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$C = 23$

77 = ? \frac{9}{5} (23) + 32

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

77 = ? \frac{9 \cdot 2 3}{5} + 32

77 = ? \frac{2 0 7}{5} + 32

Calculate quotient

77 = ? 41.4 + 32

Add terms

77 = 73.4 \times

A false statement was obtained, which suggests that Tadeo's answer is not correct. Next, substitute Magdalena's answer.

77 = \frac{9}{5} C + 32

▼

Substitute

25

for

C

and simplify

$C = 25$

77 = ? \frac{9}{5} (25) + 32

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

77 = ? \frac{9 \cdot 2 5}{5} + 32

77 = ? \frac{2 2 5}{5} + 32

Calculate quotient

77 = ? 45 + 32

Add terms

77 = 77 ✓

A true statement was obtained, which implies that Magdalena is correct in saying that

7 7^{\circ} F

is the same as

2 5^{\circ} C .

Discussion

Distributive Property of Multiplication

In addition to the commutative and associative properties, multiplication has another useful property that helps in the process of simplifying expressions with parentheses. This property is called the Distributive Property.

Rule

Distributive Property

Multiplying a number by the sum of two or more addends produces the same result as multiplying the number by each addend individually and then adding all the products together.

a (b \pm c) (b \pm c) a = a \cdot b \pm a \cdot c = a \cdot b \pm a \cdot c

Note that the factor outside the parentheses is multiplied, or distributed, to every term inside. The Distributive Property is used to simplify expressions with parentheses.

Since the Distributive Property is an axiom, it does not need a proof.

Note that, by the Symmetric Property of Equality, the Distributive Property can also be used to take out a common factor in two or more addends.

9 x + 9 \cdot 2 = 9 (x + 2)

In some cases, the right-hand side expression is more useful than the one on the left.

Discussion

Solving Multi-Step Equations

When solving equations, sometimes more than one step is needed. Both the number of steps and the operations required depend on the complexity of the given equation. For example, consider the following pair of equations.

Equation (I): Equation (II): \frac{3}{2} (x - 2) + 5 = 14 y + 2 y - 4 = 11 - 2 y

The general idea is to simplify both sides of the equation and then isolate the variable on one side of the equation. This is usually done by collecting all the variable terms on one side of the equations and combining them. Then, the operations applied to the variable are undone in reverse order.

Solving Equation (I)

Equation (I) contains a fraction and parentheses.

\frac{3}{2} (x - 2) + 5 = 14

Clear Parentheses and Combine Like Terms on Each Side

Apply the Distributive Property to clear the parentheses on the left-hand side of the equation. In this case, distribute

\frac{3}{2}

to each term inside the parentheses. Then, combine like terms on the left-hand side.

\frac{3}{2} (x - 2) + 5 = 14

Distr

Distribute $\frac{3}{2}$

\frac{3}{2} x - \frac{3}{2} \cdot 2 + 5 = 14

FracMultDenomToNumber

$\frac{a}{2} \cdot 2 = a$

\frac{3}{2} x - 3 + 5 = 14

Add terms

\frac{3}{2} x + 2 = 14

Isolate the Variable

Apply the Properties of Equality to isolate the variable on one side of the equation. In this case, first apply the Subtraction Property of Equality.

\frac{3}{2} x + 2 = 14

SubEqn

$LHS - 2 = RHS - 2$

\frac{3}{2} x + 2 - 2 = 14 - 2

SubTerms

Subtract terms

\frac{3}{2} x = 12

The coefficient of the variable is a fraction. The Multiplication Property of Equality can be used to multiply both sides of the equation by the reciprocal of the coefficient.

\frac{3}{2} x = 12

MultEqn

$LHS \cdot \frac{2}{3} = RHS \cdot \frac{2}{3}$

\frac{3}{2} x \cdot \frac{2}{3} = 12 \cdot \frac{2}{3}

CommutativePropMult

Commutative Property of Multiplication

\frac{3}{2} \cdot \frac{2}{3} \cdot x = 12 \cdot \frac{2}{3}

MultFracByInverse

$\frac{a}{b} \cdot \frac{b}{a} = 1$

1 \cdot x = 12 \cdot \frac{2}{3}

OneMult

$1 \cdot a = a$

x = 12 \cdot \frac{2}{3}

MoveLeftFacToNum

$a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$

x = \frac{1 2 \cdot 2}{3}

x = \frac{2 4}{3}

Calculate quotient

x = 8

As such,

x = 8

is the solution of Equation (I).

