1. Equations, Inequalities and Problem Solving
Chapter 9
Equations, Inequalities and Problem Solving

2. 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
Chapter 9 Outline

3. The Addition Property of Equality
§ 9.1
The Addition Property of Equality

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

5. Addition Property of Equality
a = b and a + c = b + c are equivalent equations
Example
8 + z = – 8
a.)
8 + (– 8) + z = – 8 + – 8 (Add –8 to each side)
z = – (Simplify both sides)

6. Solving Equations Example 4p – 11 – p = 2 + 2p – 20
3p – 11 = 2p – 18 (Simplify both sides)
3p + (– 2p) – 11 = 2p + (– 2p) – 18 (Add –2p to both sides)
p – 11 = – 18 (Simplify both sides)
p – = – (Add 11 to both sides)
p = – 7 (Simplify both sides)

7. Solving Equations Example 5(3 + z) – (8z + 9) = – 4z
15 + 5z – 8z – 9 = – 4z (Use distributive property)
6 – 3z = – 4z (Simplify left side)
6 – 3z + 4z = – 4z + 4z (Add 4z to both sides)
6 + z = (Simplify both sides)
6 + (– 6) + z = 0 +( – 6) (Add –6 to both sides)
z = – 6 (Simplify both sides)

8. The Multiplication Property of Equality
§ 9.2
The Multiplication Property of Equality

9. Multiplication Property of Equality
a = b and ac = bc are equivalent equations
Example
– y = 8
(– 1)(– y) = 8(– 1) (Multiply both sides by –1)
y = – 8 (Simplify both sides)

10. Solving Equations Example (Multiply both sides by 7)
(Simplify both sides)

11. Solving Equations Example (Multiply both sides by fraction)
(Simplify both sides)

12. Solving Equations Example
Recall that multiplying by a number is equivalent to dividing by its reciprocal
Example
3z – 1 = 26
3z – = (Add 1 to both sides)
3z = 27 (Simplify both sides)
(Divide both sides by 3)
z = (Simplify both sides)

13. Solving Equations Example 12x + 30 + 8x – 6 = 10
20x + 24 = (Simplify left side)
20x (– 24) = 10 + (– 24) (Add –24 to both sides)
20x = – (Simplify both sides)
(Divide both sides by 20)
(Simplify both sides)

14. Further Solving Linear Equations
§ 9.3
Further Solving Linear Equations

15. Solving Linear Equations
Solving linear equations in one variable
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)
Get variable alone (multiplication property of equality)
Check solution by substituting into original problem

16. Solving Linear Equations
Example
(Multiply both sides by 5)
(Simplify)
(Add –3y to both sides)
(Simplify; add –30 to both sides)
(Simplify; divide both sides by 7)
(Simplify both sides)

17. Solving Linear Equations
Example
5x – 5 = 2(x + 1) + 3x – 7
5x – 5 = 2x x – 7 (Use distributive property)
5x – 5 = 5x – (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.”

18. Solving Linear Equations
Example
3x – 7 = 3(x + 1)
3x – 7 = 3x (Use distributive property)
3x + (– 3x) – 7 = 3x + (– 3x) (Add –3x to both sides)
– 7 = (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.”

19. An Introduction to Problem Solving
§ 9.4
An Introduction to Problem Solving

20. Strategy for Problem Solving
General Strategy for Problem Solving
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
Translate the problem into an equation
Solve the equation
Interpret the result
Check proposed solution in problem
State your conclusion

21. 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

22. Finding an Unknown Number
Example continued
2.) Translate
The product of
the difference of –
is the same as =
twice a number 2x
5 times the number 5x
and 3 3
and ¾
Continued

23. Finding an Unknown Number
Example continued
3.) Solve
2x · 3 = 5x – ¾
6x = 5x – ¾ (Simplify left side)
6x + (– 5x) = 5x + (– 5x) – ¾ (Add –5x to both sides)
x = – ¾ (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) = – The difference of five times – ¾ and ¾ is 5(– ¾) – ¾ = – We get the same results for both portions.
State: The number is – ¾.

24. 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) (100) = = This gives us an idea of how the cost is calculated, and also know that the number of miles will be greater than If we let
x = the number of whole miles driven, then
0.29x = the cost for mileage driven
Continued

25. Solving a Problem Example continued 2.) Translate Daily costs 2(24.95)
mileage costs
0.29x
100
maximum budget
plus +
is equal to =
Continued

26. Solving a Problem Example continued 3.) Solve 2(24.95) + 0.29x = 100
x = (Simplify left side)
(Subtract from both sides)
49.90 – x = 100 – 49.90
0.29x = (Simplify both sides)
(Divide both sides by 0.29)
x  (Simplify both sides)
Continued

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 (173) = , which is over our budget. However, (172) = 99.78, which is within the budget.
State: The maximum number of whole number miles is 172.

