```5.1 Equations

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.
Cleaves/Hobbs
5.1.1 Solve Equations Using
Multiplication or Division

An equation is a mathematical statement in
which two quantities are equal.

Solving an equation means finding the value of
an unknown.

For example: 8x = 24

To solve this equation, the value of x must be
discovered.

Division is used to solve this equation.
Cleaves/Hobbs
Key Terms

The letters (x,y,z) represent unknown
amounts and are called unknowns or
variables.

The numbers are called known or given
amounts.
4x = 16
Cleaves/Hobbs
Remember!

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
“divide by 3” on the other side and so on.
Cleaves/Hobbs
How to solve an equation with
multiplication and division
8x = 24
Step one:
Isolate the unknown value.
Determine if multiplication or
division is needed.
Step two:
Use division to divide both
sides by “8.”
Step three:
Simplify: x = 3
Cleaves/Hobbs
Find the value of an unknown
using multiplication

Find the value of “a” in the following equation.

a/3 = 6

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
Cleaves/Hobbs
Do this example
Solve the following:
2b = 40
1. Determine which operation is needed.
Division
2. Perform the same operation to both sides.
Divide both sides by “2.”
3. Isolate the variable and solve the equation
b = 40/2 = 20
Cleaves/Hobbs
5.1.2 Solve an Equation with
Step one:
4 + x = 10
Isolate the unknown value.
subtraction is needed.
Step two:
Use subtraction to isolate “x.”
Step three:
Simplify: x = 6
Cleaves/Hobbs
Don’t forget!

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
Cleaves/Hobbs
Do this example
Solve the following:
b - 12 = 8
1. Determine which operation is needed.
2. Perform the same operation to both sides.
3. Isolate the variable and solve the equation.
b = 8 + 12 = 20
Cleaves/Hobbs
5.1.3 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.
Cleaves/Hobbs
Order of Operations

When two or more calculations are written
symbolically, it is agreed to perform the
operations according to a specified order of
operations.

Perform multiplication and division as they
appear from left to right.

Perform addition and subtraction as they
appear from left to right.
Cleaves/Hobbs
“Undo the operations”
To solve an equation, we undo the
operations, so we work in reverse order.
1. Undo the addition or subtraction.
2. Undo multiplication or division.
7x + 4 = 39
Cleaves/Hobbs
Try this example
7x + 4 = 39

First, undo the addition by subtracting 4 from each
side.


Next, divide each side by 7.


And that becomes 7x = 35
And that becomes x = 35/7 = 5
Verify the result by “plugging 5 in” the place of “x.”


7 (5) + 4 = 39
35 + 4 = 39
Cleaves/Hobbs
5.1.4 Equations Containing
Multiple Unknown Terms

In some equations, the unknown value may occur
more than once.

The simplest instance is when the unknown value

For example: 3a + 2a = 25


Multiply the sum by the unknown (5a = 25).

Solve for “a.” (a = 5)
Cleaves/Hobbs
Try this example
a +4a – 5 = 30

Find a if:

a + 4a = 5a
5a – 5 = 30

“Undo” the subtraction.
5a = 35

“Undo” the multiplication.
a=7

Check by replacing “a” with “7.

It is correct.
Cleaves/Hobbs
5.1.5 Solve Equations
Containing Parentheses
1. Eliminate the parentheses:
a. Multiply the number just outside the
parentheses by each addend inside the
parentheses.
b. Show the resulting products as addition
or subtraction as indicated.
2. Solve the resulting equation.
Cleaves/Hobbs
Look at this example
Solve the equation:
6(A + 2) = 24



Show the resulting products.


6 multiplied by A + 6 multiplied by 2
6A + 12 = 24
Solve the equation.


6A = 12
A=2
Cleaves/Hobbs
Tip!
Remove the parentheses first.
5 (x - 2) = 45


Do me first !

5x -10 = 45
5x = 55
x = 11
Cleaves/Hobbs
5.1.6 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.

A number written in fraction notation is also
called a ratio.

When two ratios are equal, they form a
proportion.
Cleaves/Hobbs
Cross products

An important property of proportions is that
the cross products are equal.

A cross product is the product of the
numerator of one fraction times the
denominator of another fraction.

Example: 4/6 = 6/9
Multiply 4 x 9 = 6 x 6
Cleaves/Hobbs
36 = 36
Verify that two fractions
form a proportion
Do 4/12 and 6/18 form a proportion?
1. Multiply the numerator from the first fraction by the
denominator of the second fraction.
