1. Combining Like Terms

Combining Like Terms

Combining like terms is one of the most important skills used to simplify algebraic expressions.

Consider:

3x + 5x

Both terms contain the same variable, x, so they are like terms.

We can combine them:

3x + 5x = 8x

This works because we are really saying:

3 groups of x + 5 groups of x = 8 groups of x

Just as:

3 apples + 5 apples = 8 apples

 

 

 

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What Is a Term?

A term is a number, variable, or product of numbers and variables within an expression.

Terms are separated by addition or subtraction.

For example:

4x + 7 − 2y

contains three terms:

4x

7

−2y

Notice that the negative sign belongs to the term:

−2y

Parts of a Term

Consider:

6x

The number 6 is the:

coefficient

The letter x is the:

variable

In:

−4y

the coefficient is:

−4

and the variable is:

y

Constants

A constant is a term containing no variable.

Examples:

5

−3

12

1.7

Constants are like terms with other constants.

For example:

7 + 4 = 11

and:

12 − 5 = 7

What Are Like Terms?

Like terms contain exactly the same variable part.

Examples:

3x and 8x

are like terms.

5y and −2y

are like terms.

4a² and 9a²

are like terms.

3xy and −7xy

are also like terms.

Only their coefficients are different.

Identifying Like Terms

Consider:

4x + 3 + 7x − 5

The like terms are:

4x and 7x

and:

3 and −5

Therefore:

4x + 3 + 7x − 5

can be reorganized as:

4x + 7x + 3 − 5

Then:

11x − 2

Unlike Terms

Terms are unlike terms if their variable parts are different.

Examples:

3x and 3y

are unlike.

4x and 4x²

are unlike.

2xy and 2x

are unlike.

5a² and 5a

are unlike.

These terms cannot be combined by simply adding or subtracting their coefficients.

Why Can Only Like Terms Be Combined?

Consider:

3 apples + 4 apples

These can be combined:

7 apples

But:

3 apples + 4 oranges

cannot become:

7 apples

or:

7 oranges

They are different types of objects.

Algebra works in the same way.

3x + 4x = 7x

but:

3x + 4y

cannot be simplified further because x and y represent different quantities.

 

 

 

10

Combining Coefficients

When combining like terms, we combine the coefficients but keep the variable part unchanged.

For example:

4x + 7x

Combine:

4 + 7 = 11

Keep x:

11x

Therefore:

4x + 7x = 11x

Why the Variable Does Not Change

Remember:

4x = x + x + x + x

and:

3x = x + x + x

So:

4x + 3x

means:

x + x + x + x + x + x + x

which is:

7x

The x does not become x².

Subtracting Like Terms

Like terms can also be subtracted.

Example:

9x − 4x

Subtract the coefficients:

9 − 4 = 5

Therefore:

9x − 4x = 5x

Negative Coefficients

Consider:

3x − 8x

Think of this as:

3x + (−8x)

Combine the coefficients:

3 + (−8) = −5

Therefore:

3x − 8x = −5x

Combining Two Negative Terms

Consider:

−4y − 6y

Combine:

−4 + (−6) = −10

Therefore:

−4y − 6y = −10y

When Terms Cancel

Consider:

7x − 7x

Combine coefficients:

7 − 7 = 0

So:

0x = 0

Therefore:

7x − 7x = 0

The variable terms completely cancel.

The Invisible Coefficient

A variable written without a number has a coefficient of:

1

So:

x = 1x

and:

−x = −1x

This is useful when combining terms.

Example:

x + 5x

becomes:

1x + 5x = 6x

Worked Example

Simplify:

5x + 3x

The terms are like because both contain x.

Add coefficients:

5 + 3 = 8

Therefore:

5x + 3x = 8x

Worked Example with Subtraction

Simplify:

12a − 7a

Combine coefficients:

12 − 7 = 5

Therefore:

12a − 7a = 5a

Worked Example with Constants

Simplify:

4x + 7 + 3x + 2

Group like terms:

4x + 3x + 7 + 2

Combine:

7x + 9

Therefore:

4x + 7 + 3x + 2 = 7x + 9

Rearranging Terms

The order of terms can be changed when addition is involved.

For example:

3x + 7 + 5x + 2

can be rearranged as:

3x + 5x + 7 + 2

Then:

8x + 9

This makes the like terms easier to see.

Keep the Sign with the Term

Consider:

8x − 3 + 2x − 7

The terms are:

8x

−3

2x

−7

Group them:

8x + 2x − 3 − 7

Combine:

10x − 10

Keeping the sign attached to each term helps prevent errors.

Expressions with Two Variables

Simplify:

4x + 3y + 2x + 5y

Group x terms:

4x + 2x

Group y terms:

3y + 5y

Combine:

6x + 8y

The x and y terms cannot be combined with each other.

Worked Example with Two Variables

Simplify:

7a + 2b − 3a + 5b

Group:

7a − 3a + 2b + 5b

Combine:

4a + 7b

Three Types of Terms

Consider:

3x + 4y + 5 + 2x − y − 8

There are three groups of like terms.

x terms:

3x + 2x = 5x

y terms:

4y − y = 3y

constants:

5 − 8 = −3

Therefore:

5x + 3y − 3

Powers Matter

Terms are only like terms when the variables and their exponents match.

For example:

3x² + 5x² = 8x²

But:

3x² + 5x

cannot be combined.

Even though both contain x, their variable parts are different:

x² ≠ x

 

 

 

7

Example with Powers

Simplify:

4x² + 3x + 5x² − x

Group:

4x² + 5x² + 3x − x

Combine:

9x² + 2x

The x² terms combine with each other.

The x terms combine with each other.

But x² and x do not combine.

Multiple Variables

The complete variable part must match.

For example:

3xy + 5xy = 8xy

But:

3xy + 5x

cannot be combined.

Similarly:

4x²y + 7x²y = 11x²y

but:

4x²y + 7xy²

cannot be combined because the exponents are different.

A Systematic Method

When simplifying an expression:

Identify → Group → Combine → Rewrite

Identify

Find terms with identical variable parts.

Group

Place like terms together.

Combine

Add or subtract their coefficients.

Rewrite

Write the simplified expression clearly.

Worked Example Using the Method

Simplify:

6x + 4 − 2x + 9

Identify

Like variable terms:

6x and −2x

Like constants:

4 and 9

Group

6x − 2x + 4 + 9

Combine

4x + 13

Therefore:

6x + 4 − 2x + 9 = 4x + 13

Expressions with Parentheses

Sometimes we need to use the distributive property before combining like terms.

Consider:

3(x + 4) + 2x

Distribute:

3x + 12 + 2x

Now combine:

3x + 2x = 5x

Therefore:

5x + 12

Distributive Property

The distributive property states:

a(b + c) = ab + ac

Example:

4(x + 3)

becomes:

4x + 12

Once the parentheses are removed, we can identify and combine like terms.

Worked Example with Distribution

Simplify:

2(x + 5) + 3x + 4

Distribute:

2x + 10 + 3x + 4

Group:

2x + 3x + 10 + 4

Combine:

5x + 14

Subtracting Parentheses

Be careful when a negative number is outside parentheses.

Simplify:

5x − 2(x + 3)

Distribute −2:

5x − 2x − 6

Combine:

3x − 6

Negative Distribution

Simplify:

−3(x − 4) + 2x

Distribute −3:

−3x + 12 + 2x

Combine:

−x + 12

Remember:

−3 × −4 = +12

Equivalent Expressions

Two expressions are equivalent if they have the same value for every permissible value of the variables.

For example:

3x + 5x

and:

8x

are equivalent.

We write:

3x + 5x = 8x

because both expressions always produce the same result.

Checking Equivalent Expressions

Consider:

4x + 2x + 3

and:

6x + 3

Test x = 5.

First expression:

4(5) + 2(5) + 3

= 20 + 10 + 3

= 33

Second expression:

6(5) + 3

= 30 + 3

= 33

They match.

Testing values can provide evidence that expressions are equivalent, while correct algebra shows why they are equivalent for all permissible values.

Equivalent Does Not Mean Identical Looking

Consider:

2(x + 3)

and:

2x + 6

They look different, but distributing gives:

2(x + 3) = 2x + 6

Therefore they are equivalent expressions.

Similarly:

5x + 3x − 2

and:

8x − 2

are equivalent.

Expressions Versus Equations

An expression does not contain an equals sign as part of the expression itself.

Example:

3x + 5x + 2

We simplify it to:

8x + 2

An equation states that two expressions are equal.

Example:

3x + 5x + 2 = 18

Simplify:

8x + 2 = 18

Combining like terms is often one of the first steps in solving equations.

Combining Like Terms in Equations

Solve:

3x + 5x + 4 = 28

First combine:

8x + 4 = 28

Subtract 4:

8x = 24

Divide by 8:

x = 3

Combining like terms makes the equation easier to solve.

Like Terms on Each Side

Consider:

3x + 2x + 4 = x + 16

First simplify the left:

5x + 4 = x + 16

Subtract x:

4x + 4 = 16

Subtract 4:

4x = 12

Therefore:

x = 3

Combining Like Terms in Geometry

Suppose the sides of a triangle are:

2x + 3

4x

x + 5

The perimeter can be written:

2x + 3 + 4x + x + 5

Combine x terms:

2x + 4x + x = 7x

Combine constants:

3 + 5 = 8

Therefore the perimeter is:

7x + 8

This expression is simpler and easier to use.

Combining Like Terms in Real Life

Suppose a student buys:

• 3 notebooks costing x dollars each
• 4 more notebooks costing x dollars each
• a pen costing $2

The total cost is:

3x + 4x + 2

Combine like terms:

7x + 2

This simplified expression represents the same total cost.

Algebra Tiles

Algebra tiles can help visualize why like terms combine.

 

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9

Different tiles represent different types of terms.

For example:

• long rectangular tiles can represent x
• large square tiles can represent x²
• small squares can represent 1

Three x tiles and two x tiles can combine to make:

5x

But an x tile and an x² tile represent different quantities, so they cannot be combined into a single term.

Why 3x + 4 Is Not 7x

This is one of the most common algebra mistakes.

Consider:

3x + 4

The first term means:

3 × x

The second term is simply:

4

They are different types of terms.

If x = 2:

Original expression:

3(2) + 4 = 10

If we incorrectly wrote 7x:

7(2) = 14

Since:

10 ≠ 14

the expressions are not equivalent.

Therefore:

3x + 4 cannot be simplified to 7x.

Why 3x + 4x Is Not 7x²

Another common mistake is changing the exponent.

Correct:

3x + 4x = 7x

Not:

7x²

Remember that adding terms does not multiply their variables.

For x = 2:

3(2) + 4(2) = 14

and:

7(2) = 14

But:

7(2²) = 28

So 7x² is not equivalent.

Addition and Multiplication Behave Differently

Compare:

x + x = 2x

but:

x × x = x²

Similarly:

3x + 4x = 7x

while:

(3x)(4x) = 12x²

Combining like terms applies to addition and subtraction, not multiplication in the same way.

