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Chapter 9
Equations, Inequalities and Problem Solving
Martin-Gay, Introductory Algebra, 3ed 2Martin-Gay, Developmental Mathematics 2
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
Chapter Sections
§ 9.1
The Addition Property of Equality
Martin-Gay, Introductory Algebra, 3ed 4Martin-Gay, Developmental Mathematics 4
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
Martin-Gay, Introductory Algebra, 3ed 5Martin-Gay, Developmental Mathematics 5
Addition Property of Equality
a = b and a + c = b + c are equivalent equations
Addition Property of Equality
z = – 16 (Simplify both sides)
Example
8 + (– 8) + z = – 8 + – 8 (Add –8 to each side)
8 + z = – 8a.)
Martin-Gay, Introductory Algebra, 3ed 6Martin-Gay, Developmental Mathematics 6
Example
Solving Equations
4p – 11 – p = 2 + 2p – 20
3p – 11 = 2p – 18 (Simplify both sides)
p = – 7 (Simplify both sides)
p – 11 = – 18 (Simplify both sides)
3p + (– 2p) – 11 = 2p + (– 2p) – 18 (Add –2p to both sides)
p – 11 + 11 = – 18 + 11 (Add 11 to both sides)
Martin-Gay, Introductory Algebra, 3ed 7Martin-Gay, Developmental Mathematics 7
Example
Solving Equations
5(3 + z) – (8z + 9) = – 4z
15 + 5z – 8z – 9 = – 4z (Use distributive property)
6 – 3z = – 4z (Simplify left side)
6 + z = 0 (Simplify both sides)
z = – 6 (Simplify both sides)
6 – 3z + 4z = – 4z + 4z (Add 4z to both sides)
6 + (– 6) + z = 0 +( – 6) (Add –6 to both sides)
§ 9.2
The Multiplication Property of Equality
Martin-Gay, Introductory Algebra, 3ed 9Martin-Gay, Developmental Mathematics 9
Multiplication property of equalitya = b and ac = bc are equivalent equations
Multiplication Property of Equality
Example
– y = 8
y = – 8 (Simplify both sides)
(– 1)(– y) = 8(– 1) (Multiply both sides by –1)
Martin-Gay, Introductory Algebra, 3ed 10Martin-Gay, Developmental Mathematics 10
Example
Solving Equations
9
5
7
1x
(Simplify both sides)9
35x
79
5
7
17
x (Multiply both sides by 7)
Martin-Gay, Introductory Algebra, 3ed 11Martin-Gay, Developmental Mathematics 11
Example
Solving Equations
16x (Simplify both sides)
8 63
x
3
86
8
3
3
8x (Multiply both sides by fraction)
Martin-Gay, Introductory Algebra, 3ed 12Martin-Gay, Developmental Mathematics 12
Example
Solving Equations
Recall that multiplying by a number is equivalent to dividing by its reciprocal
3z – 1 = 26
3z = 27 (Simplify both sides)
z = 9 (Simplify both sides)
3z – 1 + 1 = 26 + 1 (Add 1 to both sides)
(Divide both sides by 3)3
27
3
3
z
Martin-Gay, Introductory Algebra, 3ed 13Martin-Gay, Developmental Mathematics 13
Example
Solving Equations
12x + 30 + 8x – 6 = 10
20x + 24 = 10 (Simplify left side)
20x = – 14 (Simplify both sides)
20
14
20
20
x(Divide both sides by 20)
10
7x (Simplify both sides)
20x + 24 + (– 24) = 10 + (– 24) (Add –24 to both sides)
§ 9.3
Further Solving Linear Equations
Martin-Gay, Introductory Algebra, 3ed 15Martin-Gay, Developmental Mathematics 15
Solving Linear Equations
Solving linear equations in one variable1) Multiply to clear fractions
2) Use distributive property
3) Simplify each side of equation
4) 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
Martin-Gay, Introductory Algebra, 3ed 16Martin-Gay, Developmental Mathematics 16
301093 yy (Simplify)
7
7
7
21 y
(Simplify; divide both sides by 7)
y 3 (Simplify both sides)
Solving Linear Equations
Example3( 3) 2 6
5y y
(Multiply both sides by 5) 6255
)3(35
y
y
(Add –3y to both sides)30)3(109)3(3 yyyy
(Simplify; add –30 to both sides))30(307)30(9 y
Martin-Gay, Introductory Algebra, 3ed 17Martin-Gay, Developmental Mathematics 17
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.”
