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This pdf covers different types of properties of operations on equation. We learned that the addition property of equality tells us that if we add the same quantity to both sides of an equation, then our equation remains the same. The formula is if a = b, then a + c = b + c.

1.
MAT 056: Basic Algebra I Learning Unit 4: Handout

The Addition and Subtraction Properties of Equality

Linear equation

A linear equation in one variable can be written as follows.

ππ₯ + π = π a, b and c are any real numbers.

Example 1

The following equations are examples of linear equations.

ο· 2π₯ + 8 = 16

ο· 5π₯ = β50

ο· π₯ β 6 = 10

A solution of an equation is a replacement for the variable that makes the equation true.

Example 2

Find the solution to the following equation.

π₯ + 3 = 10

The solution of this equation is 7 since replacing π₯ with 7 will make the equation true.

π₯ + 3 = 10

7 + 3 = 10

10 = 10

The properties of equality help us find a solution to an equation.

Addition Property of Equality

Let π, π, and π represent any real numbers.

You can add the same number to both sides of an equation and get an equivalent equation.

π+π =π+π

Subtraction Property of Equality

Let π, π, and π represent any real numbers.

You can subtract the same number from both sides of an equation and get an equivalent equation.

πβπ =πβπ

Page 1 of 10

The Addition and Subtraction Properties of Equality

Linear equation

A linear equation in one variable can be written as follows.

ππ₯ + π = π a, b and c are any real numbers.

Example 1

The following equations are examples of linear equations.

ο· 2π₯ + 8 = 16

ο· 5π₯ = β50

ο· π₯ β 6 = 10

A solution of an equation is a replacement for the variable that makes the equation true.

Example 2

Find the solution to the following equation.

π₯ + 3 = 10

The solution of this equation is 7 since replacing π₯ with 7 will make the equation true.

π₯ + 3 = 10

7 + 3 = 10

10 = 10

The properties of equality help us find a solution to an equation.

Addition Property of Equality

Let π, π, and π represent any real numbers.

You can add the same number to both sides of an equation and get an equivalent equation.

π+π =π+π

Subtraction Property of Equality

Let π, π, and π represent any real numbers.

You can subtract the same number from both sides of an equation and get an equivalent equation.

πβπ =πβπ

Page 1 of 10

2.
MAT 056: Basic Algebra I Learning Unit 4: Handout

When solving an equation for π₯, we must get π₯ alone. We can do this by performing the inverse

operation. Addition and subtraction are inverse operations. We can add the same number to each side

of an equation or subtract the same number from each side of an equation.

Example 3

π₯ + 6 = 10 The inverse of adding 6 is subtracting 6.

π₯ + 6 β 6 = 10 β 6 Subtract 6 from each side of the equation.

π₯+0 = 4

Answer: π = π

Example 4

π₯ β 8 = 12 The inverse of subtracting 8 is adding 8. Add 8

to each side of the equation.

π₯ β 8 + 8 = 12 + 8 Adding opposites will give a sum of zero.

β8 + 8 = 0

π₯ + 0 = 20

Answer: π = ππ

Example 5

π₯ + 4.5 = 10.8 The inverse of adding 4.5 is subtracting 4.5.

π₯ + 4.5 β 4.5 = 10.8 β 4.5 Subtract 4.5 from each side of the equation.

π₯ + 0 = 6.3

Answer: π = π. π

Page 2 of 10

When solving an equation for π₯, we must get π₯ alone. We can do this by performing the inverse

operation. Addition and subtraction are inverse operations. We can add the same number to each side

of an equation or subtract the same number from each side of an equation.

Example 3

π₯ + 6 = 10 The inverse of adding 6 is subtracting 6.

π₯ + 6 β 6 = 10 β 6 Subtract 6 from each side of the equation.

π₯+0 = 4

Answer: π = π

Example 4

π₯ β 8 = 12 The inverse of subtracting 8 is adding 8. Add 8

to each side of the equation.

π₯ β 8 + 8 = 12 + 8 Adding opposites will give a sum of zero.

β8 + 8 = 0

π₯ + 0 = 20

Answer: π = ππ

Example 5

π₯ + 4.5 = 10.8 The inverse of adding 4.5 is subtracting 4.5.