Simplifying a Longer Expression

Simplify:

5x² + 3x − 7 + 2x² − 8x + 10

Group x² terms:

5x² + 2x² = 7x²

Group x terms:

3x − 8x = −5x

Group constants:

−7 + 10 = 3

Therefore:

7x² − 5x + 3

Worked Example with Several Variables

Simplify:

4a + 3b − 2a + 7 − 5b + 4

Group a terms:

4a − 2a = 2a

Group b terms:

3b − 5b = −2b

Group constants:

7 + 4 = 11

Therefore:

2a − 2b + 11

Worked Example with Powers

Simplify:

6x² − 2x + 4 + 3x² + 5x − 9

x² terms:

6x² + 3x² = 9x²

x terms:

−2x + 5x = 3x

constants:

4 − 9 = −5

Therefore:

9x² + 3x − 5

Worked Example with Distribution

Simplify:

3(2x + 5) − 4x + 7

Distribute:

6x + 15 − 4x + 7

Combine x terms:

6x − 4x = 2x

Combine constants:

15 + 7 = 22

Therefore:

2x + 22

Worked Example with Two Sets of Parentheses

Simplify:

2(x + 3) + 4(x − 1)

Distribute:

2x + 6 + 4x − 4

Combine:

6x + 2

Therefore:

2(x + 3) + 4(x − 1) = 6x + 2

Simplifying Does Not Mean Solving

If asked to simplify:

3x + 5x − 4

the answer is:

8x − 4

We cannot find x because there is no equation giving enough information to determine x.

Simplifying means rewriting an expression in a simpler equivalent form.

Solving means finding the value or values of a variable that make an equation true.

A Useful Organization Strategy

For longer expressions, identify categories.

For:

3x² + 4x − 7 + 2x² − x + 9

create groups:

x² terms: 3x² + 2x²

x terms: 4x − x

constants: −7 + 9

Then combine:

5x² + 3x + 2

This reduces sign and classification mistakes.

Common Mistakes

Mistake: Combining unlike variables.

Incorrect:

3x + 4y = 7xy

Correct:

3x + 4y

cannot be simplified further.

Mistake: Combining variables and constants.

Incorrect:

5x + 3 = 8x

Correct:

5x + 3

Mistake: Combining different powers.

Incorrect:

2x² + 3x = 5x³

Correct:

2x² + 3x

Mistake: Changing the exponent when adding.

Incorrect:

4x + 2x = 6x²

Correct:

4x + 2x = 6x

Mistake: Losing negative signs.

In:

7x − 3x − 5

the terms are:

7x, −3x, −5

Mistake: Forgetting an invisible coefficient.

x + 4x

means:

1x + 4x = 5x

Mistake: Distributing to only one term.

Incorrect:

3(x + 4) = 3x + 4

Correct:

3(x + 4) = 3x + 12

Mistake: Confusing addition and multiplication.

Remember:

x + x = 2x

but:

x × x = x²

Key Terms

Term — number, variable, or product of numbers and variables within an expression.

Like terms — terms with exactly the same variable part, including exponents.

Unlike terms — terms with different variable parts.

Coefficient — numerical factor multiplying a variable.

Variable — letter or symbol representing a number or quantity.

Constant — term containing no variable.

Expression — combination of numbers, variables, and mathematical operations.

Simplify — rewrite an expression in a simpler equivalent form.

Combine like terms — add or subtract the coefficients of terms with identical variable parts.

Equivalent expressions — expressions having the same value for every permissible value of their variables.

Exponent — number indicating how many times a quantity is used as a factor.

Distributive property — property represented by a(b + c) = ab + ac.

Algebra tiles — visual models representing constants, variables, and powers of variables.

Key Takeaways

• Expressions are made of terms.
• Addition and subtraction signs separate terms.
• The sign immediately before a term belongs to that term.
• Like terms have exactly the same variable part.
• The variables and their exponents must match.
• Constants are like terms with other constants.
• Coefficients may be different while terms are still like.
• Only like terms can be combined through addition or subtraction.
• Combining like terms means combining their coefficients.
• The variable part remains unchanged.
• 3x + 5x = 8x.
• 3x + 5 cannot be combined.
• 3x + 5y cannot be combined.
• 3x² + 5x² = 8x².
• 3x² + 5x cannot be combined.
• x has an implied coefficient of 1.
• −x has an implied coefficient of −1.
• Opposite like terms can cancel completely.
• The distributive property may need to be used before combining like terms.
• Simplifying does not change the value represented by an expression.
• Equivalent expressions can look different while representing the same quantity.
• Combining like terms is essential for simplifying expressions and solving equations.

Check Your Understanding

1. What is a term?
2. What makes two terms "like terms"?
3. Identify the coefficient in 7x.
4. Identify the terms in 4x − 3y + 8.
5. Which terms are like in 3x + 5y + 7x − 2y?
6. Simplify: 4x + 7x.
7. Simplify: 12y − 5y.
8. Simplify: 3a + 7 + 5a + 4.
9. Simplify: 8x − 3x + 5 − 9.
10. Simplify: 4x + 3y + 2x − 5y.
11. Simplify: 5x² + 3x² − 2x.
12. Explain why 4x² + 3x cannot be combined into one term.
13. Simplify: 3(x + 4) + 2x.
14. Simplify: 5(x − 2) − 3x + 7.
15. Simplify: −2(x − 4) + 5x.
16. Simplify: 4x² + 3x − 7 + 2x² − 5x + 10.
17. Are 3x + 5x and 8x equivalent? Explain why.
18. Explain why 3x + 4 cannot be simplified to 7x.
19. Explain the difference between x + x and x × x.
20. Describe a systematic method you can use to identify and combine like terms in a long algebraic expression.

2. The Distributive Property

The Distributive Property

The distributive property allows us to multiply a number or algebraic term by every term inside parentheses.

The basic rule is:

a(b + c) = ab + ac

For subtraction:

a(b − c) = ab − ac

For example:

3(x + 4)

means that 3 multiplies both terms inside the parentheses:

3(x) + 3(4)

Therefore:

3(x + 4) = 3x + 12

The distributive property is essential for expanding expressions, simplifying algebra, solving equations, and factoring.

 

 

 

11

Why Distribution Works

Consider:

3(x + 2)

This means three groups of:

x + 2

So we can write:

(x + 2) + (x + 2) + (x + 2)

Combine like terms:

x + x + x + 2 + 2 + 2

which gives:

3x + 6

Therefore:

3(x + 2) = 3x + 6

Distribution is not simply a rule to memorize. It follows directly from multiplication.

An Area Model

The distributive property can also be understood using the area of a rectangle.

Suppose a rectangle has:

height = 4

and:

width = x + 3

Its area is:

A = 4(x + 3)

We could divide the rectangle into two smaller rectangles.

Their areas are:

4x

and:

4(3) = 12

So the total area is:

4x + 12

Therefore:

4(x + 3) = 4x + 12

 

 

 

9

Expanding Expressions

To expand an expression means to remove parentheses by applying multiplication.

For example:

5(x + 2)

Distribute 5:

5(x) + 5(2)

Simplify:

5x + 10

Therefore:

5(x + 2) = 5x + 10

A Simple Distribution Process

When you see:

a(b + c)

use:

Multiply → Multiply → Simplify

Multiply a by the first term:

ab

Multiply a by the second term:

ac

Then write:

ab + ac

Distribution with Addition

Expand:

6(x + 3)

Multiply 6 by x:

6x

Multiply 6 by 3:

18

Therefore:

6(x + 3) = 6x + 18

Distribution with Subtraction

Expand:

4(x − 5)

Multiply 4 by x:

4x

Multiply 4 by −5:

−20

Therefore:

4(x − 5) = 4x − 20

The Sign Is Important

Consider:

7(x − 2)

The second term is:

−2

So:

7(−2) = −14

Therefore:

7(x − 2) = 7x − 14

Always keep the sign attached to the term.

Distributing a Negative Number

Negative numbers require extra care.

Expand:

−3(x + 4)

Multiply:

−3(x) = −3x

and:

−3(4) = −12

Therefore:

−3(x + 4) = −3x − 12

Negative Outside and Subtraction Inside

Expand:

−2(x − 5)

Multiply:

−2(x) = −2x

and:

−2(−5) = +10

Therefore:

−2(x − 5) = −2x + 10

Remember:

negative × negative = positive

The Invisible −1

An expression such as:

−(x + 6)

really means:

−1(x + 6)

Distribute −1:

−x − 6

Therefore:

−(x + 6) = −x − 6

Another Example

Expand:

−(3x − 7)

Multiply every term by −1:

−3x + 7

Therefore:

−(3x − 7) = −3x + 7

This is why removing parentheses preceded by a minus sign changes the signs of all terms inside.

Distributing Variables

The factor outside the parentheses does not have to be a number.

Expand:

x(x + 4)

Multiply x by each term:

x(x) + x(4)

Therefore:

x² + 4x

So:

x(x + 4) = x² + 4x

Distributing Algebraic Terms

Expand:

3x(x + 2)

Multiply:

3x(x) = 3x²

and:

3x(2) = 6x

Therefore:

3x(x + 2) = 3x² + 6x

Powers During Multiplication

Remember:

x × x = x²

and:

x² × x = x³

For example:

2x²(3x + 4)

Distribute:

2x²(3x) + 2x²(4)

= 6x³ + 8x²

More Than Two Terms

Distribution works with any number of terms inside parentheses.

For example:

3(x + 2y − 4)

Multiply 3 by every term:

3x + 6y − 12

Therefore:

3(x + 2y − 4) = 3x + 6y − 12

Worked Example

Expand:

5(2x + 3y − 4)

Multiply 5 by each term:

10x + 15y − 20

Therefore:

5(2x + 3y − 4) = 10x + 15y − 20

Distribution Followed by Combining Like Terms

Often distribution is only the first step.

Simplify:

3(x + 4) + 2x

Distribute:

3x + 12 + 2x

Combine like terms:

5x + 12

Therefore:

3(x + 4) + 2x = 5x + 12

Worked Example

Simplify:

4(x + 2) + 3x + 5

Distribute:

4x + 8 + 3x + 5

Combine variable terms:

4x + 3x = 7x

Combine constants:

8 + 5 = 13

Therefore:

7x + 13

Subtraction and Distribution

Simplify:

6x − 2(x + 5)

The factor being distributed is:

−2

So:

6x − 2x − 10

Combine:

4x − 10

Therefore:

6x − 2(x + 5) = 4x − 10

Worked Example with Negative Distribution

Simplify:

5x − 3(x − 4)

Distribute −3:

5x − 3x + 12

Combine:

2x + 12

Distribution on Both Sides

Consider:

2(x + 3) + 4(x − 1)

Distribute:

2x + 6 + 4x − 4

Combine:

6x + 2

Therefore:

2(x + 3) + 4(x − 1) = 6x + 2

Equivalent Expressions

The distributive property creates equivalent expressions.

For example:

3(x + 5)

and:

3x + 15

look different, but they have the same value for every x.

