1. Equations, Inequalities, and Problem Solving
Chapter 2
Equations, Inequalities, and Problem Solving
2.1
2.2
2.3
2.4
2.5
2.6
2.7 1

2. Linear Equations in One Variable
2.1
Linear Equations in One Variable
Objectives:
Solve linear equations using properties of equality
Solve linear equations that can be simplified by combining like terms
Solve linear equations containing fractions or decimals
Recognize identities and equations with no solutions
Key Vocabulary:
Solution, equivalent equation, conditional equation, contradiction, identity

3. Linear Equations
An algebraic equation is a statement in which two expressions have equal value.
Solving algebraic equations involves finding values for a variable that make the equation true.
Equivalent equations are equations with the same solution set.

4. Example 1 Solve for x: 3x + 7 = 16.
Check:
3x + 7 = 16
The solution or the solution set is {3}.

5. Example 3 Solve: 4p – 11 – p = 2 + 2p – 20.
3p – 11 = 2p – Combine like terms.
– 2p – 2p Subtract 2p from both sides.
p – 11 = – Simplify.
Add 11 to both sides.
p = – Simplify.
Of course you should check the solution. Let’s check this one using the sto→ button on the calculator

6. Example 3
Solve: 12x x – 6 = 10.
20x + 24 = Simplify left side.
– 24 – Subtract 24 from both sides.
20x = – Simplify both sides.
Divide both sides by 20.
Simplify both sides.
Check using the sto→ button on the calculator

7. Example 4 Solve: 5(3 + z) – (8z + 9) = – 4z.
15 + 5z – 8z – 9 = – 4z Use distributive property.
6 – 3z = – 4z Simplify left side.
+ 4z z Add 4z to both sides.
6 + z = Simplify both sides.
– – Subtract 6 from both sides.
z = – 6 Simplify both sides.
Check using the sto→ button on the calculator

8. Solving a Linear Equation in One Variable
Clear the equation of fractions by multiplying both sides of the equation by the least common denominator (LCD) of all denominators in the equation.
Use the distributive property to remove grouping symbols such as parentheses.
Combine like terms on each side of the equation.
Use the addition property of equality to rewrite the equation as an equivalent equation with variable terms on one side and numbers on the other side.
Use the multiplication property of equality to isolate the variable.
Check the proposed solution in the original equation.

9. Example 6 Solve for y: -3y -3y 9 = 7y + 30 -30 -30
Check using the sto→ button on the calculator

10. A linear equation in one variable that has exactly one solution is called a conditional equation.
An equation in one variable that has no solution is called a contradiction.
An equation in one variable that has every number (for which the equation is defined) as a solution is called an identity. (infinite solutions)

11. Example 8 Solve for x: 3x – 7 = 3(x + 1).
3x – 7 = 3x Use distributive property.
–3x –3x Subtract 3x from both sides.
–7 = Simplify both sides.
The equation 7 = 3 is false no matter what value the variable x might have. Thus, the original equation has no solution. This equation is a contradiction.

12. Example 9 Solve for x: 5x – 5 = 2(x + 1) + 3x – 7.
5x – 5 = 2x x – 7 Use distributive property.
5x – 5 = 5x – Simplify the right side.
– 5x – 5x Subtract 5x from both sides.
– 5 = – Simplify.
Since – 5 = – 5 is a true statement for every value of x, all real numbers are solutions. The solution set is the set of all real numbers. The equation is called an identity.

13. Solve: additional ex 5 4

14. Linear Equations in One Variable
2.1 summary
Linear Equations in One Variable
Objectives:
Solve linear equations using properties of equality
Solve linear equations that can be simplified by combining like terms
Solve linear equations containing fractions or decimals
Recognize identities and equations with no solutions
Key Vocabulary:
Solution, equivalent equation, conditional equation, contradiction, identity

15. An Introduction to Problem Solving
2.2
An Introduction to Problem Solving
Objectives:
Write algebraic expressions that can be simplified
Apply the steps for problem solving
Key Vocabulary:
Consecutive integers, complementary, supplementary