π₯ + 4.5 β 4.5 = 10.8 β 4.5 Subtract 4.5 from each side of the equation.

π₯ + 0 = 6.3

Answer: π = π. π

Page 2 of 10

3.
MAT 056: Basic Algebra I Learning Unit 4: Handout

Example 6

1 3 1 1 1

π₯β3=4 The inverse of subtracting 3 is adding 3. Add 3 to

each side of the equation.

1 1 3 1

π₯β3+3 = 4+3 Adding opposites will give a sum of zero.

1 1

β + = 0.

3 3

When adding fractions, get a common

3 1

denominator. 4

+3

3π₯3 9 1π₯4 4

= 12 and 3 π₯ 4 = 12

9 4

π₯+0= 12

+ 12

13

π₯=

12

π

Answer: π = π ππ

Example 7

20 = π₯ β 9 Sometimes π₯ is on the right side of the equation.

The inverse of subtracting 9 is adding 9.

20 + 9 = π₯ β 9 + 9 Add 9 to each side of the equation.

29 = π₯ + 0

Answer: ππ = π This is the same as π₯ = 29.

Example 8

5=π₯+8 Here π₯ is on the right side of the equation.

The inverse of adding 8 is subtracting 8.

5β8=π₯+8β8 Subtract 8 from each side of the equation.

β3 = π₯ + 0

Answer: βπ = π This is the same as π₯ = β3.

Page 3 of 10

Example 6

1 3 1 1 1

π₯β3=4 The inverse of subtracting 3 is adding 3. Add 3 to

each side of the equation.

1 1 3 1

π₯β3+3 = 4+3 Adding opposites will give a sum of zero.

1 1

β + = 0.

3 3

When adding fractions, get a common

3 1

denominator. 4

+3

3π₯3 9 1π₯4 4

= 12 and 3 π₯ 4 = 12

9 4

π₯+0= 12

+ 12

13

π₯=

12

π

Answer: π = π ππ

Example 7

20 = π₯ β 9 Sometimes π₯ is on the right side of the equation.

The inverse of subtracting 9 is adding 9.

20 + 9 = π₯ β 9 + 9 Add 9 to each side of the equation.

29 = π₯ + 0

Answer: ππ = π This is the same as π₯ = 29.

Example 8

5=π₯+8 Here π₯ is on the right side of the equation.

The inverse of adding 8 is subtracting 8.

5β8=π₯+8β8 Subtract 8 from each side of the equation.

β3 = π₯ + 0

Answer: βπ = π This is the same as π₯ = β3.

Page 3 of 10

4.
MAT 056: Basic Algebra I Learning Unit 4: Handout

Combine like terms

Sometimes we need to simplify one or both sides of an equation before using the addition or

subtraction property of equality. Combine like terms on each side of the equation.

Example 9

4π₯ + 6 β 3π₯ β 3 = 15 Combine like terms on the left side of the

equation.

4π₯ β 3π₯ + 6 β 3 = 15 Change the order. Combine the π₯ variables.

1π₯ + 6 β 3 = 15 Combine the numbers.

1π₯ + 3 = 15

1π₯ + 3 β 3 = 15 β 3 Subtract 3 from each side of the equation.

1π₯ + 0 = 12

1π₯ = 12

Anwer: π = ππ

Example 10

2.3 = β7π + 0.5 + 8π + 0.1 Combine like terms on the right side of the

equation.

2.3 = β7π + 8π + 0.5 + 0.1 Combine the c variables.

2.3 = 1π + 0.5 + 0.1 Combine the decimal numbers.

2.3 = 1π + 0.6 Now get π alone on the right side of the

equation.

2.3 β 0.6 = 1π + 0.6 β 0.6 Subtract 0.6 on each side of the equation.

1.7 = 1π + 0

Answer: π. π = π

Page 4 of 10

Combine like terms

Sometimes we need to simplify one or both sides of an equation before using the addition or

subtraction property of equality. Combine like terms on each side of the equation.

Example 9

4π₯ + 6 β 3π₯ β 3 = 15 Combine like terms on the left side of the

equation.

4π₯ β 3π₯ + 6 β 3 = 15 Change the order. Combine the π₯ variables.