Martin-Gay, Introductory Algebra, 3ed 18Martin-Gay, Developmental Mathematics 18
Solving Linear Equations
Example
Since no value for the variable x can be substituted into this equation that will make this a true statement, there is “no solution.”
3x – 7 = 3(x + 1)
3x – 7 = 3x + 3 (Use distributive property)
– 7 = 3 (Simplify both sides)
3x + (– 3x) – 7 = 3x + (– 3x) + 3 (Add –3x to both sides)
§ 9.4
An Introduction to Problem Solving
Martin-Gay, Introductory Algebra, 3ed 20Martin-Gay, Developmental Mathematics 20
Strategy for Problem Solving
General Strategy for Problem Solving1) 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 equation3) Solve the equation4) Interpret the result
• Check proposed solution in problem• State your conclusion
Martin-Gay, Introductory Algebra, 3ed 21Martin-Gay, Developmental Mathematics 21
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.
Finding an Unknown Number
Example
Continued
Martin-Gay, Introductory Algebra, 3ed 22Martin-Gay, Developmental Mathematics 22
The product of
·
twice a number
2x
and 3
3
is the same as
=
5 times the number
5x
and ¾
¾
the difference of
–
Finding an Unknown Number
Example continued
2.) Translate
Continued
Martin-Gay, Introductory Algebra, 3ed 23Martin-Gay, Developmental Mathematics 23
Finding an Unknown Number
Example continued3.) Solve
2x · 3 = 5x – ¾
6x = 5x – ¾ (Simplify left side)
x = – ¾ (Simplify both sides)
6x + (– 5x) = 5x + (– 5x) – ¾ (Add –5x to both sides)
4.) InterpretCheck: 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.
Martin-Gay, Introductory Algebra, 3ed 24Martin-Gay, Developmental Mathematics 24
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.) UnderstandRead 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
Solving a Problem
Example
Continued
Martin-Gay, Introductory Algebra, 3ed 25Martin-Gay, Developmental Mathematics 25
Solving a Problem
Example continued
2.) Translate
Continued
Daily costs
2(24.95)
mileage costs
0.29x
plus
+
is equal to
= 100
maximum budget
Martin-Gay, Introductory Algebra, 3ed 26Martin-Gay, Developmental Mathematics 26
Solving a Problem
Example continued3.) Solve
Continued
2(24.95) + 0.29x = 100
49.90 + 0.29x = 100 (Simplify left side)
0.29x = 50.10 (Simplify both sides)
29.0
10.50
29.0
29.0
x(Divide both sides by 0.29)
x 172.75 (Simplify both sides)
(Subtract 49.90 from both sides)49.90 – 49.90 + 0.29x = 100 – 49.90
Martin-Gay, Introductory Algebra, 3ed 27Martin-Gay, Developmental Mathematics 27
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.
§ 9.5
Formulas and Problem Solving
Martin-Gay, Introductory Algebra, 3ed 29Martin-Gay, Developmental Mathematics 29
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)
Martin-Gay, Introductory Algebra, 3ed 30Martin-Gay, Developmental Mathematics 30
Using Formulas
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
Example
Continued
Martin-Gay, Introductory Algebra, 3ed 31Martin-Gay, Developmental Mathematics 31
Using Formulas
Example continued
2.) Translate
Continued
Formula: P = a + b + c Substitute: 102 = x + 2x + x + 30
3.) Solve102 = 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)
4
4
4
72 x (Divide both sides by 4)
18 = x (Simplify both sides)
Martin-Gay, Introductory Algebra, 3ed 32Martin-Gay, Developmental Mathematics 32
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.