16. Consecutive, even, and odd integers and their representations.
+ 2
+ 1
Consecutive Integers: x
x + 1
x + 2
+ 4
Consecutive Even Integers:
+ 2 x
x + 2
x + 4
Consecutive Odd Integers:
+ 4
+ 2 x
x + 2
x + 4

17. Example 1
Write the following as an algebraic expression. Then simplify. The sum of three consecutive odd integers, if x is the first consecutive integer. In words: Translate:
first
integer
plus
next odd
integer
plus
next odd
integer x +
(x + 2) +
(x + 4)
= x + x x + 4
= 3x + 6

18. General Strategy for Problem Solving
UNDERSTAND the problem. During this step, become comfortable with the problem. Some way of doing this are:
Read and reread the problem.
Propose a solution and check.
Construct a drawing.
Choose a variable to represent the unknown.
TRANSLATE the problem into an equation.
SOLVE the equation.
INTERPRET the result. Check the proposed solution in the stated problem and state your conclusion.

19. Example 3
Find three numbers such that the second is 3 more than twice the first number, and the third number is four times the first number. The sum of the three numbers is 164. Let x = the first number, then 2x + 3 = the second number 4x = the third number
7x + 3 = 164
7x = 161
x = 23
In words:
Translate: x (2x + 3) x =
first
number
added to
second
number
added to
third
number is
164
1st # = 23
2nd # = 49
3rd # = 92

20. Example 6
Kelsey Ohleger was helping her friend Benji Burnstine study for an algebra exam. Kelsey told Benji that her last three latest art history quiz scores are three consecutive even integers whose sum is 264. Help Benji find the scores. x = the first integer. Then x + 2 = the second consecutive even integer x + 4 = the third consecutive even integer.
1st # = 86
3x + 6 = 264
3x = 258
x = 86
2nd # = 88
3rd # = 90

21. Add on: find the perimeter
The problem is miss leading on the system because they put a question mark on the picture which isn’t necessary
y + 1 2 10
20 + ?
Peri = (7y – 3)
7y – 3
= y – 6
Peri = y

22. Add on: In 2010, the population of the country was 34. 6 million
Add on: In 2010, the population of the country was 34.6 million. This represented an increase in population of 8.1% since What was the population of the country in 2001? Round to the nearest hundredth of a million.
2001population + 8.1%of(2001population) = 2010population
p p = 34.6
1.081p = 34.6
p = 32.01
If the total bill for a TV is $ and the tax rate is 8.5%, find the original price of the TV. Round to the nearest cent.
TV TV =
1.085TV =
$420.94

23. An Introduction to Problem Solving
2.2 summary
An Introduction to Problem Solving
Objectives:
Write algebraic expressions that can be simplified
Apply the steps for problem solving
Key Vocabulary:
Consecutive integers, complementary, supplementary

24. Formulas and Problem Solving
2.3
Formulas and Problem Solving
Objectives:
Solve a formula for a specified variable
Use formulas to solve problems

25. 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

26. Example 1 & 2
Solve: V = lwh for w.
Solve 4y – 5x = 9 for y.

27. Solving Equations for a Specific Variable
Clear the equations of fractions by multiplying each side of the equation by the least common denominator.
Use the distributive property to remove grouping symbols such as parentheses.
Combine like terms on each side of the equation.
Use the addition property of equality to rewrite the equation as an equivalent equation with terms containing the specified variable on one side and all other terms on the other side.
Use the distributive property and the multiplication property of equality to isolate the specified variable.

