1. CHAPTER 5
Equations

2. Solve equations using multiplication or division.
5-1
Learning Outcomes
Solve equations using multiplication or division.
Solve equations using addition or subtraction.
Solve equations using more than one operation.
Solve equations containing multiple unknown terms.
Solve equations containing parentheses.
Solve equations that are proportions.

3. Solve equations using multiplication or division
5-1-1
Section 5-1
Equations
An equation is a mathematical statement in which two quantities are equal.
Solving an equation means finding the value of an unknown.
Example:
8x = 24
To solve this equation, the value of x must be discovered.
Division is used to solve this equation.

4. The numbers are called known or given amounts.
Solve equations using multiplication or division
Section 5-1
Equations
Letters, such as (x,y,z) represent unknown amounts and are called unknowns or variables.
4x = 16
The numbers are called known or given amounts.

5. Solve equations using multiplication or division
Section 5-1
Equations
Any operation performed on one side of the equation must be performed on the other side of the equation as well.
If you “multiply by 2” on one side, you must “multiply by 2” on the other side.
If you “divide by 3” on one side, you must also “divide by 3” on the other side.

6. Solve equations using multiplication or division
HOW TO:
Section 5-1
Equations
8x = 24
Isolate the unknown value and determine if multiplication or division is needed.
STEP 1
STEP 2
Use division to divide both sides by 8.
STEP 3
Simplify: x = 3
3 x 8 = 24

7. a = 18 Find the value of a: Multiply both sides by 3 to isolate a.
Find the value of an unknown using multiplication
HOW TO:
Section 5-1
Equations
Find the value of a:
Multiply both sides by 3 to isolate a.
The left side becomes 1a or a.
The right side becomes the product of 6 x 3, or 18.
a = 18

8. 2b = 40 STEP 1 Determine which operation is needed. STEP 2
An Example…
Section 5-1
Equations
2b = 40
Determine which operation is needed.
STEP 1
Division
STEP 2
Perform the same operation to both sides.
Divide both sides by 2.
STEP 3
Isolate the variable and solve.

9. Solve equations using addition or subtraction
5-1-2
Section 5-1
Equations
Adding or subtracting any number from one side must be carried out on the other side as well.
Subtract “the given amount” from both sides.
Would solving 4 + x = 16 require addition or subtraction of “4” from each side?
Subtraction

10. Solve equations using addition or subtraction
HOW TO:
Section 5-1
Equations
4 + x = 10
Isolate the unknown value and determine if addition or subtraction is needed.
STEP 1
STEP 2
Use subtraction to isolate x.
STEP 3
Simplify: x = 6

11. b - 12 = 8 STEP 1 Determine which operation is needed. STEP 2
An Example…
Section 5-1
Equations
b - 12 = 8
Determine which operation is needed.
STEP 1
Addition
STEP 2
Perform the same operation to both sides.
Add 12 to both sides
STEP 3
Isolate the variable and solve.
b = = 20

12. Isolate the unknown value.
Solve equations using more than one operation
5-1-3
Section 5-1
Equations
Isolate the unknown value.
Add or subtract as necessary first.
Multiply or divide as necessary second.
Identify the solution.
The number on the side opposite the unknown.
Check the solution by “plugging in” the number using the original equation.

13. Order of operations
Section 5-1
Equations
When two or more calculations are written symbolically, the operations are performed according to a specified order of operations.
First — perform multiplication and division as they appear from left to right.
Second — perform addition and subtraction as they appear from left to right.

14. To solve an equation, undo the operations, working in reverse order
Order of operations
Section 5-1
Equations
To solve an equation, undo the operations, working in reverse order
First — undo the addition or subtraction.
Second — undo multiplication or division.

15. 7x + 4 = 39 STEP 1 Undo the addition by subtracting 4 from each side.
An Example…
Section 5-1
Equations
7x + 4 = 39
Undo the addition by subtracting 4 from each side.
STEP 1
7x = 35
STEP 2
Divide each side by 7.
STEP 3
Verify by plugging in 5 in place of x .
7 (5) + 4 = = 39

16. In some equations, the unknown value may occur more than once.
Solve equations containing multiple unknown terms
5-1-4
Section 5-1
Equations
In some equations, the unknown value may occur more than once.
The simplest instance is when the unknown value occurs in two addends, such as 3a + 2a = 25
Add the numbers in each addend (2+3).
Multiply the sum by the unknown (5a = 25).
Solve for a (a = 5).