Solving Equation (II)

In this equation, the variable is on both sides of the equation.

y + 2 y - 4 = 11 - 2 y

Solving the equation will require the additional step of collecting the variables on one side of the equation.

Clear Parentheses and Combine Like Terms on Each Side

Apply the Distributive Property to clear the parentheses on each side of the equation. There are no parentheses to clear in this equation, so start by combining like terms instead.

y + 2 y - 4 = 11 - 2 y

Add terms

3 y - 4 = 11 - 2 y

Collect the Variable on One Side of the Equation

Apply the Properties of Equality to collect all the variables on one side of the equation. In this case, add

2 y

to both sides of the equation to group the

y -

terms on the left-hand side.

3 y - 4 = 11 - 2 y

$LHS + 2 y = RHS + 2 y$

3 y - 4 + 2 y = 11

Commutative Property of Addition

3 y + 2 y - 4 = 11 - 2 y + 2 y

Add terms

5 y - 4 = 11

Isolate the Variable

Apply the Properties of Equality again to isolate the variable on one side of the equation. In this case, add

4

to both sides, then divide both sides of the equation by

5 .

5 y - 4 = 11

$LHS + 4 = RHS + 4$

5 y = 11 + 4

Add terms

5 y = 15

$LHS / 5 = RHS / 5$

y = \frac{1 5}{5}

Calculate quotient

y = 3

In conclusion,

y = 3

is the solution of Equation (II).

Note that the operations used to solve Equations (I) and (II) may not be the same as those required to solve a different equation.

Example

Finding the Usual Price Based on the Discounted Price

At Ali's fruit store, strawberries are on sale! Today they are $$ 0.60$ cheaper per pound than usual. Magdalena stopped at the store on her way home from school and bought eight pounds of strawberries. She paid a total of $$ 14.40 .$

Ali is inviting people to come and buy strawberries.

a Let

x

be the usual price of one pound of strawberries. Write an equation, in terms of

x,

modeling Magdalena's purchase.

b If Tadeo bought six pounds of strawberries two weeks ago, how much did he pay?

Hint

a Today, the special price per pound of strawberries is

x - 0.60

dollars. Multiply this price by the number of pounds bought by Magdalena and equate it to the total amount she paid.

b Because Tadeo bought strawberries two weeks ago, he paid the usual price for them. Solve the equation set in Part A to find the usual price. Then, multiply it by the number of pounds bought by Tadeo.

Solution

a Let

x

be the usual price of one pound of strawberries. According to Ali's offer, the special price is

$ 0.60

cheaper per pound than the usual price. Therefore, the special price equals the usual price minus

0.60 .

Special price = Usual price - 0.60 ⇓ Special price = x - 0.60

With this special price, Magdalena paid

$ 14.40

for eight pounds of strawberries. Consequently,

8

multiplied by the

special

price

is equal to the

total

amount

paid .

Using this information, an equation in terms of

x

can be established.

(Number of pounds) (Special price) = Total paid ⇓ 8 (x - 0.60) = 14.40

This equation can be used to determine the usual price of a pound of strawberries.

b Since Tadeo bought the strawberries two weeks ago, he paid the usual price. Therefore, start by finding the usual price by solving the equation set in Part A.

8 (x - 0.60) = 14.40

First, apply the Distributive Property to clear the parentheses. Then, apply the Properties of Equality to isolate the

x

variable.

8 (x - 0.60) = 14.40

Distr

Distribute $8$

8 x - 8 \cdot 0.60 = 14.40

8 x - 4.8 = 14.40

$LHS + 4.8 = RHS + 4.8$

8 x = 19.20

$LHS / 8 = RHS / 8$

x = \frac{1 9 . 2 0}{8}

Calculate quotient

x = 2.40

This means that the usual price of one pound of strawberries is

$ 2.40 .

To find how much Tadeo paid, multiply

6

— the number of pounds he bought — by the usual price.