28. Example 3
Example 4
A = P + PRT solve for T.
A = P + PRT solve for P.

29. Compound interest formula
Compound interest formula. A = amount in account after t years P = principal or amount invested t = time in years r = annual rate of interest n = number of times compounded per year
For these questions you need a scientific calculator

30. Example 5 P = r = t = n = Formula: Substitute $15,000 0.03 5 years
June Myers just received an inheritance of $15,000 and plans to place all the money in a savings account that pays 3% compounded quarterly to help her son go to college in 5 years. How much money will be in the account in 5 years?
P = r = t = n =
Formula:
Substitute
$15,000
0.03
5 years
4 times/yr
Answer: In 5 years, the account will contain $17,

31. Example 6
The fastest average speed by a cyclist across the continental United States is 15.4 mph, by Pete Penseyres. If he traveled a total distance of about miles at this speed, find his time cycling. Write the time in days, hours, and minutes. (Source: The Guinness Book of World Records) distance formula d = rt
÷ 24 = days
d =
r =
miles
15.4 mph
0.4077(24) = hours
0.7857(60) = minutes
Answer: Petes cycling time was approximately 8 days, 9 hours, 47 minutes.
hours

32. Formulas and Problem Solving
2.3 summary
Formulas and Problem Solving
Objectives:
Solve a formula for a specified variable
Use formulas to solve problems

33. Linear Inequalities and Problem Solving
2.4
Linear Inequalities and Problem Solving
Objectives:
Graph inequalities
Solve linear inequalities
Solve problems that can be modeled by linear inequalities

34. The solution set of an inequality is the set of all solutions.
A solution of an inequality is a value of the variable that makes the inequality a true statement.
The solution set of an inequality is the set of all solutions.
The inequality x > 2 has a solution set of
{x | x > 2|, read as
The set of all numbers x such that x is greater than 2

35. Example 1
Graph each on a number line. a. x ≥ 3 b. x < –3 c. 0.5 < x ≤ 4 2
– 2 1 3 4 5
– 1
– 3
– 4
– 5 2
– 2 1 3 4 5
– 1
– 3
– 4
– 5 2
– 2 1 3 4 5
– 1
– 3
– 4
– 5

36. Addition Property of Inequality
If a, b, and c are real numbers, then
a < b and a + c < b + c
are equivalent inequalities.
Multiplication Property of Inequality
If a, b, and c are real numbers and c is positive, then a < b and ac < bc
are equivalent inequalities.
If a, b, and c are real numbers and c is negative, then a < b and ac > bc

37. Example 2 Solve each and graph the solutions on a number line.
x < 13
4x + 5 ≥ 3x – 7
– 3x – 3x
x + 5 ≥ –7
– 5 – 5
x ≥ –12 12 10 11 13 14 15
– 12
– 11
– 13
– 14
– 15

38. Example 3 Solve and graph each on a number line. 2 – 2 1 3 4 5 – 1 – 31 3 4 5
– 1
– 3
– 4
– 5 2
– 2 1 3 4 5
– 1
– 3
– 4
– 5

39. Example 4 Solve and graph on a number line: 3x + 9  5(x – 1)
7(x – 2) + x > – 4(5 – x) – 12
3x + 9  5x – 5
7x – 14 + x > – x – 12
8x – 14 > 4x – 32
– 4x – 4x
4x – 14 > –32
4x > –18
– 3x – 3x
9  2x – 5
14  2x
7  x
x ≤ 7 5 4 6 7 8 2
– 2 1 3 4 5
– 1
– 3
– 4
– 5

40. Example 7 Solve: 3(x + 4) > 3x + 5. 3(x + 4) > 3x + 5
3x + 12 > 3x Distribute on the left side.
– 3x – 3x Subtract 3x from both sides.
12 > Simplify.
12 > 5 is a true statement for all values of x, so this inequality and the original inequality are true for all numbers. All real numbers are solutions.

41. Example 8
A salesperson earns $800 per month plus a commission of 25% of sales. Find the minimum amount of sales needed to receive a total income of at least $1800 per month.
Let x = amount of sales.
% commission of sales ≥ 1800
Answer: The minimum amount of sales needed for the salesperson to earn at least $1800 per month is $4000 per month.

42. Example 9
In the United States, the annual consumption of cigarettes is declining. The consumption c in billions of cigarettes per year since the year 1990 can be approximated by the formula c = –9.2t where t is the number of years after Use this formula to predict the years that the consumption of cigarettes will be less than 200 billion per year.
Answer: The annual consumption of cigarettes will be less than 200 billion more than years after 1990, or in approximately 2026.