If x = 4:

First expression:

3(4 + 5) = 27

Second expression:

3(4) + 15 = 27

Therefore they are equivalent.

Checking Distribution by Substitution

Suppose we claim:

4(x − 3) = 4x − 12

Test x = 5.

Left:

4(5 − 3) = 8

Right:

4(5) − 12 = 8

The values agree.

Correct algebra proves the expressions are equivalent for all permissible x; substitution is useful for checking our work.

Distribution and Order of Operations

Consider:

2(3 + 5)

Using parentheses first:

2(8) = 16

Using distribution:

2(3) + 2(5)

6 + 10 = 16

Both methods give the same result.

This demonstrates why the distributive property is consistent with arithmetic.

When Is Distribution Useful?

Distribution is especially useful when:

• removing parentheses
• simplifying expressions
• solving equations
• multiplying algebraic terms
• working with area models
• factoring expressions
• evaluating expressions efficiently
• rearranging scientific formulas

Distribution in Equations

Solve:

3(x + 4) = 24

Distribute:

3x + 12 = 24

Subtract 12:

3x = 12

Divide by 3:

x = 4

Check:

3(4 + 4) = 24

3(8) = 24

Correct.

Distribution Before Solving

Solve:

2(x − 5) + 6 = 18

Distribute:

2x − 10 + 6 = 18

Combine:

2x − 4 = 18

Add 4:

2x = 22

Divide by 2:

x = 11

Check:

2(11 − 5) + 6 = 18

12 + 6 = 18

Correct.

Variables on Both Sides

Solve:

3(x + 2) = 2x + 10

Distribute:

3x + 6 = 2x + 10

Subtract 2x:

x + 6 = 10

Subtract 6:

x = 4

Distribution allows us to remove the parentheses before collecting variable terms.

Parentheses on Both Sides

Solve:

2(x + 5) = 4(x − 1)

Distribute:

2x + 10 = 4x − 4

Subtract 2x:

10 = 2x − 4

Add 4:

14 = 2x

Divide by 2:

x = 7

Check:

LHS:

2(7 + 5) = 24

RHS:

4(7 − 1) = 24

Correct.

Distribution Can Sometimes Be Avoided

Distribution is useful, but it is not always the fastest method.

Consider:

5(x + 3) = 40

We could distribute:

5x + 15 = 40

and solve.

But we could also divide both sides by 5 first:

x + 3 = 8

Then:

x = 5

Both approaches are correct.

Efficient algebra involves choosing useful operations.

Another Efficient Example

Solve:

4(2x − 1) = 28

Instead of immediately distributing, divide both sides by 4:

2x − 1 = 7

Add 1:

2x = 8

Divide by 2:

x = 4

Sometimes removing a common multiplier first makes an equation simpler.

Factoring: Reversing Distribution

The distributive property also works in reverse.

Expansion:

3(x + 4) = 3x + 12

Factoring reverses this:

3x + 12 = 3(x + 4)

To factor, identify a factor shared by all terms.

 

 

 

6

Factoring a Common Number

Factor:

4x + 12

Both terms are divisible by 4.

Factor out 4:

4(x + 3)

Check by distributing:

4(x + 3) = 4x + 12

Correct.

Greatest Common Factor

The greatest common factor (GCF) is the largest factor shared by all terms.

Consider:

6x + 18

The greatest common factor is:

6

So:

6x + 18 = 6(x + 3)

Worked Factoring Example

Factor:

8x + 20

The GCF of 8 and 20 is:

4

Therefore:

8x + 20 = 4(2x + 5)

Check:

4(2x + 5)

= 8x + 20

Correct.

Factoring a Variable

Factor:

x² + 5x

Both terms contain x.

Factor out x:

x(x + 5)

Distributing confirms:

x(x + 5) = x² + 5x

Factoring a Number and Variable

Factor:

6x² + 9x

Both terms contain:

3x

Factor it out:

3x(2x + 3)

Check:

3x(2x) + 3x(3)

= 6x² + 9x

Factoring with Negative Terms

Factor:

−4x − 12

We can factor out −4:

−4(x + 3)

Check:

−4(x + 3)

= −4x − 12

Factoring out a negative can sometimes make the expression inside parentheses easier to work with.

Expanding and Factoring Are Opposites

Expansion:

5(x + 2) → 5x + 10

Factoring:

5x + 10 → 5(x + 2)

These processes are closely related.

You can often check one by performing the other.

Distribution in Mental Mathematics

The distributive property is useful even without variables.

Calculate:

7 × 23

Rewrite:

7(20 + 3)

Distribute:

7(20) + 7(3)

= 140 + 21

= 161

Another Mental Math Example

Calculate:

6 × 98

Write:

6(100 − 2)

Distribute:

600 − 12

= 588

The same property used in algebra also helps with arithmetic.

Distribution in Geometry

Suppose a rectangle has:

length = x + 5

and:

width = 3

Area:

A = 3(x + 5)

Distribute:

A = 3x + 15

Both expressions represent the same area.

 

 

 

8

Distribution in Perimeter Problems

Suppose a rectangle has:

length = x + 4

width = x + 1

Perimeter:

P = 2(x + 4) + 2(x + 1)

Distribute:

P = 2x + 8 + 2x + 2

Combine:

P = 4x + 10

The simplified expression represents the rectangle's perimeter.

Distribution in Cost Problems

Suppose three students each buy:

• one ticket costing x dollars
• one snack costing $4

The cost for one student is:

x + 4

For three students:

3(x + 4)

Distribute:

3x + 12

This represents:

• 3 tickets
• $12 total for snacks

Distribution in Science

Suppose a quantity is described by:

E = m(cΔT + L)

The mass m multiplies both terms.

Distribute:

E = mcΔT + mL

Both forms are mathematically equivalent.

The most useful form depends on what we are trying to calculate or interpret.

A Longer Simplification

Simplify:

4(2x + 3) − 2(x − 5)

Distribute:

8x + 12 − 2x + 10

Combine like terms:

6x + 22

Therefore:

4(2x + 3) − 2(x − 5) = 6x + 22

A More Challenging Example

Simplify:

−3(2x − 4) + 5(x + 2)

Distribute:

−6x + 12 + 5x + 10

Combine:

−x + 22

Therefore:

−3(2x − 4) + 5(x + 2) = −x + 22

Solving a More Complex Equation

Solve:

3(2x + 4) − 5 = 4x + 13

Distribute:

6x + 12 − 5 = 4x + 13

Combine:

6x + 7 = 4x + 13

Subtract 4x:

2x + 7 = 13

Subtract 7:

2x = 6

Divide by 2:

x = 3

Check using the original equation:

LHS:

3(2(3) + 4) − 5

= 3(10) − 5

= 25

RHS:

4(3) + 13

= 25

Correct.

Recognizing Equivalent Expressions

Which expression is equivalent to:

5(2x − 3)?

Distribute:

10x − 15

Therefore:

5(2x − 3)

and:

10x − 15

are equivalent.

This type of reasoning is useful when comparing different algebraic forms.

Checking by Factoring

Suppose we have:

12x + 18

Factor the GCF:

6(2x + 3)

Now distribute:

6(2x + 3)

= 12x + 18

This confirms the factorization.

Common Mistakes

Mistake: Multiplying only the first term.

Incorrect:

3(x + 4) = 3x + 4

Correct:

3(x + 4) = 3x + 12

The outside factor multiplies every term inside.

Mistake: Forgetting negative signs.

Incorrect:

−2(x + 5) = −2x + 10

Correct:

−2(x + 5) = −2x − 10

Mistake: Mishandling two negatives.

Incorrect:

−3(x − 4) = −3x − 12

Correct:

−3(x − 4) = −3x + 12

Mistake: Combining unlike terms after distribution.

3x + 12

cannot become:

15x

Mistake: Changing exponents during addition.

2x + 5x = 7x

not:

7x²

Mistake: Forgetting that −(x + 3) means −1(x + 3).

Correct:

−(x + 3) = −x − 3

Mistake: Thinking distribution is always required.

Sometimes dividing out a common factor before distributing makes an equation easier.

Mistake: Confusing factoring with combining like terms.

6x + 12

cannot be combined into 18x.

It can instead be factored:

6(x + 2)

Key Terms

Distributive property — property stating that multiplication can be distributed across addition or subtraction: a(b + c) = ab + ac.

Distribute — multiply an outside factor by every term inside parentheses.

Expand — rewrite an expression without parentheses by carrying out multiplication.

Factor — number or algebraic expression multiplied by another quantity.

Factoring — rewriting an expression as a product of factors.

Greatest common factor (GCF) — largest factor shared by all terms.

Equivalent expressions — expressions that have the same value for every permissible value of their variables.

Term — number, variable, or product of numbers and variables within an expression.

Coefficient — numerical factor multiplying a variable.

Like terms — terms with identical variable parts.

Expression — combination of numbers, variables, and operations.

Area model — visual representation that uses rectangle areas to demonstrate multiplication and distribution.

Key Takeaways

• The distributive property connects multiplication with addition and subtraction.
• The basic rule is a(b + c) = ab + ac.
• For subtraction, a(b − c) = ab − ac.
• The factor outside parentheses must multiply every term inside.
• Distribution removes parentheses and expands expressions.
• Positive and negative signs must be handled carefully.
• A negative outside parentheses affects every term inside.
• The expression −(x + 4) means −1(x + 4).
• Variables and algebraic terms can also be distributed.
• Distribution may create like terms that can then be combined.
• Distribution is often useful before solving equations.
• Distribution can help simplify equations containing variables on both sides.
• Sometimes dividing out a common factor before distributing is more efficient.
• Factoring is the reverse process of distribution.
• A common numerical factor can be removed from all terms.
• Variables can also be common factors.
• The GCF is often the most useful factor to remove.
• Expanding and factoring produce different-looking but equivalent expressions.
• The distributive property can be represented using rectangle area models.
• Distribution is useful in geometry, mental arithmetic, science, and real-world cost problems.
• Correct distribution is one of the foundations for later work with polynomials, factoring, and quadratic equations.

Check Your Understanding

1. State the distributive property using variables.
2. Explain in your own words what it means to distribute.
3. Expand: 3(x + 5).
4. Expand: 6(x − 4).
5. Expand: −2(x + 7).
6. Expand: −4(x − 3).
7. Expand: 5(2x + 3).
8. Expand: 3(x + 2y − 4).
9. Simplify: 2(x + 4) + 3x.
10. Simplify: 5(x − 2) + 4x + 3.
11. Simplify: 7x − 2(x + 5).
12. Simplify: −3(x − 4) + 5x.
13. Simplify: 4(2x + 3) − 2(x − 5).
14. Factor completely: 6x + 18.
15. Factor completely: 8x + 20.
16. Factor: x² + 7x.
17. Factor: 6x² + 9x.
18. Solve: 3(x + 5) = 27.
19. A rectangle has width 4 and length x + 6. Write its area in factored form and expanded form.
20. Explain why expanding 4(x + 3) and factoring 4x + 12 are reverse processes.

3. Substitution and Evaluation

Substitution and Evaluation

Substitution means replacing a variable with a known value.