1π₯ + 6 β 3 = 15 Combine the numbers.

1π₯ + 3 = 15

1π₯ + 3 β 3 = 15 β 3 Subtract 3 from each side of the equation.

1π₯ + 0 = 12

1π₯ = 12

Anwer: π = ππ

Example 10

2.3 = β7π + 0.5 + 8π + 0.1 Combine like terms on the right side of the

equation.

2.3 = β7π + 8π + 0.5 + 0.1 Combine the c variables.

2.3 = 1π + 0.5 + 0.1 Combine the decimal numbers.

2.3 = 1π + 0.6 Now get π alone on the right side of the

equation.

2.3 β 0.6 = 1π + 0.6 β 0.6 Subtract 0.6 on each side of the equation.

1.7 = 1π + 0

Answer: π. π = π

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5.
MAT 056: Basic Algebra I Learning Unit 4: Handout

The Opposite of π

A minus sign in front of π₯ can be read as the opposite of π₯.

Let π represent any real number.

ο· βπ₯ = π The opposite of π₯ is π.

ο· π₯ = βπ Therefore π₯ equals the opposite of π.

Example 11

βπ₯ = 8 The opposite of π₯ is 8.

Answer: π = βπ Therefore π₯ = β8.

Example 12

βπ₯ = β5 The opposite of π₯ is β5.

Answer: π = π Therefore π₯ = 5.

Variables on Both Sides of the Equation

Sometimes variables are on both sides of the equation. We need to get the variables on one side of the

equation and numbers on the other side. You can get the variables alone on either side of the equation.

8π₯ + 5 = 7π₯ Here the π₯ variable appears on both sides of the

equation.

You must eliminate the π₯ variable on one side of

the equation.

One way to do this is to subtract 7π₯ on each

side.

8π₯ β 7π₯ + 5 = 7π₯ β 7π₯ Notice that 7π₯ β 7π₯ = 0.

1π₯ + 5 = 0 Now π₯ is only on the left side of the equation.

Next, get π₯ alone.

1π₯ + 5 β 5 = 0 β 5 Subtract 5 on each side of the equation.

1π₯ + 0 = β5

1π₯ = β5

Answer: π = βπ

There is often more than one way to solve a problem. If you use the addition and subtraction properties

correctly, you will always get the correct answer.

Page 5 of 10

The Opposite of π

A minus sign in front of π₯ can be read as the opposite of π₯.

Let π represent any real number.

ο· βπ₯ = π The opposite of π₯ is π.

ο· π₯ = βπ Therefore π₯ equals the opposite of π.

Example 11

βπ₯ = 8 The opposite of π₯ is 8.

Answer: π = βπ Therefore π₯ = β8.

Example 12

βπ₯ = β5 The opposite of π₯ is β5.

Answer: π = π Therefore π₯ = 5.

Variables on Both Sides of the Equation

Sometimes variables are on both sides of the equation. We need to get the variables on one side of the

equation and numbers on the other side. You can get the variables alone on either side of the equation.

8π₯ + 5 = 7π₯ Here the π₯ variable appears on both sides of the

equation.

You must eliminate the π₯ variable on one side of

the equation.

One way to do this is to subtract 7π₯ on each

side.

8π₯ β 7π₯ + 5 = 7π₯ β 7π₯ Notice that 7π₯ β 7π₯ = 0.

1π₯ + 5 = 0 Now π₯ is only on the left side of the equation.

Next, get π₯ alone.

1π₯ + 5 β 5 = 0 β 5 Subtract 5 on each side of the equation.

1π₯ + 0 = β5

1π₯ = β5

Answer: π = βπ

There is often more than one way to solve a problem. If you use the addition and subtraction properties

correctly, you will always get the correct answer.

Page 5 of 10

6.
MAT 056: Basic Algebra I Learning Unit 4: Handout

Example 14

We could work the problem in example 13 another way.

We could eliminate the variable on the left side of the equation.

8π₯ + 5 = 7π₯

8π₯ β 8π₯ + 5 = 7π₯ β 8π₯ Subtract 8π₯ on each side of the equation.

0 + 5 = β1π₯ Now π₯ is just on the right side of the equation.

5 = βπ₯ This says that 5 equals the opposite of π₯.

