)
A = lw
(Area of a rectangle = length · width)
I = PRT
(Simple Interest = Principal · Rate · Time)
P=a+b+c
(Perimeter of a triangle = side a + side b + side c)
d = rt
(distance = rate · time)
V = lwh
(Volume of a rectangular solid = length · width · height)
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Using Formulas
Example
A flower bed is in the shape of a triangle with one side twice the
length of the shortest side, and the third side is 30 feet more than the
length of the shortest side. Find the dimensions if the perimeter is
102 feet.
1.) Understand
Read and reread the problem. Recall that the formula for the
perimeter of a triangle is P = a + b + c. If we let
x = the length of the shortest side, then
2x = the length of the second side, and
x + 30 = the length of the third side
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Using Formulas
Example continued
2.) Translate
Formula: P = a + b + c
Substitute: 102 = x + 2x + x + 30
3.) Solve
102 = x + 2x + x + 30
102 = 4x + 30
(Simplify right side)
102 – 30 = 4x + 30 – 30
(Subtract 30 from both sides)
72 = 4x
(Simplify both sides)
72 4 x

4
4
(Divide both sides by 4)
18 = x
(Simplify both sides)
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Using Formulas
Example continued
4.) Interpret
Check: If the shortest side of the triangle is 18 feet, then the
second side is 2(18) = 36 feet, and the third side is 18 + 30 = 48
feet. This gives a perimeter of P = 18 + 36 + 48 = 102 feet, the
correct perimeter.
State: The three sides of the triangle have a length of 18 feet,
36 feet, and 48 feet.
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Solving Formulas
It is often necessary to rewrite a formula so that it is
solved for one of the variables.
This is accomplished by isolating the designated
variable on one side of the equal sign.
Solving Equations for a Specific Variable
1)
2)
3)
4)
Multiply to clear fractions
Use distributive to remove grouping symbols
Combine like terms to simply each side
Get all terms containing specified variable on the
same time, other terms on opposite side
5) Isolate the specified variable
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Solving Equations for a Specific Variable
Example
Solve for n.
T  mnr
T
mnr

mr mr
(Divide both sides by mr)
T
n
mr
(Simplify right side)
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Solving Equations for a Specific Variable
Example
Solve for T.
A  P  PRT
A  P  P  P  PRT
(Subtract P from both sides)
A  P  PRT
(Simplify right side)
A  P PRT

PR
PR
(Divide both sides by PR)
A P
T
PR
(Simplify right side)
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Solving Equations for a Specific Variable
Example
Solve for P.
A  P  PRT
A  P(1  RT )
(Factor out P from both terms on the
right side)
A
P(1  RT )

1  RT
1  RT
(Divide both sides by 1 + RT)
A
P
1  RT
(Simplify the right side)
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§ 9.6
Percent and Mixture
Problem Solving
Solving a Percent Problem
A percent problem has three different parts:
amount = percent · base
Any one of the three quantities may be unknown.
1. When we do not know the amount:
n = 10% · 500
2. When we do not know the base:
50 = 10% · n
3. When we do not know the percent:
50 = n · 500
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Solving a Percent Problem: Amount Unknown
amount = percent · base
What is 9% of 65?
n
= 9%
· 65
n
= (0.09) (65)
n
= 5.85
5.85 is 9% of 65
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Solving a Percent Problem: Base Unknown
amount = percent · base
36 is 6% of what?
36 = 6% ·
n
36 = 0.06n
36
0.06n
=
0.06
0.06
600 = n
36 is 6% of 600
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Solving a Percent Problem: Percent Unknown
amount = percent · base
24 is what percent of 144?
24 =
n
 144
24 = 144n
24
144n
=
144
144
0.16 = n
2
16 % = n
3
2
24 is 16 % of 144
3
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Solving Markup Problems
Example
Mark is taking Peggy out to dinner. He has $66 to spend. If he wants
to tip the server 20%, how much can he afford to spend on the meal?
Let n = the cost of the meal.
Cost of meal n
100% of n
+
+
tip of 20% of the cost
20% of n
120% of n
1.2n  66
1.2n
66