For example, consider:

3x + 4

If:

x = 5

substitute 5 for x:

3(5) + 4

Then evaluate:

15 + 4 = 19

Therefore, when x = 5:

3x + 4 = 19

Substitution connects algebraic expressions with actual numerical values and is used throughout mathematics, science, engineering, finance, and everyday problem-solving.

 

 

 

8

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What Does It Mean to Evaluate?

To evaluate an expression means to calculate its numerical value for particular values of its variables.

Consider:

2x + 7

This expression does not have one fixed numerical value because its value depends on x.

If:

x = 3

then:

2(3) + 7 = 13

If:

x = 10

then:

2(10) + 7 = 27

Changing the value of the variable changes the value of the expression.

---

Substitution Versus Evaluation

These ideas are closely related but slightly different.

Substitution means replacing a variable with a known value.

Evaluation means completing the calculation after substitution.

For:

4x − 3, when x = 6:

Substitution:

4(6) − 3

Evaluation:

24 − 3 = 21

---

A Reliable Method

Use:

Identify → Substitute → Calculate → Interpret

Identify

Determine the value of each variable.

Substitute

Replace every occurrence of the variable with its value.

Calculate

Use the correct order of operations.

Interpret

Explain what the result means if the expression represents a real situation.

---

Use Parentheses When Substituting

A useful habit is to place substituted values inside parentheses.

For:

5x + 2

when:

x = 4

write:

5(4) + 2

rather than:

5 × 4 + 2

Both are mathematically correct, but parentheses make the substitution especially clear.

This becomes even more important with negative numbers.

---

Evaluating a Simple Expression

Evaluate:

6x + 5

when:

x = 3

Substitute:

6(3) + 5

Multiply:

18 + 5

Therefore:

23

---

Evaluating with Subtraction

Evaluate:

8x − 7

when:

x = 4

Substitute:

8(4) − 7

Calculate:

32 − 7 = 25

Therefore:

8x − 7 = 25

when x = 4.

---

Evaluating with Division

Evaluate:

x/5 + 6

when:

x = 20

Substitute:

20/5 + 6

Calculate division first:

4 + 6

Therefore:

10

---

Negative Values

Negative values require careful substitution.

Evaluate:

3x + 8

when:

x = −4

Substitute using parentheses:

3(−4) + 8

Multiply:

−12 + 8

Therefore:

−4

---

Why Parentheses Matter with Negative Numbers

Consider:

x²

when:

x = −3

Correct substitution:

(−3)²

This means:

(−3)(−3) = 9

Therefore:

x² = 9

The parentheses make it clear that the entire value −3 is being squared.

---

Powers and Negative Values

Compare:

(−3)² = 9

with:

−3² = −9

Using the standard order of operations:

−3²

means:

−(3²)

which is:

−9

So careful notation is extremely important during substitution.

---

Expressions with Powers

Evaluate:

x² + 3x + 2

when:

x = 4

Substitute:

(4)² + 3(4) + 2

Calculate the exponent:

16 + 12 + 2

Therefore:

30

---

Order of Operations

After substitution, evaluate using the normal order of operations:

Parentheses

Exponents

Multiplication and Division

Addition and Subtraction

For example:

Evaluate:

2x² + 3

when x = 4.

Substitute:

2(4)² + 3

Exponent first:

2(16) + 3

Multiply:

32 + 3

Therefore:

35

---

A Common Order-of-Operations Error

Evaluate:

3 + 2x²

when x = 5.

Correct:

3 + 2(5)²

3 + 2(25)

3 + 50

53

Incorrect reasoning might calculate:

(3 + 2)(5²)

but there are no parentheses telling us to add 3 and 2 first.

---

Expressions with Parentheses

Evaluate:

3(x + 4)

when:

x = 2

Substitute:

3(2 + 4)

Evaluate parentheses:

3(6)

Therefore:

18

We could also distribute first:

3x + 12

Then substitute:

3(2) + 12 = 18

Both methods produce the same result.

---

Multiple Variables

Expressions can contain more than one variable.

Evaluate:

3x + 2y

when:

x = 4

and:

y = 5

Substitute:

3(4) + 2(5)

Calculate:

12 + 10

Therefore:

22

 

 

 

7

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Keep Track of Each Variable

Evaluate:

2a + 3b − c

when:

a = 4

b = 2

c = 5

Substitute:

2(4) + 3(2) − 5

Calculate:

8 + 6 − 5

Therefore:

9

Be careful to replace each variable with the correct value.

---

Multiple Variables with Powers

Evaluate:

x² + 2xy + y²

when:

x = 3

and:

y = 2

Substitute:

(3)² + 2(3)(2) + (2)²

Calculate:

9 + 12 + 4

Therefore:

25

---

Substituting Negative Values into Multiple Variables

Evaluate:

2x − 3y

when:

x = −2

and:

y = −4

Substitute:

2(−2) − 3(−4)

Calculate:

−4 − (−12)

−4 + 12

Therefore:

8

---

Fractions

Evaluate:

x/4 + 3

when:

x = 12

Substitute:

12/4 + 3

Calculate:

3 + 3

Therefore:

6

---

Fractional Values

Variables themselves can also have fractional values.

Evaluate:

6x + 2

when:

x = 1/2

Substitute:

6(1/2) + 2

Calculate:

3 + 2

Therefore:

5

---

Decimal Values

Evaluate:

4x + 1.5

when:

x = 2.5

Substitute:

4(2.5) + 1.5

Calculate:

10 + 1.5

Therefore:

11.5

The substitution process is the same whether values are whole numbers, fractions, decimals, or negative numbers.

---

Equivalent Expressions and Substitution

Consider:

3(x + 2)

and:

3x + 6

These are equivalent expressions.

If x = 4:

First expression:

3(4 + 2) = 18

Second expression:

3(4) + 6 = 18

They give the same result.

Substitution can help us check whether our simplification is reasonable.

---

Checking Algebra Using Substitution

Suppose a student claims:

4x + 3x = 7x²

Test:

x = 2

Original:

4(2) + 3(2)

= 8 + 6

= 14

Student's expression:

7(2²)

= 7(4)

= 28

Since:

14 ≠ 28

the expressions are not equivalent.

The correct simplification is:

4x + 3x = 7x

---

Substitution into Formulas

A formula describes a mathematical relationship between quantities.

For example, the area of a rectangle is:

A = lw

where:

• A = area
• l = length
• w = width

Suppose:

l = 8 cm

and:

w = 5 cm

Substitute:

A = (8)(5)

Therefore:

A = 40 cm²

 

 

 

6

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Perimeter Formula

The perimeter of a rectangle is:

P = 2l + 2w

Suppose:

l = 12 cm

w = 7 cm

Substitute:

P = 2(12) + 2(7)

Calculate:

P = 24 + 14

Therefore:

P = 38 cm

---

Area of a Triangle

The area of a triangle is:

A = 1/2 bh

Suppose:

b = 10 cm

h = 6 cm

Substitute:

A = 1/2(10)(6)

Calculate:

A = 30 cm²

The units are square centimeters because area measures a two-dimensional region.

---

Area of a Circle

The area of a circle is:

A = πr²

Suppose:

r = 4 cm

Substitute:

A = π(4)²

A = 16π cm²

Using:

π ≈ 3.14

gives:

A ≈ 50.24 cm²

---

Circumference

The circumference of a circle is:

C = 2πr

For:

r = 5 cm

substitute:

C = 2π(5)

C = 10π cm

Approximately:

C ≈ 31.4 cm

---

Substitution in Science

Scientific formulas are algebraic relationships.

For example:

v = d/t

Suppose:

d = 150 m

t = 30 s

Substitute:

v = 150/30

Therefore:

v = 5 m/s

The result tells us the object's speed.

 

---

Force

Newton's Second Law is:

F = ma

Suppose:

m = 6 kg

and:

a = 4 m/s²

Substitute:

F = (6)(4)

Therefore:

F = 24 N

The calculated result represents the net force acting on the object.

---

Density

Density is:

ρ = m/V

Suppose:

m = 120 g

V = 40 cm³

Substitute:

ρ = 120/40

Therefore:

ρ = 3 g/cm³

The result means that each cubic centimeter of the material has a mass of 3 g.

---

Weight

Weight can be calculated using:

W = mg

Suppose:

m = 10 kg

and:

g = 9.8 N/kg

Substitute:

W = (10)(9.8)

Therefore:

W = 98 N

This represents the gravitational force acting on the object.

---

Kinetic Energy

Kinetic energy is:

Eₖ = 1/2 mv²

Suppose:

m = 4 kg

and:

v = 6 m/s

Substitute:

Eₖ = 1/2(4)(6²)

Evaluate the exponent:

Eₖ = 1/2(4)(36)

Therefore:

Eₖ = 72 J

Notice that the velocity must be squared before completing the multiplication.

---

Gravitational Potential Energy

Gravitational potential energy near Earth's surface is:

Eₚ = mgh

Suppose:

m = 3 kg

g = 9.8 N/kg

h = 5 m

Substitute:

Eₚ = (3)(9.8)(5)

Therefore:

Eₚ = 147 J

---

Electrical Power

Electrical power can be calculated using:

P = VI

Suppose:

V = 12 V

and:

I = 3 A

Substitute:

P = (12)(3)

Therefore:

P = 36 W

---

Temperature Conversion

The relationship between Celsius and Fahrenheit is:

F = (9/5)C + 32

Suppose:

C = 25°C

Substitute:

F = (9/5)(25) + 32

Calculate:

F = 45 + 32

Therefore:

F = 77°F

---

Real-World Cost Expressions

Suppose a taxi charges:

• $5 starting fee
• $3 per kilometer

The total cost can be represented by:

C = 3d + 5

where d is the distance traveled in kilometers.

If:

d = 8 km

substitute:

C = 3(8) + 5

C = 24 + 5

Therefore:

C = $29

---

Interpreting the Result

A good solution should not stop at:

C = 29

Instead say:

The taxi fare for an 8 km journey is $29.

This connects the mathematical result to the original situation.

---

Shopping Example

A store sells notebooks for $4 each and charges $6 for delivery.

Total cost:

C = 4n + 6

where n is the number of notebooks.

For:

n = 7

substitute:

C = 4(7) + 6

C = 28 + 6

Therefore:

C = $34

So seven notebooks cost $34 including delivery.

---

Earnings Example

Suppose a worker earns a fixed $50 plus $18 per hour.

Their earnings can be represented by:

E = 18h + 50

For:

h = 6

substitute:

E = 18(6) + 50

E = 108 + 50

Therefore:

E = $158

---

Distance Example

A cyclist starts 5 km from a reference point and travels at 12 km/h.

Position after t hours:

d = 12t + 5

After:

t = 3 hours

substitute:

d = 12(3) + 5

d = 36 + 5

Therefore:

d = 41 km

The cyclist is 41 km from the reference point.

---

Substitution in Tables

Expressions can be evaluated for several values and organized in a table.