17. Find a if: a + 4a – 5 = 39 Correct! Combine the unknown value addends.
An Example…
Section 5-1
Equations
Find a if: a + 4a – 5 = 39
Combine the unknown value addends.
STEP 1
a + 4a = 5a 5a – 5 = 30
STEP 2
Undo the subtraction.
STEP 3
Undo the multiplication.
5a = 35
a = 7
STEP 4
Check by replacing a with 7.
Correct!
7 + 4(7) = 35

18. Eliminate the parentheses.
5-1-5
Solve equations containing parentheses
Section 5-1
Equations
Eliminate the parentheses.
Multiply the number just outside the parentheses by each addend inside the parentheses.
Show the resulting products as addition or subtraction, as indicated
Solve the resulting equation.

19. Solve: 6(A + 2) = 24 STEP 1 Multiply 6 by each addend. STEP 2
An Example…
Section 5-1
Equations
Solve: 6(A + 2) = 24
Multiply 6 by each addend.
STEP 1
6 multiplied by A + 6 multiplied by 2
STEP 2
Show the resulting products.
6A + 12 = 24
STEP 3
Check by replacing a with 2.
6(2 + 2) = 24

20. An Example…
Section 5-1
Equations
5 (x - 2) = 45
Remove the parentheses first.
TIP:
5x -10 = 45
5x = 55
x = 11

21. A proportion is based on two pairs of related quantities.
5-1-6
Solve equations that are proportions
Section 5-1
Equations
A proportion is based on two pairs of related quantities.
The most common way to write proportions is to use fraction notation—also called a ratio.
When two ratios are equal, they form a proportion.

22. 5-1-6
Solve equations that are proportions
Section 5-1
Equations
A cross product is the product of the numerator of one fraction, times the denominator of another fraction.
An important property of proportions is that the cross products are equal.

23. Do and form a proportion?
HOW TO:
Verify that two fractions form a proportion
Section 5-1
Equations
Do and form a proportion?
Multiply the numerator from the first fraction by the denominator of the second fraction.
STEP 1
4 x 18 = 72
STEP 2
Multiply the denominator of the first fraction by the numerator of the second fraction.
6 x 12 = 72
Are they equal?
Yes, they form a proportion.

24. Use the problem-solving approach to analyze and solve word problems.
5-2
Learning Outcome
Use the problem-solving approach to analyze and solve word problems.

25. Five step problem solving approach:
Use the problem-solving approach to analyze and solve word problems.
5-2-1
Section 5-2
Using Equations to Solve Problems
Five step problem solving approach:
What you know.
Known or given facts.
What you are looking for.
Unknown or missing amounts.
Solution Plan.
Equation or relationship among known/unknown facts.
Solution.
Solve the equation.
Conclusion.
Solution interpreted within context of problem.

26. Use the problem-solving approach to analyze and solve word problems.
5-2-1
Section 5-2
Using Equations to Solve Problems
Key words in Table 5-1 will guide you in using the problem-solving approach.
These words help you interpret the information and begin to set up the equation to solve the problem.
“of” often implies multiplication.
“¼ of her salary” means “multiply her salary by ¼”
Example:

27. HOW TO: Use the solution plan
Section 5-2
Using Equations to Solve Problems
Full time employees work more hours than part-time employees.
If the difference is four per day, and part-time employees work six hours per day, how many hours per day do full-timers work?
What are we looking for?
Number of hours that FT employees work.
What do we know?
PT employees work 6 hours, and the difference between FT and PT is 4 hours.

28. HOW TO: Use the solution plan
Section 5-2
Using Equations to Solve Problems
Full time employees work more hours than part-time employees.
If the difference is four per day, and part-time employees work six hours per day, how many hours per day do full-timers work?
We also know that “difference” implies subtraction.
Set up a solution plan.
FT – PT = 4 FT = N [unknown] PT = 6 hours
N – 6 = 4
Solution plan: N = = 10
What are we looking for?
Number of hours that FT employees work.
What do we know?
PT employees work 6 hours, and the difference between FT and PT is 4 hours.
Conclusion: Full time employees work 10 hours.

29. HOW TO: Use the solution plan
Section 5-2
Using Equations to Solve Problems
Jill has three times as many trading cards as Matt. If the total number both have is 200, how many does Jill have?
1. What are you looking for?
The number of cards that Jill has.
2. What do you know?
The relationship in the number of cards is 3:1; total is 200
3. Set up a solution plan.
x(Matt’s) + 3x(Jill’s) = 200
4. Solve it.
x + 3x = 200; 4x = 200; x = 50
5. Draw the conclusion.
Jill has 3x, or 150 cards

30. HOW TO: Use the solution plan
Section 5-2
Using Equations to Solve Problems
Diane’s Card Shop spent a total of $950 ordering 600 cards from Wit’s End Co., whose humorous cards cost $1.75 each and whose nature cards cost $1.50 each. How many of each style of card did the card shop order?
What are you looking for?
How many humorous cards were ordered and how many nature cards were ordered—the total of H + N = 600 or N = 600 – H.
If we let H represent the humorous cards, Nature cards will be 600 – H, which will simplify the solution process by using only one unknown: H.