Martin-Gay, Introductory Algebra, 3ed 33Martin-Gay, Developmental Mathematics 33
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 Formulas
Solving Equations for a Specific Variable1) Multiply to clear fractions
2) Use distributive to remove grouping symbols
3) Combine like terms to simply each side
4) Get all terms containing specified variable on the same time, other terms on opposite side
5) Isolate the specified variable
Martin-Gay, Introductory Algebra, 3ed 34Martin-Gay, Developmental Mathematics 34
Solving Equations for a Specific Variable
Example
mnrT
mr
mnr
mr
T (Divide both sides by mr)
nmr
T (Simplify right side)
Solve for n.
Martin-Gay, Introductory Algebra, 3ed 35Martin-Gay, Developmental Mathematics 35
Solving Equations for a Specific Variable
Example
Solve for T.
PRTPPPA (Subtract P from both sides)
PRTPA (Simplify right side)
PR
PRT
PR
PA
(Divide both sides by PR)
TPR
PA
(Simplify right side)
A P PRT
Martin-Gay, Introductory Algebra, 3ed 36Martin-Gay, Developmental Mathematics 36
Solving Equations for a Specific Variable
Example
Solve for P.A P PRT
)1( RTPA (Factor out P from both terms on the right side)
RT
RTP
RT
A
1
)1(
1(Divide both sides by 1 + RT)
PRT
A
1(Simplify the right side)
§ 9.6
Percent and Mixture Problem Solving
Martin-Gay, Introductory Algebra, 3ed 38Martin-Gay, Developmental Mathematics 38
Solving a Percent Problem
A percent problem has three different parts:
1. When we do not know the amount:n = 10% · 500
Any one of the three quantities may be unknown.
amount = percent · base
2. When we do not know the base:50 = 10% · n
3. When we do not know the percent:50 = n · 500
Martin-Gay, Introductory Algebra, 3ed 39Martin-Gay, Developmental Mathematics 39
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
Martin-Gay, Introductory Algebra, 3ed 40Martin-Gay, Developmental Mathematics 40
Solving a Percent Problem: Base Unknown
amount = percent · base
36 is 6% of what?
n = 6% ·36
36 = 0.06n36 0.06 =
0.06 0.06n
600 = n
36 is 6% of 600
Martin-Gay, Introductory Algebra, 3ed 41Martin-Gay, Developmental Mathematics 41
Solving a Percent Problem: Percent Unknown
amount = percent · base
n = 14424
24 is what percent of 144?
24 = 144n
24 144 = 144 144
n
0.16 = n216 % = 3
n 224 is 16 % of 1443
Martin-Gay, Introductory Algebra, 3ed 42Martin-Gay, Developmental Mathematics 42
Solving Markup Problems
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 + tip of 20% of the cost = $66
100% of n + 20% of n = $66120% of n = $66
1.2 66n 1.2 66 1.2 1.2
n
55n Mark and Peggy can spend up to $55 on the meal itself.
Example
Martin-Gay, Introductory Algebra, 3ed 43Martin-Gay, Developmental Mathematics 43
Solving Discount Problems
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?
Julie paid $780 for the sofa.
Discount = discount rate list price
= 35% 1200= 420 The discount was $420.
Amount paid = list price – discount
= 1200 – 420
= 780
Example
Martin-Gay, Introductory Algebra, 3ed 44Martin-Gay, Developmental Mathematics 44
Solving Increase Problems
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
The car’s cost increased by 8%.
amount of increase = original amount
Percent of increase
= 17,280 – 16,000 = 1280
amount of increasePercent of increase = original amount
1280= 16000
= 0.08
Example
Martin-Gay, Introductory Algebra, 3ed 45Martin-Gay, Developmental Mathematics 45
Solving Decrease Problems
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
Patrick’s weight decreased by 40%.
amount of decrease = original amount
Percent of decrease
= 285 – 171 = 114
amount of decreasePercent of decrease = original amount
114= 285
= 0.4
Example
Martin-Gay, Introductory Algebra, 3ed 46Martin-Gay, Developmental Mathematics 46
Solving Mixture Problems
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.