4 x 18 = 72
2. Multiply the denominator of the first fraction by the
numerator of the second fraction.
6 x 12 = 72
3. Are they equal?
Yes, they form a proportion.
Cleaves/Hobbs
5.2 Using Equations to
Solve Problems

There is a list of key words and what operations

and begin to set up the equation to solve the
problem.

Example: “of” often implies multiplication.
“¼ of her salary” means “multiply her salary
by ¼”
Cleaves/Hobbs
Five-step problem solving
approach for equations

What you know.


What you are looking for?


Equation or relationship among known / unknown facts.
Solution


Unknown or missing amounts.
Solution Plan


Known or given facts.
Solve the equation.
Conclusion

Solution interpreted within context of problem.
Cleaves/Hobbs
Use the solution plan

Full time employees work more hours than parttime 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 work

What do we know?
PT work 6 hours;
The difference between FT and PT is 4 hours.
Cleaves/Hobbs
Use the solution plan

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:

Conclusion: Full time employees work 10 hours.
Cleaves/Hobbs
N = 4 + 6 = 10
Try this example

Jill has three times as many trading cards as
Matt. If the total number that both have is 200,
how many cards does Jill have?
Use the five-step solution plan to solve this
problem:
1. What are you looking for?
2. What do you know?
3. Set up a solution plan.
4. Solve it.
5. Draw the conclusion.
Cleaves/Hobbs
Solution Plan

What are you looking for?


What do you know?


x (Matt’s) + 3x (Jill’s) = 200
Solve


The relationship in the number of cards is 3:1; total is 200.
Solution plan


The number of cards that Jill has.
x + 3x = 200
4x = 200; x = 50
Conclusion

Jill has “3x” or 150 cards.
Cleaves/Hobbs
Solving a word problem with a
total of two types of items
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?

Use the solution plan to solve this problem.
Cleaves/Hobbs
What are you looking for?

How many humorous cards were ordered and
how many nature cards were ordered.

The total of H + N = 600

Another way to look at this is:
N = 600 – H

If we let “H” represent the humorous cards,
Nature cards will be 600- H.

This will simplify the solution process by using
only one unknown: “H.”
Cleaves/Hobbs
Organize the information

What do you
know?
Cleaves/Hobbs

A total of \$950 was
spent.

Two types of cards
were ordered.

The total number of
cards ordered was
600.

The humorous cards
cost \$1.75 each/nature
cards cost \$1.50 each.
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
Unit
prices
Cleaves/Hobbs
Total
spent
Volume
“unknowns”
Solve the equation

\$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
Cleaves/Hobbs
Conclusion

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.
Cleaves/Hobbs
Try this problem

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?

Use the solution plan to organize the
information and solve the problem.
Cleaves/Hobbs
Denise’s order

\$11.75(F) + \$9.25(75-F) = \$756.25

\$11.75 F + \$693.75 - \$9.25F = \$756.25

\$2.50F + \$693.75 = \$756.25

\$2.50F = \$62.50

F = 25

Conclusion: 25 fish dinners and 50 chicken dinners
were ordered.
Cleaves/Hobbs
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?
Cleaves/Hobbs
Use the solution plan

What are you looking for?

The number of ounces of weed killer needed for 5 gallons of
water.

What do you know?

For every 2 gallons of water, you need 3 oz. of weed killer.

Set up solution plan.

2/3 = 5/x

Solve the equation.

Cross multiply.
2x = 15; x = 7.5

Conclude

You need 7.5 ounces of weed killer for 5 gallons of water.
Cleaves/Hobbs
Proportions

Your car gets 23 miles to the gallon. How
far can you go on 16 gallons of gas?

1 gallon/23 miles = 16 gallons/ x miles

Cross multiply: 1x = 368 miles

Conclusion: You can travel 368 miles on 16
gallons of gas.
Cleaves/Hobbs
Direct Proportions

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.

In the previous example, an increase in the
amount of gas would directly and
proportionately increase the mileage yielded.
Cleaves/Hobbs
5.3 Formulas

Evaluate a formula.

Find a variation of a formula by
rearranging the formula.
Cleaves/Hobbs
How to evaluate a formula




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.
Cleaves/Hobbs
Try this problem



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
Cleaves/Hobbs
Find a variation of a formula by
rearranging the formula




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.
Cleaves/Hobbs
Try this problem

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 then: S/L = W