31. HOW TO: Use the solution plan
Section 5-2
Using Equations to Solve Problems
Diane’s Card Shop spent a total of $950 ordering 600 cards from Wit’s End Co., whose humorous cards cost $1.75 each and whose nature cards cost $1.50 each. How many of each style of card did the card shop order?
What do you know?
A total of $950 was spent.
Two types of cards were ordered.
The total number of cards ordered was 600.
Humorous cards cost $1.75 each and nature cards cost $1.50 each.

32. $1.75(H) + $1.50 (600 – H) = $950.00 HOW TO: Use the solution plan
Section 5-2
Using Equations to Solve Problems
Diane’s Card Shop spent a total of $950 ordering 600 cards from Wit’s End Co., whose humorous cards cost $1.75 each and whose nature cards cost $1.50 each. How many of each style of card did the card shop order?
Solution Plan
Set up the equation by multiplying the unit price of each by the volume, represented by the unknowns equaling the total amount spent.
$1.75(H) + $1.50 (600 – H) = $950.00
Total spent
Unit prices
Volume unknowns

33. HOW TO: Use the solution plan
Section 5-2
Using Equations to Solve Problems
Diane’s Card Shop spent a total of $950 ordering 600 cards from Wit’s End Co., whose humorous cards cost $1.75 each and whose nature cards cost $1.50 each. How many of each style of card did the card shop order?
Solution
$1.75H + $1.50(600 - H) = $950.00
$1.75H + $ $1.50H = $950.00
$0.25H + $ = $950.00
$0.25H = $50.00
H = 200

34. HOW TO: Use the solution plan
Section 5-2
Using Equations to Solve Problems
Diane’s Card Shop spent a total of $950 ordering 600 cards from Wit’s End Co., whose humorous cards cost $1.75 each and whose nature cards cost $1.50 each. How many of each style of card did the card shop order?
Conclusion
The number of humorous cards ordered is 200.
Since nature cards are 600 – H, we can conclude that 400 nature cards were ordered.
Using “200” and “400” in the original equation proves that the volume amounts are correct.

35. HOW TO: Use the solution plan
Section 5-2
Using Equations to Solve Problems
Denise ordered 75 dinners for the awards banquet. Fish dinners cost $11.75 and chicken dinners cost $9.25 each. If she spent a total of $756.25, how many of each type of dinner did she order?
$11.75(F) + $9.25(75 - F) = $756.25
$11.75F + $ $9.25F = $756.25
$2.50F + $ = $756.25
$2.50F = $62.50
F = 25
Conclusion:
25 fish dinners and 50 chicken dinners were ordered.

36. Proportions
Section 5-2
Using Equations to Solve Problems
The relationship between two factors is often described in proportions.
You can use proportions to solve for unknowns.
The label on a container of weed killer gives directions to mix three ounces of weed killer with every two gallons of water. For five gallons of water, how many ounces of weed killer should you use?
Example:

37. You need 7.5 oz of weed killer for 5 gal of water.
Proportions
Section 5-2
Using Equations to Solve Problems
The relationship between two factors is often described in proportions.
You can use proportions to solve for unknowns.
1. What are you looking for?
Number of ounces of weed killer needed for 5 gallons of water.
2. What do you know?
For every 2 gallons of water, you need 3 oz. of weed killer.
Example:
The label on a container of weed killer gives directions to mix three ounces of weed killer with every two gallons of water. For five gallons of water, how many ounces of weed killer should you use?
Set up a solution plan.
Conclusion.
You need 7.5 oz of weed killer for 5 gal of water.
Solve it.
Cross multiply: 2x = 15; x = 7.5

38. Proportions
Section 5-2
Using Equations to Solve Problems
Many business-related problems that involve pairs of numbers that are proportional involve direct proportions.
An increase (or decrease) in one amount causes an increase (or decrease) in the number that pairs with it.

39. Your car gets 23 miles to the gallon. Cross multiply: 1x = 368 miles
An Example…
Section 5-2
Using Equations to Solve Problems
Your car gets 23 miles to the gallon.
How far can you go on 16 gallons of gas?
Cross multiply: 1x = 368 miles
Conclusion:
You can travel 368 miles on 16 gallons of gas.
In this example, an increase in the amount of gas would directly and proportionately increase the mileage yielded.

