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Algebra 1 View details

2. Solving One-Step Equations in One Variable

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Algebra 1 /

Chapter 1

Solving One-Step Equations in One Variable

This lesson delves into the topic of one-step equations, emphasizing the role of inverse operations and properties of equality in finding solutions. It provides real-life examples to illustrate how these mathematical concepts are not just academic exercises but also practical tools for everyday decision-making. For instance, the material covers scenarios like calculating distances, determining ages, and budgeting. The aim is to equip learners with the skills to solve simple equations quickly and accurately, whether they are students trying to excel in algebra or adults looking to make informed decisions in their daily lives.

Lesson Settings & Tools

	13 Theory slides
	10 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

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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.

Challenge

Different Ages of Family Members

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.

Discussion

Reflexive, Symmetric, and Transitive Properties of Equality

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.

Rule

Reflexive Property of Equality

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

$a = a$

This property is an axiom, so it does not need a proof. This property is used to solve equations. For example, consider the equation below.

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 .

Rule

Symmetric Property of Equality

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 can be used together with other Properties of Equality to solve equations or to isolate a variable and express it in terms of another variable. In the example below, the Symmetric Property of Equality is used to express

x

in terms of

y .

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.

Rule

Transitive Property of Equality

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 $a = c .$

This property can be used together with other Properties of Equality to solve equations.

{x = 5 y - 30 5 y - 30 = 20 \Rightarrow 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.

Pop Quiz

Which is the Correct Property?

Select the appropriate property for each example.

Quiz Reflexivity Symmetry Transitivity Properties of Equality

Discussion

Isolating the Variable

When solving equations, certain operations usually need to be undone to isolate the variable on one side. For example, consider the following equation.

x + 5 = 7

To isolate

x,

the addition of

5

on the left-hand side has to be undone. Here is where inverse operations come into play.

Concept

Inverse Operations

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.

10 + 6 - 6 10 + 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.

x - 4 = 9

AddEqn

$LHS + 4 = RHS + 4$

x - 4 + 4 = 9 + 4

Simplify left-hand side

x - 4 + 4 = 9 + 4

AddTerms

Add terms

x = 13

In this case, the subtraction on the left-hand side of the equation can only be eliminated by adding

4

on both sides. The result of applying the Properties of Equality on an equation is an equivalent equation.

Discussion

Addition and Subtraction Properties of Equality

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.

Rule

Addition 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 $a + c = b + c .$

The Addition Property of Equality is an axiom, so it does not need a proof. This property is one of the Properties of Equality that can be used when solving equations. Consider an example.

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.

x - 3 = 5

AddEqn

$LHS + 3 = RHS + 3$

x - 3 + 3 = 5 + 3

Simplify left-hand side

x - 3 + 3 = 5 + 3

AddTerms

Add terms

x = 8

Rule

Subtraction Property of Equality

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 $a - c = b - c .$

The Subtraction Property of Equality is an axiom, so it does not need a proof. This property is one of the Properties of Equality that can be used when solving equations. Consider an example.

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.

x + 2 = 7

SubEqn

$LHS - 2 = RHS - 2$

x + 2 - 2 = 7 - 2

Simplify left-hand side

x + 2 - 2 = 7 - 2

SubTerms

Subtract terms

x = 5

Example

Finding How Much Money Davontay Had

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?

Answer

x - 20 = 140

$ 160

Hint

a Use a variable for the original amount.

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

Solution

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.

x - 20 = 140

AddEqn

$LHS + 20 = RHS + 20$

x - 20 + 20 = 140 + 20

SubTerms

Subtract terms

x = 160

Therefore, Davontay originally had

$ 160 .

Example

Finding the Number of Pairs of Pants

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?

b How many pairs of pants does Heichi have?

Hint

a Use variables to describe the situations.

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

Solution

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 : p s

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 = s s = 7 \Rightarrow 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.

p + 4 = 7

SubEqn

$LHS - 4 = RHS - 4$

p + 4 - 4 = 7 - 4

SubTerms

Subtract terms

p = 3

Therefore, Heichi has

3

pairs of pants.

Discussion

Multiplication and Division Properties 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.

Rule

Multiplication Property 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 $a \times c = b \times c .$

The Multiplication Property of Equality is an axiom, so it does not need a proof. This property is one of the Properties of Equality that can be used when solving equations. Consider the following example.

x \div 4 x \div 4 \times 4 x = 2 = 2 \times 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.

Rule

Division Property of Equality

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 $c \neq = 0,$ then $a \div c = b \div c .$

The Division Property of Equality is an axiom, so it does not need a proof to be accepted as true. This property is one of the Properties of Equality that can be used when solving equations.

5 x 5 x \div 5 x = 10 = 10 \div 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.

Example

Distances From Home

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?

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

Hint

a Use variables for each distance.

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

Solution

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 : s g

Since the school is one quarter of the way to the golf course, the value of the division of

g

4

is equal to the value of

s .

\frac{g}{4} = 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.

{\frac{g}{4} = s s = 2.5 \Rightarrow \frac{g}{4} = 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.

\frac{g}{4} = 2.5

MultEqn

$LHS \cdot 4 = RHS \cdot 4$

4 \cdot \frac{g}{4} = 4 \cdot 2.5

DenomMultFracToNumber

$4 \cdot \frac{a}{4} = a$

g = 4 \cdot 2.5

Multiply

g = 10

Therefore, the golf course is

10

miles away from LaShay's house.

Example

The Number of Chess Pieces

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?

b How many pieces does Zain have left?

