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When it comes to modeling daily life situations using equations, it is common to end with equations that require more than one step to be solved. For such cases, it is convenient to know which properties can be applied to solve the obtained equations. To learn about it, different types of equations will be set and solved along this lesson.
### Catch-Up and Review

**Here are a few recommended readings before getting started with this lesson.**

Mark wants to buy tickets online to play paintball with his friends next weekend. On the website, he finds the following price list.

Note that the $$4$ service charge is per transaction, not per ticket.

a Mark and his friends decide to choose Course $1$ and pay $$103.$ Let $t$ be the number of tickets bought. Write an equation in terms of $t$ that models this situation.

b Mark and his friends plan to go again next month but on Course $2.$ They were told that when the group has more than eight people, the processing fee of five tickets is free. Write an equation modeling the situation in which the group gets the discount and pays $$230.50.$

c How many tickets did Mark buy for his first visit? How many tickets will Mark buy next month?

Two numerical expressions are equivalent if one of the expressions is obtained from the other by applying properties of addition and multiplication. Knowing this, match each expression in the left column with the equivalent expression.

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Apart from the Properties of Equality, the properties of addition and multiplication are also heavily used when rewriting and simplifying expressions or equations. Thus, it is convenient to know these properties. The commutative property of each operation will be discussed first.

The order in which two or more terms are added does not affect the value of the sum. That is, the addends can be written in any order.

$a+b=b+a$

$4+5+1✓10 =1+5+4⇓=10✓ $

Since the Commutative Property of Addition is an axiom, it does not need a proof.The order in which two or more factors are multiplied does not affect the value of the product. That is, the multiplicands can be written in any order.

$a⋅b=b⋅a$

$3⋅2⋅6✓36 =2⋅6⋅3⇓=36✓ $

Since the Commutative Property of Multiplication is an axiom, it does not need a proof.As seen, the commutative property says that the order in which two or more terms are added or multiplied will not change the resulting sum or product. Next, the associative property of addition and multiplication will be presented.

The way three or more terms are grouped when added does not affect the value of the sum.

$(a+b)+c=a+(b+c)$

$(3+9)+412+4✓16 =3+(9+4)⇓=3+13⇓=16✓ $

Since the Associative Property of Addition is an axiom, it does not need a proof.The way three or more factors are grouped when multiplied does not affect the value of the product.

$(a⋅b)⋅c=a⋅(b⋅c) $

$(2⋅4)⋅68⋅6✓48 =2⋅(4⋅6)⇓=2⋅24⇓=48✓ $

Since the Associative Property of Multiplication is an axiom, it does not need a proof.Two expressions are equivalent when one of them is obtained by applying properties of addition and multiplication to the other. For each pair of given expressions, determine whether they are equivalent. In the affirmative case, name the property that transforms one expression into the other.

When solving equations, there are instances in which some parts of an expression can be substituted with an equivalent expression. As an example, consider the following equation.
*Substitution Property of Equality*, which also guarantees that the resulting equation and the initial are equivalent. ### Rule

## Substitution Property of Equality

Since the Substitution Property of Equality is an axiom, it does not need a proof.

$5t+(9−3)=30 $

Since $9−3$ equals $6,$ the second term on the left-hand side can be substituted with $6.$
$5t+(9−3) 6 =30⇓5t+6=30 $

The previous simplification is able thanks to the If two real numbers are equal, then one can be substituted for another in any expression.

If $a=b,$ then $a$ can be substituted for $b$ in any expression.

This morning in science class, Tadeo and Magdalena learned how to convert degrees Celsius to Fahrenheit using the following formula.
### Hint

### Solution

$Celsius to Fahrenheit: F=59 C+32 $

Later, while eating lunch in the schoolyard, they saw that the thermometer indicated a temperature of $77_{∘}F.$ They wondered what the temperature would be in Celsius. To make the conversion, they used the formula they learned in the morning. However, they got different results.
Who is correct? {"type":"choice","form":{"alts":["Magdalena","Tadeo","None"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":0}

Apply the Substitution Property of Equality to determine who got the correct answer. Start by substituting $77$ for $F$ into the conversion formula. Then, substitute the values Tadeo and Magdalena found for $C.$ If a true statement is obtained, the corresponding value is a solution; otherwise, it is not.