28. Formulas and Problem Solving
§ 9.5
Formulas and Problem Solving

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)

30. 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
Continued

31. Using Formulas Example continued 2.) Translate Formula: P = a + b + c
Substitute: = x + 2x + x + 30
3.) Solve
102 = x + 2x + x + 30
102 = 4x (Simplify right side)
102 – 30 = 4x + 30 – (Subtract 30 from both sides)
72 = 4x (Simplify both sides)
(Divide both sides by 4)
18 = x (Simplify both sides)
Continued

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 = 48 feet. This gives a perimeter of P = = 102 feet, the correct perimeter.
State: The three sides of the triangle have a length of 18 feet, 36 feet, and 48 feet.

33. 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
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
Isolate the specified variable

34. Solving Equations for a Specific Variable
Example
Solve for n.
(Divide both sides by mr)
(Simplify right side)

35. Solving Equations for a Specific Variable
Example
Solve for T.
(Subtract P from both sides)
(Simplify right side)
(Divide both sides by PR)
(Simplify right side)

36. Solving Equations for a Specific Variable
Example
Solve for P.
(Factor out P from both terms on the right side)
(Divide both sides by 1 + RT)
(Simplify the right side)

37. Percent and Mixture Problem Solving
§ 9.6
Percent and Mixture Problem Solving

38. 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

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

40. Solving a Percent Problem: Base Unknown
amount = percent · base
36 is 6% of what? n = 6% 36
36 = 0.06n
36 is 6% of 600

41. Solving a Percent Problem: Percent Unknown
amount = percent · base
24 is what percent of 144? n =
144  24

42. 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 +
tip of 20% of the cost =
$66
100% of n +
20% of n =
$66
120% of n =
$66
Mark and Peggy can spend up to $55 on the meal itself.

43. Solving Discount Problems
Example
Julie bought a leather sofa that was on sale for 35% off the original price of $ What was the discount? How much did Julie pay for the sofa?
Discount = discount rate  list price
= 35%  1200
= 420
The discount was $420.
Amount paid = list price – discount
= 1200 – 420
= 780
Julie paid $780 for the sofa.

44. Solving Increase Problems
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
The car’s cost increased by 8%.

45. Solving Decrease Problems
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
Patrick’s weight decreased by 40%.

46. 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

47. Solving Mixture Problems
Example continued
2.) Translate
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
Continued

48. Solving Mixture Problems
Example continued
3.) Solve
6n + 8(144  n) = 144(7.5)
6n  8n = 1080
(Eliminate the parentheses)
1152  2n = 1080
(Combine like terms)
2n = 72
(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
Continued

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) ?
= 1080 ?
1080 = 1080 ? 

50. Solving Linear Inequalities
§ 9.7
Solving Linear Inequalities

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 

52. 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}
Represents the set {xx > – 4}

53. 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
Represents the set (– , 7]
Interval Notation
Represents the set (– 4, )

54. 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

55. Solving Linear Inequalities
Solving linear inequalities in one variable
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)
Isolate variable (multiplication property of equality)

56. Solving Linear Inequalities
Example
3x + 9  5(x – 1)
3x + 9  5x – (Use distributive property on right side)
3x – 3x + 9  5x – 3x – (Subtract 3x from both sides)
9  2x – (Simplify both sides)
9 + 5  2x – (Add 5 to both sides)
14  2x (Simplify both sides)
7  x (Divide both sides by 2)
Graph of solution (– ,7]

57. Solving Linear Inequalities
Example
7(x – 2) + x > – 4(5 – x) – 12
7x – 14 + x > – x – (Use distributive property)
8x – 14 > 4x – (Simplify both sides)
8x – 4x – 14 > 4x – 4x – (Subtract 4x from both sides)
4x – 14 > – (Simplify both sides)
4x – > – (Add 14 to both sides)
4x > – (Simplify both sides)
(Divide both sides by 4 and simplify)
Graph of solution ( ,)

58. 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).

59. Solving Compound Inequalities
Example
0  4(5 – x) < 8
0  20 – 4x < (Use the distributive property)
0 – 20  20 – 20 – 4x < 8 – (Subtract 20 from each part)
– 20  – 4x < – (Simplify each part)
5  x > (Divide each part by –4)
Remember that the sign direction changes when you divide by a number < 0!
Graph of solution (3,5]