Consider:

y = 2x + 1

x	y = 2x + 1
−2	−3
−1	−1
0	1
1	3
2	5

For example, when x = −2:

y = 2(−2) + 1

y = −3

Tables help show how one variable changes in response to another.

---

Substitution and Graphs

The points from a table can be plotted.

For:

y = 2x + 1

some points are:

(−2, −3)

(−1, −1)

(0, 1)

(1, 3)

(2, 5)

Substitution therefore helps us generate coordinates for graphs.

 

 

 

6

---

Checking Solutions to Equations

Substitution is also used to determine whether a value is a solution to an equation.

Is:

x = 4

a solution to:

3x + 2 = 14?

Substitute:

3(4) + 2 = 14

12 + 2 = 14

14 = 14

The statement is true.

Therefore:

x = 4 is a solution.

---

When the Value Is Not a Solution

Is:

x = 5

a solution to:

2x + 3 = 15?

Substitute:

2(5) + 3 = 15

13 = 15

This is false.

Therefore:

x = 5 is not a solution.

---

Checking Multi-Step Equation Solutions

Suppose we solve:

4x + 7 = 31

and obtain:

x = 6

Check by substitution:

4(6) + 7 = 31

24 + 7 = 31

31 = 31

Therefore the solution is verified.

Substitution is one of the best ways to detect algebraic errors.

---

Substitution Does Not Mean Solving

Suppose:

3x + 5

and we are told:

x = 7

We are evaluating an expression:

3(7) + 5 = 26

We are not solving for x because its value was already provided.

By contrast:

3x + 5 = 26

requires us to solve for x.

---

Formula Substitution Versus Formula Rearrangement

Suppose:

F = ma

and m and a are known.

We can substitute directly:

m = 5 kg

a = 3 m/s²

F = (5)(3) = 15 N

But if F and m are known and a is unknown, we first need to rearrange or solve:

F = ma

a = F/m

Then substitute.

Recognizing what is known and what is unknown is an important problem-solving skill.

---

Dimensional Meaning

When substituting into formulas, units help us understand the result.

For:

v = d/t

if:

d = 100 m

and:

t = 20 s

then:

v = 100 m / 20 s

v = 5 m/s

The units show that the answer represents speed.

---

Consistent Units

Before substitution, quantities sometimes need to be converted into compatible units.

Suppose:

v = d/t

with:

d = 500 m

and:

t = 2 minutes

If we want speed in m/s, convert:

2 min = 120 s

Then:

v = 500/120

v ≈ 4.17 m/s

Using incompatible units without conversion can produce misleading results.

---

Estimating Before Calculating

Estimation can help detect mistakes.

Suppose:

C = 7.95n + 4

and:

n = 10

Before calculating exactly, estimate:

8 × 10 + 4 ≈ 84

Exact calculation:

C = 7.95(10) + 4

C = 79.50 + 4

C = 83.50

The exact answer is close to the estimate, so it appears reasonable.

---

Calculator Use

Calculators are useful for complicated substitutions, but the substitution should still be shown clearly.

For example:

Eₖ = 1/2 mv²

with:

m = 7.5 kg

and:

v = 4.8 m/s

Write:

Eₖ = 1/2(7.5)(4.8²)

before entering the calculation.

This makes it possible to check whether the values were substituted correctly.

---

A Longer Algebraic Example

Evaluate:

3x² − 2xy + y²

when:

x = −2

and:

y = 3

Substitute:

3(−2)² − 2(−2)(3) + (3)²

Evaluate powers:

3(4) − 2(−2)(3) + 9

Multiply:

12 + 12 + 9

Therefore:

33

Careful use of parentheses prevents sign errors.

---

A Real-World Formula Example

Suppose the distance traveled by an object is modeled by:

d = 5t² + 2t

where d is in meters and t is in seconds.

Find the distance at:

t = 3 s

Substitute:

d = 5(3)² + 2(3)

d = 5(9) + 6

d = 45 + 6

Therefore:

d = 51 m

Interpretation:

After 3 seconds, the object has traveled 51 m according to the model.

---

Common Mistakes

Mistake: Forgetting multiplication.

If:

x = 4

then:

3x

means:

3(4)

not:

34

---

Mistake: Forgetting parentheses around negative values.

For x = −3:

x² = (−3)² = 9

---

Mistake: Ignoring the order of operations.

For:

2x²

with x = 3:

2(3²) = 18

not:

(2 × 3)² = 36

---

Mistake: Substituting into only one occurrence of a variable.

For:

x² + 3x

with x = 2:

both x values must be replaced:

(2)² + 3(2)

---

Mistake: Mixing up variables.

If x = 3 and y = 5, substitute each value into the correct location.

---

Mistake: Forgetting units in formulas.

A science calculation should usually end with an appropriate unit.

---

Mistake: Giving only a number in a real-world problem.

Interpret what the result represents.

---

Mistake: Using inconsistent units.

Convert quantities when necessary before substituting.

---

Key Terms

Substitution — replacing a variable with a known value.

Evaluate — calculate the numerical value of an expression.

Variable — letter or symbol representing a quantity.

Expression — combination of numbers, variables, and mathematical operations.

Formula — equation expressing a relationship between quantities.

Order of operations — agreed sequence for performing mathematical operations.

Coefficient — numerical factor multiplying a variable.

Exponent — number showing how many times a quantity is used as a factor.

Equivalent expressions — expressions that have the same value for every permissible value of their variables.

Solution — value that makes an equation true.

Verification — checking a solution, often through substitution.

Interpretation — explaining the meaning of a calculated value in context.

Model — mathematical representation of a real situation.

Unit — standard measurement associated with a quantity.

Estimate — approximate calculation used to judge the expected size of an answer.

---

Key Takeaways

• Substitution means replacing a variable with a known value.
• Evaluation means calculating the result after substitution.
• Use parentheses around substituted values, especially negative numbers.
• Every occurrence of the variable must be replaced.
• After substitution, follow the normal order of operations.
• Exponents must be evaluated correctly.
• Negative values require particular care when powers are involved.
• Expressions may contain several variables with different assigned values.
• Fractions and decimals can be substituted just like whole numbers.
• Equivalent expressions should produce the same value for the same variable values.
• Substitution can help check algebraic simplification.
• Substitution can verify whether a value solves an equation.
• Mathematical formulas are evaluated by substituting known quantities.
• Geometry formulas use substitution to calculate areas, perimeters, and other measurements.
• Physics formulas use substitution to calculate quantities such as speed, force, density, energy, and power.
• Real-world expressions can model costs, earnings, distances, and other quantities.
• Units should be included when interpreting calculated results.
• Units must sometimes be converted before substitution.
• Estimates can help determine whether an answer is reasonable.
• A final answer should explain what the calculated number means in its original context.
• Substitution is a foundational skill used throughout algebra, geometry, science, statistics, and many real-world applications.

Check Your Understanding

1. What does substitution mean in algebra?
2. What is the difference between substitution and evaluation?
3. Evaluate 3x + 5 when x = 4.
4. Evaluate 7x − 2 when x = 6.
5. Evaluate x² + 4 when x = 5.
6. Evaluate x² + 3x − 2 when x = −2.
7. Evaluate 2x + 3y when x = 4 and y = 5.
8. Evaluate x² + xy + y² when x = 3 and y = 2.
9. Why should parentheses be used when substituting negative values?
10. Explain the difference between (−4)² and −4².
11. Use A = lw to calculate the area of a rectangle with length 9 cm and width 6 cm.
12. Use A = 1/2 bh to calculate the area of a triangle with base 12 cm and height 7 cm.
13. Use v = d/t to calculate the speed of an object that travels 240 m in 30 s.
14. Use F = ma to calculate the force on a 7 kg object accelerating at 3 m/s².
15. Use ρ = m/V to calculate the density of a 150 g object with a volume of 50 cm³.
16. Use Eₖ = 1/2 mv² to calculate the kinetic energy of a 5 kg object moving at 4 m/s.
17. A taxi fare is modeled by C = 4d + 6. Calculate and interpret C when d = 8 km.
18. Determine whether x = 5 is a solution of 4x + 3 = 23 by substitution.
19. Explain why units may need to be converted before values are substituted into a formula.
20. Describe a systematic method for substituting values into a complicated expression and evaluating it accurately.

 

 

 

4. Rearranging Simple Formulas

Rearranging Simple Formulas

A formula is an equation that shows the relationship between two or more quantities.

For example:

v = d/t

connects:

• v = speed
• d = distance
• t = time

If we know distance and time, we can calculate speed directly. But sometimes we need to find distance or time instead.

We can rearrange the formula so the variable we want is isolated.

From:

v = d/t

we can rearrange to:

d = vt

or:

t = d/v

These are not new formulas. They are equivalent versions of the same relationship.

 

 

 

6

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Variables in Formulas

A variable is a symbol representing a quantity that can have different values.

Consider:

A = lw

The variables are:

• A = area
• l = length
• w = width

The formula tells us that the area of a rectangle equals its length multiplied by its width.

In formulas, letters often have specific meanings rather than simply representing an unknown number.

---

Identifying Variables

Consider:

F = ma

The variables are:

• F = force
• m = mass
• a = acceleration

Consider:

P = E/t

The variables are:

• P = power
• E = energy transferred
• t = time

Before rearranging a formula, make sure you understand what each variable represents.

---

Constants and Variables

Not every symbol in a formula necessarily changes.

For example:

C = 2πr

The quantities C and r are variables.

The number 2 is a constant.

π is also a constant:

π ≈ 3.14159

Recognizing the difference between variables and constants helps us understand formulas correctly.

---

What Does Rearranging Mean?

To rearrange a formula means to rewrite it so that a different variable is isolated.

For example:

F = ma

already has F isolated.

If we want m instead:

m = F/a

If we want a:

a = F/m

The relationship between force, mass, and acceleration has not changed.

Only the form of the equation has changed.

---

The Goal: Isolate the Required Variable

When rearranging a formula, ask:

Which variable do I need by itself?

For:

P = E/t

if the question asks for E, we want:

E = ?

If the question asks for t, we want:

t = ?

This helps us decide which operations need to be undone.

---

The Balance Principle

A formula is an equation, so both sides must remain equal.

If we perform an operation on one side, we must perform the same operation on the other side.

For example:

x + 5 = 12

Subtract 5 from both sides:

x = 7

The same principle applies when rearranging formulas.

 

 

 

12

---

Inverse Operations

Inverse operations undo each other.

Operation	Inverse operation
+	−
−	+
×	÷
÷	×
square	square root

For example, if a variable is multiplied by 5:

y = 5x

divide by 5 to isolate x:

x = y/5

---

Rearranging Addition Formulas

Consider:

y = x + 4

Make x the subject.

Subtract 4 from both sides:

y − 4 = x

Rewrite:

x = y − 4

Therefore:

y = x + 4

can be rearranged as:

x = y − 4

---

Rearranging Subtraction Formulas

Consider:

y = x − 7

Make x the subject.

Add 7 to both sides:

y + 7 = x

Therefore:

x = y + 7

---

Be Careful with Subtraction Order

Consider:

y = 10 − x

Make x the subject.

