BUSINESS

MATHEMATICS
(BBQT 1013)
Chapter 3 :
Equations
 LEARNING OUTCOME
In the end of this chapter, students will be
able to:
1. Solve equations using multiplication or division.
2. Solve equations using addition or subtraction.
3. Solve equations using more than one
operation.
4. Solve equations containing multiple unknown
terms.
5. Solve equations containing parentheses.
6. Solve equations that are proportions.
7. Use the problem-solving approach to analyze
and solve word problems.
 EQUATION
◼ 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.
 EQUATION
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.
 EQUATION
◼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.
– etc
 SOLVE EQUATIONS USING MULTIPLICATION
OR DIVISION
8x = 24
STEP 1
Isolate the unknown value and determine
if multiplication or division is needed.
STEP 2
Use division to divide both sides by 8.

STEP 3
Simplify: x = 3 3 x 8 = 24
FIND THE VALUE OF AN UNKNOWN USING
MULTIPLICATION
a
Find the value of a: =6
3
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
 EXAMPLE
2b = 40
STEP 1
Determine which operation is needed.
Division

STEP 2
Perform the same operation to both sides.
Divide both sides by 2.

STEP 3
40
Isolate the variable and solve. b= = 20
2
 EXERCISE
1. Solve for A: 3 A = 24

2. Solve for B:
B
8=
6
 SOLVE EQUATIONS USING ADDITION OR
SUBTRACTION
◼ 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
 EXAMPLE
4 + x = 10
STEP 1
Isolate the unknown value and determine
if addition or subtraction is needed.
STEP 2
Use subtraction to isolate x.

STEP 3
Simplify: x = 6
 EXAMPLE
b - 12 = 8
STEP 1
Determine which operation is needed.
Addition

STEP 2
Perform the same operation to both sides.
Add 12 to both sides

STEP 3
Isolate the variable and solve. b = 8 + 12 = 20
 EXERCISE
Solve for the variable.
a) A + 12 = 20

b) 15 = A + 3

c) N − 7 = 10

d) 28 = M − 5
 SOLVE EQUATIONS USING MORE THAN
ONE OPERATION
◼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.
◼To solve an equation, undo the operations, working in
reverse order
– First — undo the addition or subtraction.
– Second — undo multiplication or division.
 EXAMPLE
7x + 4 = 39
STEP 1
Undo the addition by subtracting 4 from each side.
7x = 35

STEP 2 35
Divide each side by 7. x= =5
7
STEP 3
Verify by plugging in 5 in place of x .
7 (5) + 4 = 39 35 + 4 = 39
 EXERCISE
Solve:

a) 5N − 7 = 13
B
b) −2=2
8
c) A
12 = − 8
5
SOLVE EQUATIONS CONTAINING MULTIPLE
UNKNOWN TERMS
◼ 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).
 EXAMPLE
Find a if: a + 4a – 5 = 30
STEP 1
Combine the unknown value addends.
a + 4a = 5a 5a – 5 = 30

STEP 2 STEP 3
Undo the subtraction. Undo the multiplication.
5a = 35 a=7
STEP 4
Check by replacing a with 7. Correct!
7 + 4(7) = 35
 EXERCISE
Solve:

a) 6B − 2B − 7 = 13

b) 7 + 3B + 2B = 17

c) 5 A − 3 + 2 A = 18
 SOLVE EQUATIONS CONTAINING
PARENTHESES
◼ 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.
 EXAMPLE
Solve: 6(A + 2) = 24
STEP 1
Multiply 6 by each addend.
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
 EXAMPLE

5 (x - 2) = 45
TIP: Remove the parentheses first.
5x -10 = 45
5x = 55
x = 11
 EXERCISE
Solve:

a) 3(N − 30) = 45

b) 30 = 6(2 A + 3)
SOLVE EQUATIONS THAT ARE PROPORTIONS
◼ 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.
SOLVE EQUATIONS THAT ARE PROPORTIONS
◼ 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.
 HOW TO VERIFY THAT TWO FRACTIONS
FORM A PROPORTION
4 6
Do and form a proportion?
12 18
STEP 1
Multiply the numerator from the first fraction by the
denominator of the second fraction.
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.

 EXERCISE
5 3
1. Which of the fractions or is proportional
to 20 ? 7 4
28
2. Solve:

3 N
a) =
4 8
b) 5 15
=
12 N
 USE THE PROBLEM-SOLVING APPROACH
TO ANALYZE AND SOLVE WORD 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.
 USE THE PROBLEM-SOLVING APPROACH
TO ANALYZE AND SOLVE WORD PROBLEMS
These words help you interpret the information and
begin to set up the equation to solve the problem.

“of” often implies multiplication.

Example:
“¼ of her salary” means “multiply her salary by ¼”
 EXAMPLE
Full-time employees at Charlie’s Steakhouse work
more hours per day than part-time employees. If
the difference of working hours is 4 hours per day,
and if part-timers work 6 hours per day, how many
hours per day do full-timers work?
 SOLUTION
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.

We also know that

Set up a solution plan. “difference” implies
FT – PT = 4 subtraction.
FT = N [unknown] PT = 6 hours
N–6=4 Conclusion:
Solution plan: N = 4 + 6 = 10 Full time employees
work 10 hours.
 EXAMPLE
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
 EXAMPLE
1
Wanda plans to save of her salary each week.
10
If her weekly salary is $350, how much will she be
save each week?
 SOLUTION
What are we looking for?
Amount to be saved: S
What do we know?
Salary = $350
1
Rate of saving: 10

We also know that “of”

Set up a solution plan.
implies multiplication.
Amount to rate of
=  salary
be saved saving

Conclusion:
Solution plan: S = 110 ( 350 ) = 35 Wanda will save $35
per week.
 EXAMPLE
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
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.
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.
 SOLUTION
Set up the equation by multiplying the unit
Solution Plan price of each by the volume, represented by
the unknowns equaling the total amount spent.

$1.75(H) + $1.50 (600 – H) = $950.00

Unit Volume Total

prices unknowns spent
 SOLUTION
$1.75H + $1.50(600 - H) = $950.00

$1.75H + $900.00 - $1.50H = $950.00

$0.25H + $900.00 = $950.00

$0.25H = $50.00

H = 200
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.
 EXERCISE
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?
 EXERCISE
In planning for a conference on Successful Small
Business Practices, the organizers are anticipating
that twice as many men as women will attend the
conference. If they are expecting 600 to attend the
conference, how many men and how many
women are likely to attend?
 PROPORTIONS
◼ The relationship between two factors is often
described in proportions.
– You can use proportions to solve for unknowns.
 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?
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.
2 5
Set up a solution plan. = Conclusion.
3 x
You need 7.5 oz of weed
Solve it.
killer for 5 gal of water.
Cross multiply: 2x = 15; x = 7.5
 EXAMPLE
Your car gets 23 miles to the gallon.
How far can you go on 16 gallons of gas?

1 gallon 16 gallons
=
23 miles x miles
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.
 EXERCISE
The label on a container of concentrated weed
killer gives directions to mix 3 ounces of weed killer
with every 2 gallons of water. For 5 gallons of water,
how many ounces of weed killer should you use?
 EXERCISE
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?
 EXERCISE
Wallpaper costs $12.97 per roll and a kitchen
requires 9 rolls. What is the cost of the wallpaper
needed to paper the kitchen?
 EXERCISE
If 5 dozen roses can be purchased for $62.50, how
much will 8 dozen cost?
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