Example
Continued
Martin-Gay, Introductory Algebra, 3ed 47Martin-Gay, Developmental Mathematics 47
Solving Mixture Problems
Example continued
2.) Translate
Continued
Use a table to summarize the information.
Number of Pounds Price per Pound Value of Candy$6 candy n 6 6n$8 candy 144 n 8 8(144 n)$7.50 candy 144 7.50 144(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
Martin-Gay, Introductory Algebra, 3ed 48Martin-Gay, Developmental Mathematics 48
Solving Mixture Problems
Example continued
Continued
3.) Solve
6n + 8(144 n) = 144(7.5)
6n + 1152 8n = 1080
1152 2n = 1080
2n = 72
(Eliminate the parentheses)
(Combine like terms)
(Subtract 1152 from both sides)
n = 36 (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
Martin-Gay, Introductory Algebra, 3ed 49Martin-Gay, Developmental Mathematics 49
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?
State: She should use 36 pounds of the $6 per pound candy and 108 pounds of the $8 per pound candy.
6(36) + 8(108) = 144(7.5)?
216 + 864 = 1080 ?
1080 = 1080?
§ 9.7
Solving Linear Inequalities
Martin-Gay, Introductory Algebra, 3ed 51Martin-Gay, Developmental Mathematics 51
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
Martin-Gay, Introductory Algebra, 3ed 52Martin-Gay, Developmental Mathematics 52
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
7Represents the set {xx 7}
-4Represents the set {xx > – 4}
Graphing Solutions
Martin-Gay, Introductory Algebra, 3ed 53Martin-Gay, Developmental Mathematics 53
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
7]
-4( Represents the set (– 4, )
Represents the set (– , 7]
Interval Notation
Linear Inequalities
Martin-Gay, Introductory Algebra, 3ed 54Martin-Gay, Developmental Mathematics 54
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
Properties of Inequality
Martin-Gay, Introductory Algebra, 3ed 55Martin-Gay, Developmental Mathematics 55
Solving linear inequalities in one variable1) Multiply to clear fractions
2) Use distributive property
3) Simplify each side of equation
4) 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)
Solving Linear Inequalities
Martin-Gay, Introductory Algebra, 3ed 56Martin-Gay, Developmental Mathematics 56
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)
14 2x (Simplify both sides)
7 x (Divide both sides by 2)
9 + 5 2x – 5 + 5 (Add 5 to both sides)
Graph of solution (– ,7]7]
Solving Linear Inequalities
Example
Martin-Gay, Introductory Algebra, 3ed 57Martin-Gay, Developmental Mathematics 57
7(x – 2) + x > – 4(5 – x) – 12 7x – 14 + x > – 20 + 4x – 12 (Use distributive property) 8x – 14 > 4x – 32 (Simplify both sides)8x – 4x – 14 > 4x – 4x – 32 (Subtract 4x from both sides) 4x – 14 > –32 (Simplify both sides)4x – 14 + 14 > –32 + 14 (Add 14 to both sides) 4x > –18 (Simplify both sides)
2
9x (Divide both sides by 4 and simplify)
-92
(Graph of solution ( ,)-92
Solving Linear Inequalities
Example
Martin-Gay, Introductory Algebra, 3ed 58Martin-Gay, Developmental Mathematics 58
A compound inequality is two inequalities joined together.
To solve the compound inequality, perform operations simultaneously to all three parts of the inequality (left, middle and right).
Compound Inequalities
0 4(5 – x) < 8
Martin-Gay, Introductory Algebra, 3ed 59Martin-Gay, Developmental Mathematics 59
0 4(5 – x) < 8
0 20 – 4x < 8 (Use the distributive property)
3 5](
Graph of solution (3,5]
– 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!
0 – 20 20 – 20 – 4x < 8 – 20 (Subtract 20 from each part)
Solving Compound Inequalities
Example
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