Hint

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.

Solution

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 : z o

Since Zain's opponent has twice the pieces that Zain has, multiplying

z

2

is equal to

o .

2 z = 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.

{2 z = o o = 6 \Rightarrow 2 z = 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.

2 z = 6

DivEqn

$LHS / 2 = RHS / 2$

\frac{2 z}{2} = \frac{6}{2}

CancelCommonFac

Cancel out common factors

\frac{2 z}{2} = \frac{6}{2}

SimpQuot

Simplify quotient

z = \frac{6}{2}

CalcQuot

Calculate quotient

z = 3

Therefore, Zain has

3

pieces left.

Pop Quiz

Solving Equations Using the Properties of Equality

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

Closure

Finding the Ages of Family Members

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.

Answer

{b - 10 = 16 \frac{f}{2} = b

b Brother's Age:

26

years
Father's Age:

52

years

Hint

a Use variables for the ages of each person.

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

Solution

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 : z b f

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 = z z = 16 \Rightarrow b - 10 = 16

Finally, since Zosia's brother's age is half of their father's age, the dividing

f

2

is equal to

b .

\frac{f}{2} = b

Therefore, these are the two equations for the ages of Zosia's brother and father.

{b - 10 = 16 \frac{f}{2} = 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.

b - 10 = 16

AddEqn

$LHS + 10 = RHS + 10$

b - 10 + 10 = 16 + 10

AddTerms

Add terms

b = 26

Therefore, Zosia's brother is

26

years old. Then, using the Transitive Property of Equality again, the second equation from Part A can be rewritten.

{\frac{f}{2} = b b = 26 \Rightarrow \frac{f}{2} = 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.

\frac{f}{2} = 26

MultEqn

$LHS \cdot 2 = RHS \cdot 2$

2 \cdot \frac{f}{2} = 2 \cdot 26

DenomMultFracToNumber

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

f = 2 \cdot 26

Multiply

f = 52

Consequently, Zosia's father is

52

years old.

Level 1

Level 2

Level 3

Solving One-Step Equations in One Variable

Exercises

The price of candy is

$ 0.40

per gram. How much candy can Tiffaniqua buy if she has

$ 2

to spend?

Let x be the amount of candy in grams that we can buy for $2. If we multiply the price of a gram of candy $0.40 by the amount of candy that we can buy x, the resulting product is $2. This can be written as an equation. 0.40x = 2 To find the value of x, we can divide both sides of the equation by 0.40.

Therefore, Tiffaniqua can buy 5 grams of candy for $2.

Solve the following equations.

x - 10 = 27

5 y = 20

\frac{z}{3} = 7

We can solve the equation by adding 10 to both sides. Doing this isolates x on the left-hand side of the equation.

Therefore, x=37 is the solution to the equation!

On the left-hand side we have 5 times the value of y. Since we want to find the value of y, we need to divide both sides of the equation by 5.

The value of y is 4.

In this equation, we have one third of z. We will multiply both sides of the equation by 3 to isolate z and find its value.

The value of z is 21.

Tearrik worked

2.5

hours and earned

$ 9 .

How much would he earn if he worked

4.5

hours at the same salary?

To find out how much money Tearrik would earn working 4.5 hours, we need to find out how much he earns per hour. Let x be Tearrik's salary. The result of multiplying 2.5 by x is equal to 9. This can be written as an equation. 2.5x = 9 To find Tearrik's salary, we will divide both sides of the equation by 2.5.

We found that Tearrik earns $ 3.60 each hour he works. Now we can multiply his salary by 4.5 to find how much he would earn if he worked for 4.5 hours. 4.5 * 3.60 = 16.20 Tearrik would earn $16.20 if he worked 4.5 hours.

Solve the following equations. Check your answer.

7 = x + 3

- 5 + a = 19

- 17 = k + 6

To solve the equation we need to isolate x on one side of the equation. We will do this by subtracting 3 from both sides of the equation.

We found that x=4 is the solution to the equation. To check the answer, we will substitute 4 for x into the original equation.

Since both sides of the equation are equal, the solution is correct.

We will isolate a on one side of the equation to find the solution. To do so, we will add 5 to both sides of the equation.

The value of a is 24. To check if this value is correct, we will substitute 24 for a into the original equation.

Since both sides of the equation are equal, the solution is correct.

To solve the equation, we have to isolate k on one side of the equation. We will do this by subtracting 6 from both sides of the equation.

We found that k=-23. To verify if this is a correct solution, we will substitute -23 for k into the original equation.

Both sides of the equation are equal. Now we can be sure that we found the correct value.

Solve the following equations. Check your answer.

- 8 m = - 56

15 = 2.5 t

\frac{q}{6} = - 12

To solve the equation, we need to isolate m on one side of the equation. We will do this by dividing both sides of the equation by -8.

We found that m=7 is the solution to the equation. To check the answer, we will substitute 7 for m into the original equation.

Since both sides of the equation are equal, the solution is correct.

To find the solution, we will isolate t on one side of the equation. We will do this by dividing both sides of the equation by 2.5.

The value of t is 6. To check if this solution is correct, we will substitute 6 for t into the original equation.

Since both sides of the equation are equal, the solution is correct.

To solve the equation, we have to isolate q on one side of the equation. We will do this by multiplying both sides of the equation by 6.

We found the value of q is -72. To verify if this is the correct solution, we will substitute -72 for q in the equation.

Both sides of the equation are equal, so we can be sure that we found the correct value.

Solving One-Step Equations in One Variable

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