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After writing an equation modeling some real-life situation, the next step is to solve it. To do so, it is necessary to apply different Properties of Equality. Along this lesson, these properties will be introduced and applied to equations that can be solved in one step.
### Catch-Up and Review

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

Zosia's brother is $10$ years older than she is. Also, Zosia's brother is half their father's age.

a If Zosia is $16$ years old, write equations for the ages of her brother and father.

b Solve these equations to find their ages.

The Properties of Equality are rules that allow manipulation of an equation in such a way that an equivalent equation is obtained. These properties will be reviewed in sets. The first set of properties is shown below.

For any real number, the number is equal to itself.

$a=a$

$x+3=8 $

The sum on the left-hand side can be interpreted as a number, so this sum has to equal $8.$ This implies that $x$ must be equal to $5.$For all real numbers, the order of an equality does not matter. Let $a$ and $b$ be real numbers.

If $a=b,$ then $b=a.$

This property is an axiom, so it does not need a proof to be accepted as true. The Symmetric Property of Equality also holds true if $a$ and $b$ are complex numbers.

For all real numbers, if two numbers are equal to the same number, then they are equal to each other. Let $a,$ $b,$ and $c$ be real numbers.

If $a=b$ and $b=c,$ then $a=c.$

${x=5y−305y−30=20 ⇒x =20 $

Since this property is an axiom, it does not need a proof to be accepted as true. The Transitive Property of Equality also holds true if $a,$ $b,$ and $c$ are complex numbers.Select the appropriate property for each example.

When solving equations, certain operations usually need to be undone to isolate the variable on one side. For example, consider the following equation.
*inverse operations* come into play. ### Concept

## Inverse Operations

$x+5=7 $

To isolate $x,$ the addition of $5$ on the left-hand side has to be undone. Here is where
Inverse operations are two operations that *undo* one another. For example, adding $6$ and subtracting $6$ are inverse operations because they cancel each other out. This means that adding $6$ to *any* number and then subtracting $6$ results in the original number.
*both sides*. The result of applying the Properties of Equality on an equation is an equivalent equation.

$ 10+6−610+6 −6 10 $

Equations are solved by using inverse operations. By the Properties of Equality, any operation performed on one side of an equation must also be performed on the other side of the equation to maintain equality. Consider an example.
$x−4=9 $

This equation can be solved by adding $4$ to both sides.
In this case, the subtraction on the left-hand side of the equation can only be eliminated by adding $4$ on Some of the most commonly used inverse operations are addition and subtraction. These operations fall under the Addition Property of Equality and the Subtraction Property of Equality.

Adding the same number to both sides of an equation results in an equivalent equation. Let $a,$ $b,$ and $c$ be real numbers.

If $a=b,$ then $a+c=b+c.$

$x−3=5 $

By adding $3$ to both sides of the equation, the variable $x$ can be isolated and the solution to the equation can be found. Subtracting the same number from both sides of an equation results in an equivalent equation. Let $a,$ $b,$ and $c$ be real numbers.

If $a=b,$ then $a−c=b−c.$

$x+2=7 $

By subtracting $2$ from both sides of the equation, the variable $x$ can be isolated and the solution to the equation can be found. One of Davontay's hobbies is playing the saxophone. He plays in the school band and wants to join a local community band as well. He goes to the music store to buy more reeds for his saxophone since he will be spending more time playing. After spending $$20$ on reeds, he is left with $$140.$

a Write an equation for this situation.

b How much money did Davontay originally have?

a $x−20=140$

b $$160$

a Use a variable for the original amount.

b Use the Properties of Equality to solve the equation set in Part A.

a It is possible write this situation as an equation using variables. Let $x$ be the original amount of money that Davontay had. It is given that Davontay spent $$20.$ Therefore, after buying the reeds, Davontay had $x−20$ dollars. This expression is the first part of the desired equation.

$x−20 $

Then, it is given that Davontay has $$140$ left after spending $$20.$ The equation can be completed by equating the previous expression with $140.$ $x−20=140 $

b To find how much money Davontay had initially, the equation set in Part A needs to be solved for $x.$ To do so, $$20$ will be added to both sides of the equation. This is valid by the Addition Property of Equality.
Therefore, Davontay originally had $$160.$

Heichi like collecting a particular brand of clothing. He is inspecting his wardrobe before going to the mall. He notices that he has $7$ shirts and $4$ more shirts than pairs of pants.

a If $p$ is the number of pairs of pants that Heichi has, which equation correctly describes the given situation?

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b How many pairs of pants does Heichi have?

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a Use variables to describe the situations.

b Use the Properties of Equality to solve the equation set in Part A.

a To write an equation that describes the given situation, variables need to be assigned to each item.

$Number of Pairs of Pants:Number of Shirts: ps $

Since Heichi has $4$ more shirts than pants, the sum of $p$ and $4$ is equal to $s.$
$p+4=s $

Also, it is given that Heichi has $7$ shirts. Therefore, by the Transitive Property of Equality, the equation can be rewritten using this information.
${p+4=ss=7 ⇒p+4=7 $

b To find out how many pairs of pants Heichi has, the equation set in Part A needs to be solved for $p.$ To do so, $4$ will be subtracted from both sides of the equation. This is valid by the Subtraction Property of Equality.

The other most common type of inverse operations are the multiplication and division operations. These are valid by the following properties of equality.