1.2
1.2
n  55
=
=
=
$66
$66
$66
Mark and Peggy can spend
up to $55 on the meal itself.
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Solving Discount Problems
Example
Julie bought a leather sofa that was on sale for 35% off the original
price of $1200. What was the discount? How much did Julie pay
for the sofa?
Discount = discount rate  list price
= 35%  1200
The discount was $420.
= 420
Amount paid = list price – discount
= 1200 – 420
= 780
Julie paid $780 for the sofa.
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Solving Increase Problems
Percent of increase =
amount of increase
original amount
Example
The cost of a certain car increased from $16,000 last year to
$17,280 this year. What was the percent of increase?
Amount of increase = original amount – new amount
= 17,280 – 16,000 = 1280
amount of increase
original amount
1280
=
= 0.08 The car’s cost increased by 8%.
16000
Percent of increase =
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Solving Decrease Problems
Percent of decrease =
amount of decrease
original amount
Example
Patrick weighed 285 pounds two years ago. After dieting, he reduced
his weight to 171 pounds. What was the percent of decrease in his
weight?
Amount of decrease = original amount – new amount
= 285 – 171 = 114
amount of decrease
Percent of decrease =
original amount
114
=
= 0.4
285
Patrick’s weight
decreased by 40%.
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Solving Mixture Problems
Example
The owner of a candy store is mixing candy worth $6 per pound with
candy worth $8 per pound. She wants to obtain 144 pounds of candy
worth $7.50 per pound. How much of each type of candy should she
use in the mixture?
1.) Understand
Let n = the number of pounds of candy costing $6 per pound.
Since the total needs to be 144 pounds, we can use 144  n for
the candy costing $8 per pound.
Continued
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Solving Mixture Problems
Example continued
2.) Translate
Use a table to summarize the information.
$6 candy
$8 candy
$7.50 candy
Number of Pounds
n
144  n
144
Price per Pound
6
8
7.50
Value of Candy
6n
8(144  n)
144(7.50)
6n + 8(144  n) = 144(7.5)
# of
pounds of
$6 candy
# of
pounds of
$8 candy
# of
pounds of
$7.50
candy
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Solving Mixture Problems
Example continued
3.) Solve
6n + 8(144  n) = 144(7.5)
6n + 1152  8n = 1080
1152  2n = 1080
2n = 72
n = 36
(Eliminate the parentheses)
(Combine like terms)
(Subtract 1152 from both sides)
(Divide both sides by 2)
She should use 36 pounds of the $6 per pound candy.
She should use 108 pounds of the $8 per pound candy.
(144  n) = 144  36 = 108
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Solving Mixture Problems
Example continued
4.) Interpret
Check: Will using 36 pounds of the $6 per pound candy and
108 pounds of the $8 per pound candy yield 144 pounds of
candy costing $7.50 per pound?
?
6(36) + 8(108) = 144(7.5)
?
216 + 864 = 1080
?
1080 = 1080 
State: She should use 36 pounds of the $6 per pound candy and
108 pounds of the $8 per pound candy.
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§ 9.7
Solving Linear
Inequalities
Linear Inequalities
A linear inequality in one variable is an equation
that can be written in the form ax + b < c
• a, b, and c are real numbers, a  0
• < symbol could be replaced by > or  or 
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Graphing Solutions
Graphing solutions to linear inequalities in one
variable (using circles)
• Use a number line
• Use a closed circle at the endpoint of a interval
if you want to include the point
• Use an open circle at the endpoint if you DO
NOT want to include the point
Represents the set {xx  7}
7
Represents the set {xx > – 4}
-4
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Linear Inequalities
Graphing solutions to linear inequalities in one
variable (using interval notation)
• Use a number line
• Use a bracket at the endpoint of a interval if you want
to include the point
• Use a parenthesis at the endpoint if you DO NOT want
to include the point
Interval Notation
]
Represents the set (– , 7]
7
(
Represents the set (– 4, )
-4
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Properties of Inequality
Addition Property of Inequality
• a < b and a + c < b + c are equivalent
inequalities
Multiplication Property of Inequality
• a < b and ac < bc are equivalent inequalities,
if c is positive
• a < b and ac > bc are equivalent inequalities,
if c is negative
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Solving Linear Inequalities
Solving linear inequalities in one variable
1)
2)
3)
4)
Multiply to clear fractions
Use distributive property
Simplify each side of equation
Get all variable terms on one side and
numbers on the other side of equation
(addition property of equality)
5) Isolate variable (multiplication property of
equality)
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Solving Linear Inequalities
Example
3x + 9  5(x – 1)
3x + 9  5x – 5
(Use distributive property on right side)
3x – 3x + 9  5x – 3x – 5
(Subtract 3x from both sides)
9  2x – 5
(Simplify both sides)
9 + 5  2x – 5 + 5
(Add 5 to both sides)
14  2x
(Simplify both sides)
7x
(Divide both sides by 2)
Graph of solution (– ,7]
]
7
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Solving Linear Inequalities
Example
7(x – 2) + x > – 4(5 – x) – 12
7x – 14 + x > – 20 + 4x – 12
8x – 14 > 4x – 32
8x – 4x – 14 > 4x – 4x – 32
4x – 14 > –32
4x – 14 + 14 > –32 + 14
4x > –18
x
9
2
(Use distributive property)
(Simplify both sides)
(Subtract 4x from both sides)
(Simplify both sides)
(Add 14 to both sides)
(Simplify both sides)
(Divide both sides by 4 and simplify)
Graph of solution ( -9
,)
2
(
-9
2
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Compound Inequalities
A compound inequality is two inequalities joined
together.
0  4(5 – x) < 8
To solve the compound inequality, perform operations
simultaneously to all three parts of the inequality (left,
middle and right).
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Solving Compound Inequalities
Example
0  4(5 – x) < 8
0  20 – 4x < 8
(Use the distributive property)
0 – 20  20 – 20 – 4x < 8 – 20
(Subtract 20 from each part)
– 20  – 4x < – 12
(Simplify each part)
5x>3
(Divide each part by –4)
Remember that the sign direction changes when you divide by a
number < 0!
Graph of solution (3,5]
(
]
3
5
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