Tadeo and Magdalena want to convert $77_{∘}F$ into degrees Celsius. Therefore, by the Substitution Property of Equality, $77$ can be substituted for $F$ in the conversion formula.
A false statement was obtained, which suggests that Tadeo's answer is not correct. Next, substitute Magdalena's answer.
A true statement was obtained, which implies that Magdalena is correct in saying that $77_{∘}F$ is the same as $25_{∘}C.$

$77=59 C+32 $

Again, by the Substitution Property of Equality, the values found by Tadeo and Magdalena will be substituted for $C$ in the above equation one at a time.
$Tadeo:Magdalena: 23_{∘}C25_{∘}C $

After substituting each value and simplifying, if a true statement is obtained, the substituted value is a solution of the equation. Otherwise, it is not. First, $C=23$ will be substituted.
$77=59 C+32$

Substitute $23$ for $C$ and simplify

Substitute

$C=23$

$77=?59 (23)+32$

MoveRightFacToNum

$ca ⋅b=ca⋅b $

$77=?59⋅23 +32$

Multiply

Multiply

$77=?5207 +32$

CalcQuot

Calculate quotient

$77=?41.4+32$

AddTerms

Add terms

$77=73.4×$

$77=59 C+32$

Substitute $25$ for $C$ and simplify

Substitute

$C=25$

$77=?59 (25)+32$

MoveRightFacToNum

$ca ⋅b=ca⋅b $

$77=?59⋅25 +32$

Multiply

Multiply

$77=?5225 +32$

CalcQuot

Calculate quotient

$77=?45+32$

AddTerms

Add terms

$77=77✓$

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*.

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±c)(b±c)a =a⋅b±a⋅c=a⋅b±a⋅c $

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

$9x+9⋅2=9(x+2) $

In some cases, the right-hand side expression is more useful than the one on the left.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): 23 (x−2)+5=14y+2y−4=11−2y $

The general idea is to simplify both sides of an equation and then isolate the variable on one side of the equation.

1

Multiply to Clear Any Fractions

If there are fractions in the equation, start by multiplying the entire equation by the least common denominator. In this case, multiply both sides of Equation (I) by $2.$

$2⋅(23 (x−2)+5)=2⋅14 $

By applying the Distributive Property, distribute $2$ one the left-hand side.
$2⋅23 (x−2)+2⋅5=2⋅14 $

Then, cross out the common factors and perform the required multiplications. After this, Equation (I) is simplified into an equivalent equation with no fractions.
$2⋅23 (x−2)+2⋅5=2⋅14$

CrossCommonFac

Cross out common factors

$2 ⋅2 3 (x−2)+2⋅5=2⋅14$

CancelCommonFac

Cancel out common factors

$3(x−2)+2⋅5=2⋅14$

Multiply

Multiply

$3(x−2)+10=28$

2

Clear Parentheses and Combine Like Terms on Each Side

Apply the Distributive Property again to clear the parentheses on the left-hand side of the equation. In this case, distribute $3$ to each term inside the parentheses. Then, combine like terms.

3

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 and then apply the Division Property of Equality.
Consequently, $x=8$ is the solution of Equation (I).

$3x+4=28$

SubEqn

$LHS−4=RHS−4$

$3x=28−4$

SubTerms

Subtract terms

$3x=24$

DivEqn

$LHS/3=RHS/3$

$x=324 $

CalcQuot

Calculate quotient

$x=8$

1

Multiply to Clear Any Fractions

Since Equation (II) has no fractions, this step can be omitted.

2

Clear Parentheses and Combine Like Terms on Each Side

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

3

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 $2y$ to both sides of the equation to group the $y-$terms on the left-hand side.

$3y−4=11−2y$

AddEqn

$LHS+2y=RHS+2y$

$3y−4+2y=11$

CommutativePropAdd

Commutative Property of Addition

$3y+2y−4=11$

AddTerms

Add terms

$5y−4=11$

4

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 and then divide both sides of the equation by $5.$
In conclusion, $y=3$ is the solution of Equation (II).