Answer: βπ = π Therefore β5 equals π₯.

This is the same solution we found in example 13. You can eliminate the variable from either side of the

equation and still get the same answer.

Eliminate the Parentheses in an Equation

5(π₯ β 1) = 6π₯ + 4 Distribute to get rid of the parentheses.

5π₯ β 5 = 6π₯ + 4

5π₯ β 6π₯ β 5 = 6π₯ β 6π₯ + 4 Eliminate the x variable on the right side of the

equation.

β1π₯ β 5 = 0 + 4

β1π₯ β 5 = 4 Now get x alone on the left side of the equation.

Add 5 to each side of the equation. Donβt try to

eliminate the coefficient of x which is β1 before

adding 5 to each side of the equation.

β1π₯ β 5 + 5 = 4 + 5 Add 5 to each side of the equation.

β1π₯ + 0 = 9

β1π₯ = 9 Remember: β1π₯ = βπ₯

βπ₯ = 9 This means the opposite of π₯ equals 9.

Answer: π = βπ Therefore π₯ equals β9.

Page 6 of 10

Example 14

We could work the problem in example 13 another way.

We could eliminate the variable on the left side of the equation.

8π₯ + 5 = 7π₯

8π₯ β 8π₯ + 5 = 7π₯ β 8π₯ Subtract 8π₯ on each side of the equation.

0 + 5 = β1π₯ Now π₯ is just on the right side of the equation.

5 = βπ₯ This says that 5 equals the opposite of π₯.

Answer: βπ = π Therefore β5 equals π₯.

This is the same solution we found in example 13. You can eliminate the variable from either side of the

equation and still get the same answer.

Eliminate the Parentheses in an Equation

5(π₯ β 1) = 6π₯ + 4 Distribute to get rid of the parentheses.

5π₯ β 5 = 6π₯ + 4

5π₯ β 6π₯ β 5 = 6π₯ β 6π₯ + 4 Eliminate the x variable on the right side of the

equation.

β1π₯ β 5 = 0 + 4

β1π₯ β 5 = 4 Now get x alone on the left side of the equation.

Add 5 to each side of the equation. Donβt try to

eliminate the coefficient of x which is β1 before

adding 5 to each side of the equation.

β1π₯ β 5 + 5 = 4 + 5 Add 5 to each side of the equation.

β1π₯ + 0 = 9

β1π₯ = 9 Remember: β1π₯ = βπ₯

βπ₯ = 9 This means the opposite of π₯ equals 9.

Answer: π = βπ Therefore π₯ equals β9.

Page 6 of 10

7.
MAT 056: Basic Algebra I Learning Unit 4: Handout

Example 16

β6π₯ + 8 = 10 β 5π₯ + 14

β6π₯ + 8 = 10 + 14 β 5π₯ Combine like terms on the right side of the

equation.

β6π₯ + 8 = 24 β 5π₯ Eliminate the x variable on the right side of the

equation.

β6π₯ + 5π₯ + 8 = 24 β 5π₯ + 5π₯ Add 5π₯ to each side of the equation.

β1π₯ + 8 = 24 + 0

β1π₯ + 8 = 24 Now get x alone.

β1π₯ + 8 β 8 = 24 β 8 Subtract 8 from each side of the equation.

β1π₯ + 0 = 16

β1π₯ = 16 Remember: β1π₯ = βπ₯

βπ₯ = 16 This means the opposite of π₯ equals 16.

Answer: π = βππ Therefore π₯ equals β16.

Example 17

9π₯ β 3 β 5π₯ = β3π₯ + 8 + 6π₯

9π₯ β 5π₯ β 3 = β3π₯ + 8 + 6π₯ Combine like terms on the left side of the

equation.

4π₯ β 3 = β3π₯ + 6π₯ + 8 Combine like terms on the right side of the

equation.

4π₯ β 3 = 3π₯ + 8 Eliminate 3π₯ on the right side of the equation.

4π₯ β 3π₯ β 3 = 3π₯ β 3π₯ + 8 Subtract 3π₯ on each side of the equation.

1π₯ β 3 = 0 + 8

1π₯ β 3 = 8

1π₯ β 3 + 3 = 8 + 3 Add 3 to each side of the equation.