40. Find an equivalent formula by rearranging the formula.
5-3
Learning Outcomes
Evaluate a formula.
Find an equivalent formula by rearranging the formula.

41. Interpret the solution within the context of the formula.
5-3-1
Evaluate the formula
Section 5-3
Formulas
Write the formula.
Rewrite the formula substituting known values for the letters of the formula.
Solve the equation for the unknown letter or perform the indicated operations, applying the order of operations.
Interpret the solution within the context of the formula.

42. An Example…
Section 5-3
Formulas
A plasma TV that costs $2,145 is marked up $854. What is the selling price of the TV?
Use the formula S = C + M where S is the selling price, C is the cost, and M is Markup.
S = $2,145 + $854
S or Selling Price = $2,999

43. Find an Equivalent Formula by Rearranging the Formula
5-3-2
Section 5-3
Formulas
Determine which variable of the formula is to be isolated (solved for).
Highlight or mentally locate all instances of the variable to be isolated.
Treat all other variables of the formula as you would treat numbers in an equation, and perform normal steps for solving an equation.
If the isolated variable is on the right side of the equation, interchange the sides so that it appears on the left side.

44. The formula for Square Footage = Length x Width or S = L x W.
An Example…
Section 5-3
Formulas
The formula for Square Footage = Length x Width or S = L x W.
Solve the formula for W or width. Isolate W by dividing both sides by L.
The new formula is:

45. Exercises Set A

46. Find the value of the variable: 2. 4.
EXERCISE SET A
Find the value of the variable:

47. Find the value of the variable: 6. 8.
EXERCISE SET A
Find the value of the variable:

48. Find the value of the variable: 10.
EXERCISE SET A
Find the value of the variable:
10.

49. Solve the proportion for N: 12.
EXERCISE SET A
Solve the proportion for N:
12.

50. 12N = 60 N = number of cards of each type
EXERCISE SET A
14. Bottletree Bakery and Card Shop ordered an equal number of 12 different cards. If a total of 60 cards were ordered, how many of each type of card were ordered?
12N = 60 N = number of cards of each type
Five of each type of card were ordered.

51. 1,584 + N = 2,000 N = number to be ordered 1584 1584 N = 416
EXERCISE SET A
16. An inventory clerk is expected to have 2,000 fan belts in stock. If the current count is 1,584 fan belts, how many more should be ordered?
1,584 + N = 2,000 N = number to be ordered
1584 1584
N = 416
416 fan belts should be ordered

52. 12.97(9) = N N = total cost of wallpaper 116.73 = N
EXERCISE SET A
18. Wallpaper costs $12.97 per roll and a kitchen requires 9 rolls. What is the cost of the wallpaper needed to paper the kitchen?
12.97(9) = N N = total cost of wallpaper
= N
Wallpaper for the kitchen will cost $

53. Eight dozen roses cost $100.
EXERCISE SET A
20. If 5 dozen roses can be purchased for $62.50, how much will 8 dozen cost?
Eight dozen roses cost $100.

54. EXERCISE SET A
22. In the formula T = P - D, T represents total installment payments, P represents installment price, and D represents down payment amount. Find the installment price if the total of installment payments is $15, and the down payment is $3,
T = P - D
15, = P - 3,973.16
+ 3, ,973.16
$19, = P

55. 24. Solve the installment payment formula for N.
EXERCISE SET A
24. Solve the installment payment formula
for N.

56. Practice Test

57. PRACTICE TEST
Solve.

58. PRACTICE TEST
Solve.

59. PRACTICE TEST
Solve.
10.

60. N = number of containers
PRACTICE TEST
12. A container of oil holds 585 gallons. How many containers each holding 4.5 gallons will be needed if all the oil is to be transferred to the smaller containers?
N = number of containers
130 containers are needed

61. 14. Find the cost of 200 suits if 75 suits cost $10,200.
PRACTICE TEST
14. Find the cost of 200 suits if 75 suits cost $10,200.
N = cost of 200 suits
The cost of 200 suits is $27,200.

62. PRACTICE TEST
18. The formula for the installment price of an item purchased with financing is Installment price = Total of installment payments + Down payment. The formula can be written in symbols as I = T + D. Find the installment price I if T = $24, and D = $2,500.
I = T + D
I = $24, $2,500
I = +27,346.38

63. 20. Rearrange the formula I = T + D to solve for D . I = T + D
PRACTICE TEST
20. Rearrange the formula I = T + D to solve for D .
I = T + D