43. Linear Inequalities and Problem Solving
2.4 Summary
Linear Inequalities and Problem Solving
Objectives:
Graph inequalities
Solve linear inequalities
Solve problems that can be modeled by linear inequalities

44. Compound Inequalities
2.5
Compound Inequalities
Objectives:
Find the intersection and union of two sets
Solve compound inequalities containing and and or

45. The solution set of a compound inequality formed by the word and is the intersection of the solution sets of the two inequalities. We use the symbol  to represent “intersection.”
Intersection of Two Sets
The intersection of two sets, A and B, is the set of all elements common to both sets.
A intersect B is denoted by A B
A  B

46. The solution set of a compound inequality formed by the word or is the union of the solution sets of the two inequalities. We use the symbol  to denote “union.”
Union of Two Sets
The union of two sets, A and B, is the set of elements that belong to either of the sets. A union B is denoted by A
A  B B

47. Graphs Or (υ) Union And (∩) Intersection x > 2 x > 8
All real #’s
2 < x < 8 2 8
x < 2 or x > 8
No solution 2 8

48. Example 1
If A = {x | x is an odd number greater than 0 but less than 9} and B = {4, 5, 6, 7, 8}, find A  B. List the elements of A. A = {1, 3, 5, 7} List the elements of B. A = {4, 5, 6, 7, 8} The numbers 5 and 7 are in sets A and B. The intersection is {5, 7}.

49. Example 2
Solve: x + 4 > 0 and 4x > 0. Solve each inequality separately. x + 4 > 0 and 4x > 0 x > 4 and x > 0 Graph the two inequalities on a number line and find their intersection. x > 4 x > 0 x > 4 and x > 0 Solution is x > 0 2
– 2 1 3 4 5
– 1
– 3
– 4
– 5 2
– 2 1 3 4 5
– 1
– 3
– 4
– 5

50. Example 3
Solve: 5x > 0 and 3x  4 ≤ 13. Solve each inequality separately. 5x > 0 and 3x – 4 ≤ 13 x > 0 and 3x ≤ 9 x ≤ 3 Graph the two inequalities and find their intersection. x > 0 x > 0 and x ≤ 3 There is no number that is greater than 0 and less than or equal to 3. The answer is no solution. 2
– 2 1 3 4 5
– 1
– 3
– 4
– 5 2
– 2 1 3 4 5
– 1
– 3
– 4
– 5

51. To solve a compound inequality written in compact form, such as 3 < 5 – x < 9, we get x alone in the “middle part.” We must perform the same operations on all three parts of the inequality.

52. Example 4
Solve and graph on a number line: 3 < 5 – x < 9. 3 < 5 – x < 9 –5 –5 –5 2
– 2 1 3 4 5
– 1
– 3
– 4
– 5

53. Example 6
If A = {x | x is an odd number greater than 0 but less than 9} and B = {4, 5, 6, 7, 8}, find A  B. List the elements of A. A = {1, 3, 5, 7} B = {4, 5, 6, 7, 8} The numbers that are in either set or both sets are {1, 3, 4, 5, 6, 7, 8} This set is the union.

54. Example 7 Solve: 6x – 4 ≤ 12 or x + 2 ≥ 8.
Solve each inequality separately.
Find the union of the graphs. 4 2 3 5 6 7 1
– 1
– 2
– 3 4 2 3 5 6 7 1
– 1
– 2
– 3
The solutions are x ≤ 8/3 or x ≥ 6.

55. Example 8 Solve: 2x – 6 < 2 or 8x < 0. Find the union.
Solve each inequality separately.
Find the union. 2
– 2 1 3 4 5
– 1
– 3
– 4
– 5 2
– 2 1 3 4 5
– 1
– 3
– 4
– 5
The solutions are all real numbers.