Add x to both sides:

y + x = 10

Subtract y:

x = 10 − y

The answer is:

x = 10 − y

not:

x = y − 10

Subtraction order matters.

---

Rearranging Multiplication Formulas

Consider:

y = 4x

Make x the subject.

Divide both sides by 4:

y/4 = x

Therefore:

x = y/4

---

Rearranging Division Formulas

Consider:

y = x/5

Make x the subject.

Multiply both sides by 5:

5y = x

Therefore:

x = 5y

---

A Reliable Strategy

For simple formulas, use:

Identify → Target → Undo → Simplify → Check

Identify

Identify all variables and what they represent.

Target

Decide which variable needs to be isolated.

Undo

Use inverse operations to remove everything attached to that variable.

Simplify

Write the rearranged formula clearly.

Check

Confirm that the new formula is equivalent to the original.

---

Distance, Speed, and Time

One of the most useful examples is:

v = d/t

where:

• v = speed
• d = distance
• t = time

To make d the subject, multiply both sides by t:

vt = d

Rewrite:

d = vt

To make t the subject:

Start with:

v = d/t

Multiply by t:

vt = d

Divide by v:

t = d/v

Therefore:

v = d/t

d = vt

t = d/v

All three represent the same relationship.

 

 

 

7

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Applying the Rearranged Formula

Suppose:

v = 20 m/s

and:

t = 15 s

Find d.

Use:

d = vt

Substitute:

d = (20)(15)

Therefore:

d = 300 m

---

Finding Time

Suppose:

d = 500 km

and:

v = 100 km/h

Use:

t = d/v

Substitute:

t = 500/100

Therefore:

t = 5 h

---

Newton's Second Law

Newton's Second Law is:

F = ma

where:

• F = net force
• m = mass
• a = acceleration

To make m the subject:

Divide by a:

m = F/a

To make a the subject:

Divide by m:

a = F/m

Therefore:

F = ma

can also be written:

m = F/a

or:

a = F/m

---

Applying F = ma

Suppose:

F = 60 N

m = 12 kg

Find acceleration.

Use the rearranged formula:

a = F/m

Substitute:

a = 60/12

Therefore:

a = 5 m/s²

---

Density

Density is:

ρ = m/V

where:

• ρ = density
• m = mass
• V = volume

To make m the subject:

Multiply by V:

ρV = m

Therefore:

m = ρV

To make V the subject:

Start with:

ρ = m/V

Multiply by V:

ρV = m

Divide by ρ:

V = m/ρ

 

---

Density Example

A material has:

ρ = 8 g/cm³

and:

m = 200 g

Find its volume.

Use:

V = m/ρ

Substitute:

V = 200/8

Therefore:

V = 25 cm³

---

Weight

Weight is calculated using:

W = mg

where:

• W = weight
• m = mass
• g = gravitational field strength

Make m the subject:

m = W/g

Make g the subject:

g = W/m

---

Weight Example

An object weighs:

196 N

where:

g = 9.8 N/kg

Find its mass.

Use:

m = W/g

Substitute:

m = 196/9.8

Therefore:

m = 20 kg

---

Pressure

Pressure is:

p = F/A

where:

• p = pressure
• F = force
• A = area

Make F the subject:

F = pA

Make A the subject:

A = F/p

---

Pressure Example

A pressure of:

500 Pa

acts over an area of:

0.20 m²

Find the force.

Use:

F = pA

Substitute:

F = (500)(0.20)

Therefore:

F = 100 N

---

Electrical Relationships

Electrical formulas also require rearrangement.

For example:

V = IR

where:

• V = voltage
• I = current
• R = resistance

Make I the subject:

I = V/R

Make R the subject:

R = V/I

 

 

 

5

---

Ohm's Law Example

Suppose:

V = 12 V

and:

R = 4 Ω

Find current.

Use:

I = V/R

Substitute:

I = 12/4

Therefore:

I = 3 A

---

Electrical Power

Electrical power can be calculated using:

P = VI

Make V the subject:

V = P/I

Make I the subject:

I = P/V

Example:

A device uses:

P = 120 W

at:

V = 12 V

Find current:

I = P/V

I = 120/12

I = 10 A

---

Energy, Power, and Time

Power is:

P = E/t

Make E the subject:

E = Pt

Make t the subject:

t = E/P

Example:

A 100 W device operates for 30 s.

E = Pt

E = (100)(30)

Therefore:

E = 3000 J

---

Rectangle Area

The area of a rectangle is:

A = lw

Make l the subject:

l = A/w

Make w the subject:

w = A/l

 

---

Rectangle Example

A rectangle has:

A = 60 cm²

and:

w = 5 cm

Find its length.

Use:

l = A/w

Substitute:

l = 60/5

Therefore:

l = 12 cm

---

Triangle Area

The area of a triangle is:

A = 1/2 bh

Make b the subject.

Start with:

A = 1/2 bh

Multiply both sides by 2:

2A = bh

Divide by h:

b = 2A/h

To make h the subject:

h = 2A/b

---

Triangle Example

A triangle has:

A = 36 cm²

and:

b = 9 cm

Find its height.

Use:

h = 2A/b

Substitute:

h = 2(36)/9

h = 72/9

Therefore:

h = 8 cm

---

Circle Circumference

Circumference is:

C = 2πr

Make r the subject.

Divide both sides by 2π:

r = C/(2π)

If:

C = 31.4 cm

then:

r = 31.4/(2π)

Using π ≈ 3.14:

r = 5 cm

---

Temperature Conversion

The Celsius-to-Fahrenheit formula is:

F = (9/5)C + 32

Suppose we want C.

Start:

F = (9/5)C + 32

Subtract 32:

F − 32 = (9/5)C

Multiply by 5/9:

C = (5/9)(F − 32)

This is a multi-step rearrangement.

---

Temperature Example

Convert:

86°F

to Celsius.

Use:

C = (5/9)(F − 32)

Substitute:

C = (5/9)(86 − 32)

C = (5/9)(54)

Therefore:

C = 30°C

---

Undo Operations in Reverse Order

Consider:

y = 3x + 5

To isolate x, ask:

What has happened to x?

First:

x was multiplied by 3

Then:

5 was added

Undo these operations in reverse order.

Subtract 5:

y − 5 = 3x

Divide by 3:

x = (y − 5)/3

This reverse-order idea is extremely useful.

---

Another Two-Step Formula

Consider:

y = 4x − 7

Make x the subject.

Add 7:

y + 7 = 4x

Divide by 4:

x = (y + 7)/4

---

Formula with a Divided Expression

Consider:

y = (x + 3)/5

Make x the subject.

Multiply by 5:

5y = x + 3

Subtract 3:

x = 5y − 3

Notice that we undo the outermost operation first.

---

Formula with Subtraction in the Numerator

Consider:

y = (x − 4)/3

Multiply by 3:

3y = x − 4

Add 4:

x = 3y + 4

---

Variables on Both Sides

Some formulas contain the required variable more than once or on both sides.

For example:

y = ax + bx

Factor x:

y = x(a + b)

Divide by a + b:

x = y/(a + b)

Factoring can sometimes be an important part of rearranging formulas.

---

Factoring to Isolate a Variable

Consider:

P = 2l + 2w

Suppose we want l.

Subtract 2w:

P − 2w = 2l

Divide by 2:

l = (P − 2w)/2

This can also be written:

l = P/2 − w

Both forms are equivalent.

---

Checking a Rearranged Formula

One way to check rearrangement is to reverse the algebra.

Suppose:

v = d/t

was rearranged to:

d = vt

Starting from:

d = vt

divide by t:

d/t = v

Therefore:

v = d/t

We have returned to the original formula.

This supports that the rearrangement is correct.

---

Checking with Numerical Values

We can also test a rearranged formula using numbers.

Original:

F = ma

Let:

m = 4 kg

a = 5 m/s²

Then:

F = 20 N

Now test:

m = F/a

m = 20/5

m = 4 kg

The rearranged formula gives the expected value.

---

Checking by Units

Units can also help detect mistakes.

For:

v = d/t

speed has units:

m/s

Rearrange:

d = vt

Units:

(m/s)(s) = m

Correct.

For:

t = d/v

units:

m ÷ (m/s) = s

Correct.

This is called dimensional checking or dimensional analysis.

---

Formula Triangles

Some simple formulas are sometimes remembered using formula triangles.

 

 

 

For example, for:

v = d/t

a triangle may show d above v and t.

However, formula triangles are memory aids rather than a replacement for algebra.

Learning to rearrange equations using inverse operations works for much more complicated formulas.

---

Practical Example: Travel

A car travels:

360 km

at an average speed of:

90 km/h

Find the time.

Start with:

v = d/t

Make t the subject:

t = d/v

Substitute:

t = 360/90

Therefore:

t = 4 h

Interpretation:

The journey takes 4 hours at an average speed of 90 km/h.

---

Practical Example: Density

A liquid has:

density = 0.8 g/cm³

and:

volume = 250 cm³

Find its mass.

Start:

ρ = m/V

Rearrange:

m = ρV

Substitute:

m = (0.8)(250)

Therefore:

m = 200 g

---

Practical Example: Electricity

A resistor has:

R = 15 Ω

and carries:

I = 2 A

Find voltage.

Use:

V = IR

Substitute:

V = (2)(15)

Therefore:

V = 30 V

---

Practical Example: Geometry

A triangle has:

area = 54 cm²

and:

height = 9 cm

Find the base.

Start:

A = 1/2 bh

Rearrange:

b = 2A/h

Substitute:

b = 2(54)/9

Therefore:

b = 12 cm

---

Practical Example: Cost

Suppose the total cost of renting equipment is:

C = 15h + 25

where:

• C = total cost
• h = number of hours
• $25 = fixed fee
• $15 = hourly charge

Make h the subject.

Subtract 25:

C − 25 = 15h

Divide by 15:

h = (C − 25)/15

If the total bill is $100:

h = (100 − 25)/15

h = 75/15

Therefore:

h = 5 hours

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Choosing Whether to Rearrange First

Sometimes we can substitute values before rearranging.

For:

F = ma

if:

F = 40

and:

m = 8

we could write:

40 = 8a

and solve:

a = 5

Or we could rearrange first:

a = F/m

then substitute:

a = 40/8 = 5

Both methods are correct.

However, rearranging first is especially useful when the same type of calculation will be repeated several times.

---

Common Mistakes

Mistake: Performing an operation on only one side.

An equation must remain balanced.

---

Mistake: Moving terms without understanding the operation.

Terms do not simply "jump across" an equals sign.

For:

y = x + 5

we subtract 5 from both sides to obtain:

x = y − 5

---

Mistake: Reversing subtraction incorrectly.

From:

y = 10 − x

the correct rearrangement is:

x = 10 − y

not:

x = y − 10

---

Mistake: Multiplying when division is required.

From:

F = ma

the correct rearrangement for mass is:

m = F/a

---

Mistake: Dividing only part of an expression.

From:

y = 3x + 6

we get:

x = (y − 6)/3

The entire expression y − 6 is divided by 3.