Total paid = 6 x

$x = 2.40$

Total paid = 6 (2.40)

Total paid = 14.40

Consequently, last week Tadeo paid

$ 14.40

for six pounds of strawberries. Note that Tadeo paid the same as Magdalena, but he bought

2

pounds fewer.

Example

Solving an Equation With Variables on Both Sides

After dinner, Magdalena works on a $3 D$ model of a castle using cardboard for a history project. She wants the walls of the ramparts to be $7$ inches wide. Additionally, she wants the value of the perimeter of the wall, in inches, to be the same as the value of its area, in square inches.

The rampart of a castle. The rampart is a rectangle with dimensions 7 by x, and at the top, two small rectangles are cut. The two small rectangles are 2 by 1.

How high should the wall be?

Hint

The area of the wall is $7 x$ minus the area of the two small rectangles at the top. Equate the area expression with the perimeter of the wall. Then, solve the equation for $x .$

Solution

Since Magdalena wants the wall to have the same perimeter and area, regardless of the units of measure, start by finding the perimeter of the wall. To do so, add all the side lengths of the wall. Notice that the shorter sides of the small rectangles have a length of

1

inch. Knowing this, the length of the top of the wall can be found.

Top Length = 1 + 1 + 2 + 1 + 1 + 1 + 2 + 1 + 1

Add terms

Top Length = 11

The right-hand side of the wall has the same length as the left-hand side of the wall — that is, both have a length of

x

inches. Adding these two lengths to the base length,

7

inches, and to the length of the top of the wall, the perimeter of the wall will be obtained.

P = x + 7 + x + Top Length

▼

Substitute

11

for

Top Length

and simplify

$Top Length = 11$

P = x + 7 + x + 11

Commutative Property of Addition

P = x + x + 7 + 11

AssociativePropAdd

Associative Property of Addition

P = (x + x) + (7 + 11)

Add terms

P = 2 x + 18

To find the wall's area, note that it is made by cutting two small rectangles from the bigger rectangle, whose dimensions are

7

x .

Consequently, the wall's area is equal to the area of the big rectangle minus the area of the two small rectangles. The big rectangle area is

7 x,

and the area of each small rectangle is

2

square inches.

A = A_{Big} - 2 \cdot A_{Small}

SubstituteII

$A_{Big} = 7 x$ , $A_{Small} = 2$

A = 7 x - 2 \cdot 2

A = 7 x - 4

Since the perimeter and area of the wall must be the same, equate

P

and

A .

P 2 x + 18 = A ⇓ = 7 x - 4

To find the wall's height, solve the equation for

x .

Since the equation has variables on both sides, start by collecting them on one side of the equation.

2 x + 18 = 7 x - 4

SubEqn

$LHS - 2 x = RHS - 2 x$

18 = 7 x - 4 - 2 x

Commutative Property of Addition

18 = 7 x - 2 x - 4

SubTerms

Subtract terms

18 = 5 x - 4

$LHS + 4 = RHS + 4$

18 + 4 = 5 x

Add terms

22 = 5 x

$LHS / 5 = RHS / 5$

\frac{2 2}{5} = x

Calculate quotient

4.4 = x

RearrangeEqn

Rearrange equation

x = 4.4

Therefore, the wall should be

4.4

inches high.

Pop Quiz

Solving Different Equations

Solve the given equation for the indicated variable. If necessary, round the answer to two decimal places.

Random Equations. Some include variables on both sides.

After learning about the properties covered in this lesson, consider again the challenge presented at the beginning.

Closure

Finding the Number of Tickets Bought

At the beginning of the lesson, Mark was buying some paintball tickets. He found the following prices online, where the

$ 4

service charge is per transaction, not per ticket.

a Write an equation modeling the situation in which Mark and his friends chose Course

1,

got no discount, and paid

$ 103 .

b Write an equation modeling the situation in which Mark and his friends chose Course

2,

got the discount, and paid

$ 230.50 .

c How many tickets did Mark buy in the first situation? How many tickets did Mark buy in the second?

Answer

a Example Equation:

15 t + 1.5 t + 4 = 103

b Example Equation:

18 t + 1.5 (t - 5) + 4 = 230.5

c Mark bought

6

tickets in the first situation and

12

in the second.