1π₯ + 0 = 11

1π₯ = 11

Answer: π = ππ

Page 7 of 10

Example 16

β6π₯ + 8 = 10 β 5π₯ + 14

β6π₯ + 8 = 10 + 14 β 5π₯ Combine like terms on the right side of the

equation.

β6π₯ + 8 = 24 β 5π₯ Eliminate the x variable on the right side of the

equation.

β6π₯ + 5π₯ + 8 = 24 β 5π₯ + 5π₯ Add 5π₯ to each side of the equation.

β1π₯ + 8 = 24 + 0

β1π₯ + 8 = 24 Now get x alone.

β1π₯ + 8 β 8 = 24 β 8 Subtract 8 from each side of the equation.

β1π₯ + 0 = 16

β1π₯ = 16 Remember: β1π₯ = βπ₯

βπ₯ = 16 This means the opposite of π₯ equals 16.

Answer: π = βππ Therefore π₯ equals β16.

Example 17

9π₯ β 3 β 5π₯ = β3π₯ + 8 + 6π₯

9π₯ β 5π₯ β 3 = β3π₯ + 8 + 6π₯ Combine like terms on the left side of the

equation.

4π₯ β 3 = β3π₯ + 6π₯ + 8 Combine like terms on the right side of the

equation.

4π₯ β 3 = 3π₯ + 8 Eliminate 3π₯ on the right side of the equation.

4π₯ β 3π₯ β 3 = 3π₯ β 3π₯ + 8 Subtract 3π₯ on each side of the equation.

1π₯ β 3 = 0 + 8

1π₯ β 3 = 8

1π₯ β 3 + 3 = 8 + 3 Add 3 to each side of the equation.

1π₯ + 0 = 11

1π₯ = 11

Answer: π = ππ

Page 7 of 10

8.
MAT 056: Basic Algebra I Learning Unit 4: Handout

Example 18

8(π + 2) β 10 β π = 6(π + 1) Distribute to eliminate the parentheses.

Multiply by the number outside the

parentheses.

8π + 16 β 10 β π = 6π + 6 Combine terms on the left side of the equation.

8π β π + 16 β 10 = 6π + 6 Combine the π variables.

8π β 1π + 16 β 10 = 6π + 6 Remember: βπ = β1π

7π + 16 β 10 = 6π + 6

7π + 6 = 6π + 6 Eliminate the variable on the right side of the

equation.

7π β 6π + 6 = 6π β 6π + 6 Subtract 6π on each side of the equation.

1π + 6 = 0 + 6

1π + 6 = 6 Now get π alone on the left side of the equation.

1π + 6 β 6 = 6 β 6 Subtract 6 on each side of the equation.

1π + 0 = 0

1π = 0

Answer: π = π

Page 8 of 10

Example 18

8(π + 2) β 10 β π = 6(π + 1) Distribute to eliminate the parentheses.

Multiply by the number outside the

parentheses.

8π + 16 β 10 β π = 6π + 6 Combine terms on the left side of the equation.

8π β π + 16 β 10 = 6π + 6 Combine the π variables.

8π β 1π + 16 β 10 = 6π + 6 Remember: βπ = β1π

7π + 16 β 10 = 6π + 6

7π + 6 = 6π + 6 Eliminate the variable on the right side of the

equation.

7π β 6π + 6 = 6π β 6π + 6 Subtract 6π on each side of the equation.

1π + 6 = 0 + 6

1π + 6 = 6 Now get π alone on the left side of the equation.

1π + 6 β 6 = 6 β 6 Subtract 6 on each side of the equation.

1π + 0 = 0

1π = 0

Answer: π = π

Page 8 of 10

9.
MAT 056: Basic Algebra I Learning Unit 4: Handout

Some equations have no solution and some have an infinite number of solutions. If the variable term is

eliminated on both sides of the equation and you end up with a false statement, there is no solution to

the equation. If you end up with a true statement, there are infinite solutions to the equation and we

say the equation has all real numbers for the solution.

Example 19

3π₯ + 7 + 1 = 2π₯ + 1π₯ + 4 + 4

3π₯ + 7 + 1 = 2π₯ + 1π₯ + 4 + 4 Combine like terms.