---

Mistake: Ignoring parentheses.

(y − 6)/3

is not the same as:

y − 6/3

---

Mistake: Substituting too early and losing track of the algebra.

For unfamiliar formulas, rearranging symbolically first can make the relationship clearer.

---

Mistake: Forgetting units.

After applying a rearranged formula to a practical problem, include the appropriate unit.

---

Key Terms

Formula — equation representing a relationship between quantities.

Variable — symbol representing a quantity that may change or be unknown.

Constant — quantity with a fixed value.

Subject of a formula — variable isolated on one side of the formula.

Rearrange — rewrite a formula so that a different variable becomes the subject.

Isolate — get a variable alone on one side.

Inverse operations — operations that undo each other.

Equivalent formulas — different forms of a formula representing the same mathematical relationship.

Substitution — replacing variables with known values.

Factoring — rewriting an expression as a product of factors.

Dimensional analysis — using units to examine whether a mathematical relationship is consistent.

Practical application — use of mathematics to solve a real-world problem.

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Key Takeaways

• A formula is an equation relating two or more quantities.
• Variables in formulas usually represent specific physical or mathematical quantities.
• Rearranging changes the form of a formula but not the relationship it represents.
• The required variable should be isolated on one side.
• Use inverse operations to undo operations affecting the required variable.
• Addition is undone by subtraction.
• Subtraction is undone by addition.
• Multiplication is undone by division.
• Division is undone by multiplication.
• Operations are often undone in reverse order.
• Both sides of a formula must remain balanced.
• Parentheses are important when an entire expression is divided.
• Subtraction order must be handled carefully.
• Formulas such as v = d/t can be rearranged into several equivalent forms.
• F = ma can become m = F/a or a = F/m.
• ρ = m/V can become m = ρV or V = m/ρ.
• Factoring may help isolate a variable that appears in multiple terms.
• Rearranged formulas can be checked by reversing the algebra.
• Numerical substitution can also help check a rearrangement.
• Units provide another useful way to check formulas.
• Formula triangles may help with simple relationships, but algebraic rearrangement is more powerful.
• Formula rearrangement is widely used in mathematics, physics, chemistry, engineering, finance, and everyday problem-solving.

Check Your Understanding

1. What does it mean to make a variable the subject of a formula?
2. Identify all variables in F = ma.
3. Rearrange y = x + 7 to make x the subject.
4. Rearrange y = x − 5 to make x the subject.
5. Rearrange y = 4x to make x the subject.
6. Rearrange y = x/6 to make x the subject.
7. Rearrange v = d/t to make d the subject.
8. Rearrange v = d/t to make t the subject.
9. Rearrange F = ma to make m the subject.
10. Rearrange F = ma to make a the subject.
11. Rearrange ρ = m/V to make m the subject.
12. Rearrange ρ = m/V to make V the subject.
13. Rearrange V = IR to make R the subject.
14. Rearrange P = E/t to make t the subject.
15. Rearrange A = 1/2 bh to make h the subject.
16. Rearrange y = 5x + 3 to make x the subject.
17. A car travels 450 km at 90 km/h. Rearrange and use v = d/t to determine the travel time.
18. A material has a density of 6 g/cm³ and a volume of 20 cm³. Rearrange and use ρ = m/V to determine its mass.
19. Explain two different ways you could check whether a rearranged formula is correct.
20. Explain why learning algebraic rearrangement is more useful than relying only on formula triangles.

 

 

 

5. Algebra and Measurement Formulas

Algebra and Measurement Formulas

A measurement formula is an equation that describes a relationship between measurable quantities such as length, area, perimeter, volume, speed, mass, density, or temperature.

For example, the perimeter of a rectangle is:

P = 2l + 2w

where:

• P = perimeter
• l = length
• w = width

If the length is 8 cm and the width is 5 cm:

P = 2(8) + 2(5)

P = 16 + 10

P = 26 cm

Using formulas successfully involves more than calculation. We must choose the correct formula, substitute values accurately, rearrange formulas when necessary, and interpret the result using appropriate units.

 

 

 

5

A Formula as a Mathematical Model

A formula describes a relationship between quantities.

For example:

A = lw

tells us that the area of a rectangle depends on its length and width.

If either dimension changes, the area may also change.

Formulas allow us to:

• calculate unknown measurements
• compare objects
• solve practical problems
• make predictions
• represent geometric and scientific relationships

A Reliable Formula Method

For measurement problems, use:

Identify → Choose → Substitute → Calculate → Interpret

Identify

Determine what information is known and what must be found.

Choose

Select the appropriate formula.

Substitute

Replace the variables with known values.

Calculate

Perform the mathematical operations accurately.

Interpret

State the answer with the correct units and explain what it represents.

If the required variable is not already isolated, add another step:

Rearrange

Perimeter

Perimeter is the total distance around the outside of a two-dimensional shape.

Because perimeter measures distance, its units are ordinary length units such as:

• mm
• cm
• m
• km

 

 

 

8

Perimeter of a Rectangle

For a rectangle:

P = 2l + 2w

This works because a rectangle has:

• two sides of length l
• two sides of width w

So:

P = l + w + l + w

which simplifies to:

P = 2l + 2w

Rectangle Perimeter Example

A rectangle has:

l = 12 cm

w = 7 cm

Substitute:

P = 2(12) + 2(7)

P = 24 + 14

Therefore:

P = 38 cm

Perimeter of a Square

All four sides of a square have equal length.

If s is the side length:

P = 4s

For:

s = 9 cm

substitute:

P = 4(9)

Therefore:

P = 36 cm

Perimeter of a Triangle

For a triangle with side lengths a, b, and c:

P = a + b + c

If:

a = 5 cm

b = 8 cm

c = 10 cm

then:

P = 5 + 8 + 10

Therefore:

P = 23 cm

Area

Area measures the amount of two-dimensional space inside a shape.

Area uses square units.

Examples:

• mm²
• cm²
• m²
• km²

This is because area represents a two-dimensional measurement.

Area of a Rectangle

The area of a rectangle is:

A = lw

For:

l = 10 cm

and:

w = 6 cm

substitute:

A = (10)(6)

Therefore:

A = 60 cm²

 

 

 

9

Why A = lw Works

Imagine dividing a rectangle into 1 cm × 1 cm squares.

If the rectangle is:

6 cm × 4 cm

there are:

6 × 4 = 24

unit squares.

Therefore:

A = 24 cm²

The formula A = lw counts the number of square units covering the rectangle.

Area of a Square

For a square:

A = s²

If:

s = 7 cm

then:

A = 7²

Therefore:

A = 49 cm²

Notice the difference:

Perimeter:

P = 4s

Area:

A = s²

These measure different properties.

Area of a Triangle

The area of a triangle is:

A = 1/2 bh

where:

• b = base
• h = perpendicular height

For:

b = 12 cm

h = 8 cm

substitute:

A = 1/2(12)(8)

A = 48 cm²

 

 

 

7

Why the Height Must Be Perpendicular

The height of a triangle is the perpendicular distance from the base to the opposite vertex.

It is not necessarily one of the triangle's sides.

The height must meet the base at:

90°

Using a sloping side instead of the perpendicular height will usually produce an incorrect area.

Area of a Parallelogram

For a parallelogram:

A = bh

where:

• b = base
• h = perpendicular height

If:

b = 14 cm

h = 6 cm

then:

A = (14)(6)

Therefore:

A = 84 cm²

Again, use the perpendicular height rather than the sloping side.

Area of a Trapezoid

For a trapezoid with parallel sides a and b:

A = 1/2(a + b)h

Suppose:

a = 8 cm

b = 14 cm

h = 5 cm

Substitute:

A = 1/2(8 + 14)(5)

A = 1/2(22)(5)

Therefore:

A = 55 cm²

Circumference of a Circle

The distance around a circle is called its circumference.

The formulas are:

C = 2πr

or:

C = πd

where:

• r = radius
• d = diameter
• π ≈ 3.14159

Since:

d = 2r

the two circumference formulas are equivalent.

 

 

 

6

Circumference Example

A circle has:

r = 6 cm

Use:

C = 2πr

Substitute:

C = 2π(6)

C = 12π cm

Using π ≈ 3.14:

C ≈ 37.68 cm

Area of a Circle

The area of a circle is:

A = πr²

If:

r = 5 cm

then:

A = π(5²)

A = 25π cm²

Using π ≈ 3.14:

A ≈ 78.5 cm²

Radius Versus Diameter

A common mistake is confusing radius and diameter.

The relationship is:

d = 2r

Therefore:

r = d/2

If a circle has diameter:

d = 12 cm

then:

r = 6 cm

For area:

A = π(6²)

not:

π(12²)

unless the formula has first been rewritten in terms of diameter.

Composite Shapes

Some objects can be divided into simpler shapes.

 

 

 

10

Suppose a shape consists of:

• a rectangle with area 40 cm²
• a triangle with area 12 cm²

Total area:

A_total = 40 + 12

Therefore:

A_total = 52 cm²

For more complicated shapes, we can:

divide → calculate → combine

Subtracting Areas

Sometimes an area is found by subtracting one shape from another.

For example, suppose a rectangular frame has:

outer area = 120 cm²

and:

inner opening = 72 cm²

Frame area:

A = 120 − 72

Therefore:

A = 48 cm²

Algebraic Dimensions

Measurement formulas can contain algebraic expressions.

Suppose a rectangle has:

length = x + 3

width = x

Its area is:

A = x(x + 3)

Using the distributive property:

A = x² + 3x

Both expressions represent the same area.

Algebraic Perimeter

Suppose:

length = x + 4

width = x + 1

Use:

P = 2l + 2w

Substitute:

P = 2(x + 4) + 2(x + 1)

Distribute:

P = 2x + 8 + 2x + 2

Combine like terms:

P = 4x + 10

The perimeter is represented by:

4x + 10

Finding an Unknown Side

Suppose a rectangle has:

P = 34 cm

and:

w = 6 cm

Find l.

Start:

P = 2l + 2w

Substitute:

34 = 2l + 2(6)

34 = 2l + 12

Subtract 12:

22 = 2l

Divide by 2:

l = 11 cm

Rearranging Before Substituting

We could instead rearrange:

P = 2l + 2w

Subtract 2w:

P − 2w = 2l

Divide by 2:

l = (P − 2w)/2

Now substitute:

l = (34 − 2(6))/2

l = 22/2

Therefore:

l = 11 cm

Rearranging the Rectangle Area Formula

Start:

A = lw

Make l the subject:

l = A/w

Make w the subject:

w = A/l

Example:

A rectangle has:

A = 72 cm²

w = 8 cm

Then:

l = 72/8

Therefore:

l = 9 cm

Rearranging the Triangle Area Formula

Start:

A = 1/2 bh

Make h the subject.

Multiply by 2:

2A = bh

Divide by b:

h = 2A/b

Similarly:

b = 2A/h

Triangle Rearrangement Example

A triangle has:

A = 60 cm²

and:

b = 15 cm

Find h.