56. Compound Inequalities
2.5 summary
Compound Inequalities
Objectives:
Find the intersection and union of two sets
Solve compound inequalities containing and and or

57. Absolute Value Equations
2.6
Absolute Value Equations
Objectives:
Solve absolute value equations

58. Example 1 Solve: 6 + 2n = 4. 6 + 2n = 4 or 6 + 2n = 4
Check: To check, let n = 1 and then n = 5 in the original equation.
|6 + 2(1)| = 4 |6 + 2(5)| = 4
|6 – 2| = |6 – 10| = 4
|4| = | –4| = 4
4 = 4 True = 4 True
The solutions are 1 and 5.

59. Example 2 Solve: 2x 6 = 4 2x 6 = 4 2x= 10 2x = 10 or 2x = 10
We want the absolute value expression alone on one side of the equation, so begin by adding 6 to both sides. Then apply the absolute value property.
2x 6 = 4
2x= 10
2x = 10 or 2x = 10
x = 5 or x = 5
The solutions are 5 and 5.

60. Example 3 Example 4 Solve: 7x+2 = 0. 7x +2= 0 x = –2/7
Solve: 3z  2+ 8 = 1.
First, isolate the absolute value.
3z  2+ 8 = 1
3z  2 = 7
The absolute value of a number is NEVER negative, so this equation has no solution.

61. Example 5
Solve: The absolute value of any expression is NEVER negative, so no solution exists. This equation has no solution.

62. Example 6 Solve: The solutions are 1/3 and 1.
This equation is true if the expressions inside the absolute value bars are equal to or are opposites of each other.
The solutions are 1/3 and 1.

63. Example 7
Solve:
Solution: x = 5

64. Absolute Value Equations
2.6 summary
Absolute Value Equations
Objectives:
Solve absolute value equations

65. Absolute Value Inequalities
2.7
Absolute Value Inequalities
Objectives:
Solve absolute value inequalities

66. Example 1
Solve: |x| ≤ 7. The solution set of this inequality contains all numbers whose distance from 0 is less than or equal to 7. Thus 7, 7, and all numbers between 7 and 7 are in the solution set. The solutions are 7 ≤ x ≤ 7.

67. Example 2
Solve for x: x + 4 < 6. |x + 4| < 6 is equivalent to 6 < x + 4 < 6 6 < x + 4 < 6 –4 –4 –4 10 < x < 2

68. Example 3 Solve for x: |x – 3| + 6 ≤ 7. +3 +3 +3
First, isolate the absolute value expression by subtracting 6 from both sides.
Solve for x: |x – 3| + 6 ≤ 7.

69. Example 4
Solve for x: 8x  3 < 2. The absolute value of a number is always nonnegative and can never be less than 2. Thus this absolute value inequality has no solution.

70. Example 5
Solve for y: |y – 5| > 8. Solve the compound inequality. y – 5 < – 8 or y – 5 > y < –3 or y > 13
The solutions are all numbers y such that y < 3 or y > 13.

71. Example 6 Solve for x: 4x + 5 + 6 > 2. 4x + 5 > –4
Isolate the absolute value expression by subtracting 6 from both sides.
4x + 5 > –4
The absolute value of any number is always nonnegative and thus is always greater than 4. This inequality and the original inequality are true for all values of x. The solutions are all real numbers.

72. Example 7 Solve 10 + 3x + 1 > 2. 10 + 3x > 1
Isolate the absolute value expression by subtracting 1 from both sides.
10 + 3x > 1
Write the absolute value inequality as an equivalent compound inequality and solve.
3x < or x > -9
x < 11/3 or x > 3

73. Example 7 Solve The solutions are x ≤ 17 or x ≥ 3.
Isolate the absolute value expression by adding 1 to both sides.
Write the absolute value inequality as an equivalent compound inequality and solve.
The solutions are x ≤ 17 or x ≥ 3.

74. Example 8 Solve The solution is –2.
Recall that “≤” means “less than or equal to.” The absolute value of any expression will never be less than 0, but it may be equal to 0. Thus, to solve set the expression equal to 0.
Solve
The solution is –2.

75. Absolute Value Inequalities
2.7 summary
Absolute Value Inequalities
Objectives:
Solve absolute value inequalities