3π₯ + 8 = 2π₯ + 1π₯ + 8 Combine like terms.

3π₯ + 8 = 3π₯ + 8

3π₯ β 3π₯ + 8 = 3π₯ β 3π₯ + 8 Subtract 3π₯ on each side of the equation.

8=8 The variables are eliminated on each side of the

equation.

8 = 8 is a true statement so the solution is all real numbers.

Example 20

3(π₯ + 4) = 3π₯ + 10 + 1

3(π₯ + 4) = 3π₯ + 10 + 1 Distribute on the left side of the equation.

Combine like terms on the right side of the

equation.

3π₯ + 12 = 3π₯ + 11

3π₯ β 3π₯ + 12 = 3π₯ β 3π₯ + 11 Subtract 3π₯ from each side of the equation.

12 = 11 The variables are eliminated from each side of

the equation.

12 = 11 is a false statement. There is no solution.

Page 9 of 10

Some equations have no solution and some have an infinite number of solutions. If the variable term is

eliminated on both sides of the equation and you end up with a false statement, there is no solution to

the equation. If you end up with a true statement, there are infinite solutions to the equation and we

say the equation has all real numbers for the solution.

Example 19

3π₯ + 7 + 1 = 2π₯ + 1π₯ + 4 + 4

3π₯ + 7 + 1 = 2π₯ + 1π₯ + 4 + 4 Combine like terms.

3π₯ + 8 = 2π₯ + 1π₯ + 8 Combine like terms.

3π₯ + 8 = 3π₯ + 8

3π₯ β 3π₯ + 8 = 3π₯ β 3π₯ + 8 Subtract 3π₯ on each side of the equation.

8=8 The variables are eliminated on each side of the

equation.

8 = 8 is a true statement so the solution is all real numbers.

Example 20

3(π₯ + 4) = 3π₯ + 10 + 1

3(π₯ + 4) = 3π₯ + 10 + 1 Distribute on the left side of the equation.

Combine like terms on the right side of the

equation.

3π₯ + 12 = 3π₯ + 11

3π₯ β 3π₯ + 12 = 3π₯ β 3π₯ + 11 Subtract 3π₯ from each side of the equation.

12 = 11 The variables are eliminated from each side of

the equation.

12 = 11 is a false statement. There is no solution.

Page 9 of 10

10.
MAT 056: Basic Algebra I Learning Unit 4: Handout

Example 21

5π₯ = 5π₯

5π₯ β 5π₯ = 5π₯ β 5π₯ Subtract 5π₯ from each side of the equation.

0=0 The variables are eliminated from each side of

the equation.

0 = 0 is a true statement so the solution is all real numbers.

Example 22

5 + 2(π₯ + 2) = 2π₯ + 9

5 + 2(π₯ + 2) = 2π₯ + 9 Remember the order of operations.

Multiply (distribute) before adding.

5 + 2π₯ + 4 = 2π₯ + 9 Collect like terms.

2π₯ + 9 = 2π₯ + 9

2π₯ β 2π₯ + 9 = 2π₯ β 2π₯ + 9 Subtract 2π₯ from each side of the equation.

9=9 The variables are eliminated from each side of

the equation.

9 = 9 is a true statement so the solution is all real numbers.

Β© Brenda Moore and Indian Hills Community College

Page 10 of 10

Example 21

5π₯ = 5π₯

5π₯ β 5π₯ = 5π₯ β 5π₯ Subtract 5π₯ from each side of the equation.

0=0 The variables are eliminated from each side of

the equation.

0 = 0 is a true statement so the solution is all real numbers.

Example 22

5 + 2(π₯ + 2) = 2π₯ + 9

5 + 2(π₯ + 2) = 2π₯ + 9 Remember the order of operations.

Multiply (distribute) before adding.

5 + 2π₯ + 4 = 2π₯ + 9 Collect like terms.

2π₯ + 9 = 2π₯ + 9

2π₯ β 2π₯ + 9 = 2π₯ β 2π₯ + 9 Subtract 2π₯ from each side of the equation.

9=9 The variables are eliminated from each side of

the equation.

9 = 9 is a true statement so the solution is all real numbers.

Β© Brenda Moore and Indian Hills Community College

Page 10 of 10