Use:

h = 2A/b

Substitute:

h = 2(60)/15

h = 120/15

Therefore:

h = 8 cm

Rearranging Circle Formulas

For circumference:

C = 2πr

Make r the subject:

r = C/(2π)

For circle area:

A = πr²

Divide by π:

A/π = r²

Take the square root:

r = √(A/π)

This requires a square root because r is squared in the original formula.

Circle Example

A circular garden has area:

A = 314 m²

Using π ≈ 3.14:

r = √(314/3.14)

r = √100

Therefore:

r = 10 m

Units Are Part of the Answer

Measurements are incomplete without appropriate units.

Length:

8 cm

Perimeter:

24 cm

Area:

48 cm²

Volume:

60 cm³

Speed:

15 m/s

Density:

2.7 g/cm³

Force:

30 N

Units communicate what the calculated number represents.

Linear, Square, and Cubic Units

It is important to distinguish between dimensions.

Length

One-dimensional:

cm

Area

Two-dimensional:

cm²

Volume

Three-dimensional:

cm³

For example:

A rectangle may have perimeter:

24 cm

and area:

32 cm²

These units cannot be interchanged.

Unit Conversion

Measurements sometimes need to be converted before using a formula.

For length:

1 m = 100 cm

But for area:

1 m² = 10,000 cm²

because:

1 m² = (100 cm)(100 cm)

For volume:

1 m³ = 1,000,000 cm³

because:

1 m³ = (100 cm)³

This is an important distinction.

Example with Unit Conversion

A rectangle has:

length = 2 m

width = 50 cm

Before calculating area, use consistent units.

Convert:

2 m = 200 cm

Then:

A = lw

A = (200)(50)

Therefore:

A = 10,000 cm²

This is equivalent to:

1 m²

Science Measurement Formulas

The same algebraic skills used in geometry apply to science.

For example:

v = d/t

ρ = m/V

F = ma

p = F/A

P = E/t

These formulas involve measured quantities and can be substituted into or rearranged just like geometry formulas.

 

Speed

Speed is:

v = d/t

where:

• v = speed
• d = distance
• t = time

Suppose:

d = 180 km

t = 3 h

Substitute:

v = 180/3

Therefore:

v = 60 km/h

Rearranging Speed

From:

v = d/t

we can obtain:

d = vt

and:

t = d/v

Suppose a vehicle travels at:

80 km/h

for:

2.5 h

Then:

d = vt

d = (80)(2.5)

Therefore:

d = 200 km

Density

Density is:

ρ = m/V

where:

• ρ = density
• m = mass
• V = volume

Suppose:

m = 240 g

V = 80 cm³

Then:

ρ = 240/80

Therefore:

ρ = 3 g/cm³

Rearranging Density

From:

ρ = m/V

we can obtain:

m = ρV

and:

V = m/ρ

Suppose:

ρ = 2.5 g/cm³

V = 40 cm³

Then:

m = ρV

m = (2.5)(40)

Therefore:

m = 100 g

Force

Newton's Second Law is:

F = ma

Suppose:

m = 8 kg

a = 3 m/s²

Substitute:

F = (8)(3)

Therefore:

F = 24 N

Pressure

Pressure is:

p = F/A

Suppose:

F = 600 N

and:

A = 0.20 m²

Substitute:

p = 600/0.20

Therefore:

p = 3000 Pa

Notice that area must be measured in appropriate square units.

Rearranging Pressure

From:

p = F/A

we can obtain:

F = pA

and:

A = F/p

Suppose:

p = 5000 Pa

and:

A = 0.04 m²

Then:

F = pA

F = (5000)(0.04)

Therefore:

F = 200 N

Real-World Example: Fencing

A rectangular garden is:

12 m long

and:

8 m wide

How much fencing is needed?

Because fencing goes around the outside, we need perimeter, not area.

P = 2l + 2w

P = 2(12) + 2(8)

P = 24 + 16

Therefore:

40 m of fencing is required.

Real-World Example: Flooring

A room is:

6 m long

and:

4.5 m wide

How much floor area must be covered?

Use:

A = lw

A = (6)(4.5)

Therefore:

A = 27 m²

At least 27 m² of flooring is needed before allowing for waste or extra material.

Real-World Example: Painting a Wall

A wall measures:

5 m × 2.8 m

Its area is:

A = (5)(2.8)

A = 14 m²

If a window occupies:

2 m²

then the area to paint is:

14 − 2 = 12 m²

Real-World Example: Circular Garden

A circular garden has radius:

4 m

Area:

A = πr²

A = π(4²)

A = 16π

Therefore:

A ≈ 50.3 m²

The circumference is:

C = 2πr

C = 8π

Therefore:

C ≈ 25.1 m

So approximately:

• 50.3 m² of ground is inside the garden
• 25.1 m of edging is needed around it

This demonstrates the difference between area and perimeter.

Estimating Before Calculating

Estimation can help detect mistakes.

Suppose a rectangle measures:

9.8 m × 5.1 m

Estimate:

10 × 5 = 50 m²

Exact calculation:

9.8 × 5.1 = 49.98 m²

The exact result is close to 50 m², so it appears reasonable.

Checking with Units

Units can help determine whether a formula has been used correctly.

For:

A = lw

units are:

m × m = m²

For:

v = d/t

units are:

m/s

For:

ρ = m/V

units might be:

g/cm³

This type of check can reveal mistakes.

Choosing the Correct Formula

A major part of solving measurement problems is deciding what is actually being asked.

If a question asks:

How much fencing?

Think:

perimeter

If it asks:

How much carpet?

Think:

area

If it asks:

How much liquid can the container hold?

Think:

volume

If it asks:

How fast?

Think:

speed

If it asks:

How tightly matter is packed?

Think:

density

A Full Problem-Solving Example

A rectangular sports court has an area of:

180 m²

and width:

12 m

Find its length and perimeter.

Find the length

Start:

A = lw

Rearrange:

l = A/w

Substitute:

l = 180/12

l = 15 m

Find the perimeter

Use:

P = 2l + 2w

Substitute:

P = 2(15) + 2(12)

P = 30 + 24

P = 54 m

Therefore:

The court is 15 m long and has a perimeter of 54 m.

Algebraic Measurement Problem

A rectangle has:

length = x + 3

width = x

and perimeter:

30 cm

Use:

P = 2l + 2w

Substitute:

30 = 2(x + 3) + 2x

Distribute:

30 = 2x + 6 + 2x

Combine:

30 = 4x + 6

Subtract 6:

24 = 4x

Divide by 4:

x = 6

Therefore:

Width:

6 cm

Length:

9 cm

Check:

2(9) + 2(6) = 30 cm

Correct.

Common Mistakes

Mistake: Confusing perimeter and area.

Perimeter measures distance around a shape.

Area measures space inside a shape.

Mistake: Using incorrect units.

Area should use square units such as cm², not cm.

Mistake: Forgetting to convert units.

Do not directly multiply 2 m by 50 cm without first deciding on consistent units.

Mistake: Using diameter as radius.

For:

A = πr²

the value substituted must be the radius.

Mistake: Using the sloping side as a triangle's height.

The height must be perpendicular to the chosen base.

Mistake: Substituting into the wrong variable.

Identify what every letter represents before substituting.

Mistake: Rearranging incorrectly.

From:

A = lw

the correct formula for width is:

w = A/l

not:

w = Al

Mistake: Forgetting to square the radius.

Circle area is:

A = πr²

not:

A = πr

Mistake: Giving a numerical answer without interpretation.

Instead of only writing:

27

write:

The floor area is 27 m².

Key Terms

Measurement — process of determining the size or amount of a quantity.

Formula — equation showing a mathematical relationship between quantities.

Variable — symbol representing a measurable or changing quantity.

Substitution — replacing a variable with a known value.

Rearrangement — rewriting a formula to isolate a different variable.

Perimeter — total distance around a two-dimensional shape.

Circumference — perimeter of a circle.

Area — amount of two-dimensional space inside a shape.

Length — one-dimensional measurement of distance.

Radius — distance from the center of a circle to its edge.

Diameter — distance across a circle through its center; equal to twice the radius.

Perpendicular height — shortest distance from a base to the opposite point or parallel side, measured at 90°.

Composite shape — shape constructed from two or more simpler shapes.

Square unit — unit used for area, such as cm² or m².

Unit conversion — changing a measurement from one unit to an equivalent measurement in another.

Density — mass per unit volume.

Speed — distance traveled per unit time.

Pressure — force per unit area.

Key Takeaways

• Measurement formulas describe relationships between measurable quantities.
• A formula can be treated like any other algebraic equation.
• Begin by identifying what is known and what must be calculated.
• Select the correct formula before substituting values.
• Perimeter measures the distance around a shape.
• Area measures the two-dimensional space inside a shape.
• Perimeter uses linear units such as cm or m.
• Area uses square units such as cm² or m².
• Rectangle perimeter is P = 2l + 2w.
• Rectangle area is A = lw.
• Triangle area is A = 1/2 bh.
• Circle circumference is C = 2πr or C = πd.
• Circle area is A = πr².
• Algebraic expressions can represent unknown dimensions.
• Measurement formulas can be rearranged using inverse operations.
• Units should be consistent before values are substituted.
• Converting area units requires squaring the length conversion factor.
• Science formulas use the same algebraic skills as geometry formulas.
• Examples include v = d/t, ρ = m/V, F = ma, and p = F/A.
• Units help us interpret calculated values and check whether an answer is reasonable.
• Estimates provide another useful way to detect calculation errors.
• A complete answer includes the numerical result, correct units, and its meaning in context.

Check Your Understanding

1. Explain the difference between perimeter and area.
2. Calculate the perimeter of a rectangle measuring 9 cm by 4 cm.
3. Calculate the area of the same 9 cm by 4 cm rectangle.
4. Find the perimeter of a square with side length 7 m.
5. Find the area of a square with side length 7 m.
6. Calculate the area of a triangle with base 14 cm and perpendicular height 8 cm.
7. Calculate the area of a parallelogram with base 12 m and perpendicular height 5 m.
8. Calculate the area of a trapezoid with parallel sides 6 cm and 10 cm and height 4 cm.
9. Find the circumference of a circle with radius 5 cm.
10. Find the area of a circle with radius 5 cm.
11. A circle has diameter 16 cm. Find its radius and then calculate its area.
12. Rearrange A = lw to make w the subject.
13. Rearrange A = 1/2 bh to make h the subject.
14. A rectangle has area 96 cm² and width 8 cm. Find its length.
15. A triangle has area 45 cm² and base 10 cm. Find its perpendicular height.
16. A car travels 280 km in 4 hours. Use v = d/t to calculate its average speed.
17. A material has mass 270 g and volume 90 cm³. Use ρ = m/V to calculate its density.
18. A rectangular garden measures 15 m by 8 m. Calculate both the amount of fencing needed around it and the area of ground inside it.
19. Explain why 1 m² = 10,000 cm² even though 1 m = 100 cm.
20. Describe a systematic method for choosing, rearranging, substituting into, and interpreting a measurement formula.