$5y−4=11$

AddEqn

$LHS+4=RHS+4$

$5y=11+4$

AddTerms

Add terms

$5y=15$

DivEqn

$LHS/5=RHS/5$

$y=515 $

CalcQuot

Calculate quotient

$y=3$

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.$

a Let $x$ be the usual price of one pound of strawberries. Write an equation, in terms of $x,$ modeling Magdalena's purchase.

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

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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.

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 ofpounds )(Specialprice )=Totalpaid ⇓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 $8$

$8x−8⋅0.60=14.40$

Multiply

Multiply

$8x−4.8=14.40$

AddEqn

$LHS+4.8=RHS+4.8$

$8x=19.20$

DivEqn

$LHS/8=RHS/8$

$x=819.20 $

CalcQuot

Calculate quotient

$x=2.40$

After dinner, Magdalena works on a $3D$ 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.

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The area of the wall is $7x$ 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 $x.$

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.
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.
To find the wall's area, note that it is made by cutting two small rectangles from the bigger rectangle, whose dimensions are $7$ by $x.$
Therefore, the wall should be $4.4$ inches high.

$P=x+7+x+Top Length$

Substitute $11$ for $Top Length$ and simplify

Substitute

$Top Length=11$

$P=x+7+x+11$

CommutativePropAdd

Commutative Property of Addition

$P=x+x+7+11$

AssociativePropAdd

Associative Property of Addition

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

AddTerms

Add terms

$P=2x+18$

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 $7x,$ and the area of each small rectangle is $2$ square inches.
Since the perimeter and area of the wall must be the same, equate $P$ and $A.$

$P2x+18 =A⇓=7x−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.
$2x+18=7x−4$

SubEqn

$LHS−2x=RHS−2x$

$18=7x−4−2x$

CommutativePropAdd

Commutative Property of Addition

$18=7x−2x−4$

SubTerms

Subtract terms

$18=5x−4$

AddEqn

$LHS+4=RHS+4$

$18+4=5x$

AddTerms

Add terms

$22=5x$

DivEqn

$LHS/5=RHS/5$

$522 =x$

CalcQuot

Calculate quotient

$4.4=x$

RearrangeEqn

Rearrange equation

$x=4.4$

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

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

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.
### Answer

### Hint

### Solution

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 $t.$

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 $t.$
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 $t.$
In conclusion, Mark bought $12$ tickets the second time.

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?

a **Example Equation:** $15t+1.5t+4=103$

b **Example Equation:** $18t+1.5(t−5)+4=230.5$

c Mark bought $6$ tickets in the first situation and $12$ in the second.

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.

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$ |

$15t $

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 $t.$
$1.5t $

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 $1.$
$15t+1.5t+4 $

Finally, it is said that they paid a total of $$103.$ Therefore, equate the previous expression to $103.$ $15t+1.5t+4=103$

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$ |

$18t $

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.
$18t+1.5(t−5)+4 $

Finally, it is said that they paid a total of $$230.50.$ Therefore, equate the previous expression to $230.5.$ $18t+1.5(t−5)+4=230.5$

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 $t.$

$15t+1.5t+4=103$

Solve for $t$

AssociativePropAdd

Associative Property of Addition

$(15t+1.5t)+4=103$

AddTerms

Add terms

$16.5t+4=103$

SubEqn

$LHS−4=RHS−4$

$16.5t=99$

DivEqn

$LHS/16.5=RHS/16.5$

$t=16.599 $

CalcQuot

Calculate quotient

$t=6$

$18t+1.5(t−5)+4=230.5$

Solve for $t$

Distr

Distribute $1.5$

$18t+1.5t−1.5⋅5+4=230.5$

Multiply

Multiply

$18t+1.5t−7.5+4=230.5$

CommutativePropAdd

Commutative Property of Addition

$18t+1.5t+4−7.5=230.5$

AssociativePropAdd

Associative Property of Addition

$(18t+1.5t)+(4−7.5)=230.5$

AddSubTerms

Add and subtract terms

$19.5t+(-3.5)=230.5$

AddNeg

$a+(-b)=a−b$

$19.5t−3.5=230.5$

AddEqn

$LHS+3.5=RHS+3.5$

$19.5t=234$

DivEqn

$LHS/19.5=RHS/19.5$

$t=19.5234 $

CalcQuot

Calculate quotient

$t=12$

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