Hint

a Find the amount paid only for the tickets, without fees. Next, find the amount paid in fees. Then, add these two costs to the service charge. Finally, equate the resulting expression with the actual money paid.

b Find the amount paid only for the tickets, without fees. Next, find the amount paid in fees. Remember, five of the tickets paid no fee. Then, add these two costs to the service charge. Finally, equate the resulting expression with the actual money paid.

c Solve the corresponding equation for

t .

Use the Distributive Property, the Associative Property of Addition, the Substitution Property of Equality, and the Properties of Equality.

Solution

a Start by listing the prices corresponding to Course

1 .

Course $1$
Ticket price	$$ 15.00$
Processing fee per ticket	$$ 1.50$
Service charge	$$ 4.00$

Let

t

be the number of purchased tickets. Then, the amount paid only for the tickets without any fees will be the product of

15

and

t .

15 t

Since the group did not get a discount, they paid the processing fee for all the tickets bought. To find the amount paid only in fees, multiply the price of the processing fee by

t .

1.5 t

They also had to pay a service charge, which is

$ 4

no matter the number of tickets bought. The sum of the three costs together will give an expression representing the total cost of purchasing

t

tickets for Course

1 .

15 t + 1.5 t + 4

Finally, it is said that they paid a total of

$ 103 .

Therefore, equate the previous expression to

103 .

$15 t + 1.5 t + 4 = 103$

This equation models the situation in which Mark and his friends chose Course $1,$ got no discount, and paid $$ 103 .$

b This time, the group decided to play on Course

2 .

Recall the prices corresponding to Course

2 .

Course $2$
Ticket price	$$ 18.00$
Processing fee per ticket	$$ 1.50$
Service charge	$$ 4.00$

Let

t

be the number of purchased tickets. Since the price per ticket is

$ 18.00,

the amount paid only for the tickets — without any fees — will be the product of the price per ticket and

t .

18 t

Since the group received a discount this time, they did not pay the fee for five of the tickets bought. Thus, to find the amount paid in fees, multiply the processing fee by

(t - 5) .

1.5 (t - 5)

They also paid a service charge, which is

$ 4

no matter the number of tickets bought. By adding the three costs, an expression representing the money paid for the purchase of

t

tickets for Course

2

will be found.

18 t + 1.5 (t - 5) + 4

Finally, it is said that they paid a total of

$ 230.50 .

Therefore, equate the previous expression to

230.5 .

$18 t + 1.5 (t - 5) + 4 = 230.5$

This equation models the situation in which Mark and his friends chose Course $2,$ got a discount, and paid $$ 230.50 .$

c To find the number of tickets bought by Mark the first time, the equation set in Part A will be solved for

t .

15 t + 1.5 t + 4 = 103

▼

Solve for

t

AssociativePropAdd

Associative Property of Addition

(15 t + 1.5 t) + 4 = 103

Add terms

16.5 t + 4 = 103

SubEqn

$LHS - 4 = RHS - 4$

16.5 t = 99

$LHS / 16.5 = RHS / 16.5$

t = \frac{9 9}{1 6 . 5}

Calculate quotient

t = 6

Consequently, the first time Mark bought

6

tickets. This confirms why he did not get the discount. To find the number of tickets he bought the second time, the equation set in Part B will be solved for

t .

18 t + 1.5 (t - 5) + 4 = 230.5

▼

Solve for

t

Distr

Distribute $1.5$

18 t + 1.5 t - 1.5 \cdot 5 + 4 = 230.5

18 t + 1.5 t - 7.5 + 4 = 230.5

Commutative Property of Addition

18 t + 1.5 t + 4 - 7.5 = 230.5

AssociativePropAdd

Associative Property of Addition

(18 t + 1.5 t) + (4 - 7.5) = 230.5

AddSubTerms

Add and subtract terms

19.5 t + (- 3.5) = 230.5

AddNeg

$a + (- b) = a - b$

19.5 t - 3.5 = 230.5

$LHS + 3.5 = RHS + 3.5$

19.5 t = 234

$LHS / 19.5 = RHS / 19.5$

t = \frac{2 3 4}{1 9 . 5}