Given an equation, multiplying each side of the equation by the same number yields an equivalent equation. Let $a,$ $b,$ and $c$ be real numbers.

If $a=b,$ then $a×c=b×c.$

$x÷4x÷4×4x =2=2×4=8 $

Here, by multiplying both sides of the equation by $4,$ the variable $x$ was isolated and the solution of the equation was found. Note that the Multiplication Property of Equality also holds true if $a,$ $b,$ and $c$ are complex numbers.Dividing each side of an equation by the same nonzero number yields an equivalent equation. Let $a,$ $b,$ and $c$ be real numbers.

If $a=b$ and $c =0,$ then $a÷c=b÷c.$

$5x5x÷5x =10=10÷5=2 $

As can be observed, by dividing both sides of the equation by $5,$ the variable $x$ was isolated and the solution of the equation was found. Note that the Division Property of Equality also holds true if $a,$ $b,$ and $c$ are complex numbers.LaShay's hobbies include playing golf with her father. She wants to go to the golf course to practice. She knows that her school is one quarter of the way from her house to the golf course.

Consider that the school is about $2.5$ miles from LaShay's house.

a If $g$ is the distance from LaShay's house to the golf course, which equation correctly describes the situation?

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b How far is the golf course from LaShay's house?

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a Use variables for each distance.

b Use the Properties of Equality to solve the equation set in Part A.

a To write an equation that describes the given situation, variables need to be assigned to each distance.

$Distance to School:Distance to the Golf Course: sg $

Since the school is one quarter of the way to the golf course, the value of the division of $g$ by $4$ is equal to the value of $s.$
$4g =s $

Also, it is given that the distance to school is $2.5$ miles. $s=2.5 $

Using the Transitive Property of Equality, it is possible to rewrite the first equation.
${4g =ss=2.5 ⇒4g =2.5 $

b To find out how far is the golf course from LaShay's house, the equation set in Part A needs to be solved. To do so, both sides of the equation will be multiplied by $4.$ This is valid by the Multiplication Property of Equality.
Therefore, the golf course is $10$ miles away from LaShay's house.

$4g =2.5$

MultEqn

$LHS⋅4=RHS⋅4$

$4⋅4g =4⋅2.5$

DenomMultFracToNumber

$4⋅4a =a$

$g=4⋅2.5$

Multiply

Multiply

$g=10$

Zain is in the chess club at their school. During one game, their opponent has double the pieces that Zain has. Zain's opponent has $6$ pieces left on the board.

a If $z$ is the number of pieces that Zain has, which equation correctly describes the situation?

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b How many pieces does Zain have left?

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a Use variables for the number of pieces of each player.

b Use the Properties of Equality to solve the equation set in Part A.

a An equation can be written if variables are used for the number of pieces each player has left.

$Number of Zain’s Pieces:Number of Opponent’s Pieces: zo $

Since Zain's opponent has twice the pieces that Zain has, multiplying $z$ by $2$ is equal to $o.$
$2z=o $

It is given that the opponent has $6$ pieces left. Therefore, $o$ is equal to $6.$ Using the Transitive Property of Equality, it is possible to rewrite the equation.
${2z=oo=6 ⇒2z=6 $

b To find the number of pieces that Zain has left, both sides of the equation will be divided by $2.$ This is valid by the Division Property of Equality.

$2z=6$

DivEqn

$LHS/2=RHS/2$

$22z =26 $

CancelCommonFac

Cancel out common factors

$2 2 z =26 $

SimpQuot

Simplify quotient

$z=26 $

CalcQuot

Calculate quotient

$z=3$

Find the value of the variable on each equation using the Properties of Equality.

The challenge at the beginning of the lesson gave some information about Zosia's family.

- Zosia is $16$ years old.
- Zosia's brother is $10$ years older than she is.
- Zosia's brother is half the age of their father.

Then, the following exercises were presented.

a Write equations for the ages of Zosia's brother and father.

b Solve these equations to find their ages.

a ${b−10=162f =b $

b **Brother's Age: ** $26$ years

**Father's Age: ** $52$ years

a Use variables for the ages of each person.

b Use the Properties of Equality to solve each equation set in Part A.

a To write equations with the given information, it is convenient to use variables to represent each person's age.

$Zosia’s Age:Zosia’s Brother’s Age:Zosia’s Father’s Age: zbf $

Zosia's brother is $10$ years older than Zosia. This means that subtracting $10$ from $b$ is equal to $z.$
$b−10=z $

Also, it is given that Zosia is $16$ years old. Using the Transitive Property of Equality, it is possible to rewrite this equation.
${b−10=zz=16 ⇒b−10=16 $

Finally, since Zosia's brother's age is half of their father's age, the dividing $f$ by $2$ is equal to $b.$
$2f =b $

Therefore, these are the two equations for the ages of Zosia's brother and father.
${b−10=162f =b $

b To find the age of Zosia's brother, $10$ will be added to both sides of the first equation. This is valid by the Addition Property of Equality.

${2f =bb=26 ⇒2f =26 $

To find their father's age, both sides of the equation have to be multiplied by $2.$ This is valid by the Multiplication Property of Equality.
$2f =26$

MultEqn

$LHS⋅2=RHS⋅2$

$2⋅2f =2⋅26$

DenomMultFracToNumber

$2⋅2a =a$

$f=2⋅26$

Multiply

Multiply

$f=52$