Algebraic Expressions and Properties

Anatomy of an Algebraic Expression

When you first learn algebra, expressions can look like a new language. The good news is that algebraic expressions are built from a few important parts. Once you know the parts, the whole expression becomes much easier to understand.

In this lesson, you will learn how to identify the variables, constants, coefficients, and terms in an algebraic expression. These are the basic building blocks of algebra.

What is an algebraic expression?

An algebraic expression is a math phrase that can include numbers, variables, and operation symbols such as addition, subtraction, multiplication, and division. An expression does not have an equals sign.

Examples of algebraic expressions are:

• \(3x + 5\)
• \(7n - 2\)
• \(4a + 3b\)
• \(12 - y\)

Notice that none of these have an equals sign. If there were an equals sign, it would be an equation instead of just an expression.

Main parts of an algebraic expression

Let’s look at the expression \(5x + 3\). This expression has several parts, and each part has a name.

1. Variable

A variable is a letter that stands for a number. The value of the variable can change.

In \(5x + 3\), the variable is \(x\).

2. Coefficient

A coefficient is the number multiplied by a variable.

In \(5x + 3\), the coefficient of \(x\) is \(5\), because \(5x\) means \(5 \times x\).

If a variable appears by itself, like \(x\), its coefficient is \(1\), because \(x = 1x\).

3. Constant

A constant is a number all by itself. It does not have a variable attached to it.

In \(5x + 3\), the constant is \(3\).

4. Term

A term is one part of an expression. Terms are separated by addition or subtraction signs.

In \(5x + 3\), the terms are \(5x\) and \(3\).

This means:

• Variable: \(x\)
• Coefficient: \(5\)
• Constant: \(3\)
• Terms: \(5x\), \(3\)

How to spot terms

Terms are separated by plus or minus signs. This is very important.

For example, in the expression

the terms are:

• \(8m\)
• \(-4\)
• \(2n\)

The minus sign belongs with the term after it, so \(-4\) is one term.

More than one variable

Some expressions have more than one variable. For example:

In this expression:

• The variables are \(a\) and \(b\).
• The coefficient of \(a\) is \(3\).
• The coefficient of \(b\) is \(2\).
• The constant is \(-7\).
• The terms are \(3a\), \(2b\), and \(-7\).

Important note about subtraction

In algebra, subtraction can make a term negative. For example, in \(x - 6\), the terms are \(x\) and \(-6\), not just \(6\).

In \(4p - 9q + 2\), the terms are:

• \(4p\)
• \(-9q\)
• \(2\)

That means the coefficient of \(q\) is \(-9\), not \(9\).

Worked Example 1

Find the variables, coefficients, constant, and terms in:

Step 1: Identify the terms.

The terms are separated by the plus sign, so the terms are \(6x\) and \(10\).

Step 2: Find the variable.

The variable is \(x\).

Step 3: Find the coefficient.

The coefficient is the number multiplying the variable. In \(6x\), the coefficient is \(6\).

Step 4: Find the constant.

The constant is the number without a variable, so it is \(10\).

Answer:

• Variable: \(x\)
• Coefficient: \(6\)
• Constant: \(10\)
• Terms: \(6x\), \(10\)

Worked Example 2

Find the variables, coefficients, constant, and terms in:

Step 1: Identify the terms.

The terms are \(9y\) and \(-4\).

Step 2: Find the variable.

The variable is \(y\).

Step 3: Find the coefficient.

The coefficient of \(y\) is \(9\).

Step 4: Find the constant.

The constant is \(-4\).

Answer:

• Variable: \(y\)
• Coefficient: \(9\)
• Constant: \(-4\)
• Terms: \(9y\), \(-4\)

Worked Example 3

Find the variables, coefficients, constant, and terms in:

Step 1: Identify the terms.

The terms are \(3a\), \(5b\), and \(-12\).

Step 2: Find the variables.

The variables are \(a\) and \(b\).

Step 3: Find the coefficients.

The coefficient of \(a\) is \(3\), and the coefficient of \(b\) is \(5\).

Step 4: Find the constant.

The constant is \(-12\).

Answer:

• Variables: \(a\), \(b\)
• Coefficients: \(3\) for \(a\), \(5\) for \(b\)
• Constant: \(-12\)
• Terms: \(3a\), \(5b\), \(-12\)

Worked Example 4

Find the variables, coefficients, constant, and terms in:

Step 1: Identify the terms.

The terms are \(x\), \(-7\), and \(4z\).

Step 2: Find the variables.

The variables are \(x\) and \(z\).

Step 3: Find the coefficients.

The coefficient of \(x\) is \(1\), because \(x = 1x\). The coefficient of \(z\) is \(4\).

Step 4: Find the constant.

The constant is \(-7\).

Answer:

• Variables: \(x\), \(z\)
• Coefficients: \(1\) for \(x\), \(4\) for \(z\)
• Constant: \(-7\)
• Terms: \(x\), \(-7\), \(4z\)

Common mistakes to avoid

• Forgetting that subtraction creates a negative term: In \(5x - 2\), the constant is \(-2\), not \(2\).
• Forgetting the invisible 1: In \(n + 8\), the coefficient of \(n\) is \(1\).
• Mixing up terms and factors: In \(4x\), this is one term, not two terms. The number \(4\) and the variable \(x\) are multiplied together.
• Calling every number a constant: A number attached to a variable, like the \(5\) in \(5x\), is a coefficient, not a constant.

Quick check

Try identifying the parts of this expression:

You should find:

• Variables: \(m\), \(n\)
• Coefficients: \(7\) and \(2\)
• Constant: \(-5\)
• Terms: \(7m\), \(2n\), \(-5\)

Summary

An algebraic expression is made of different parts. A variable is a letter that stands for a number. A coefficient is the number multiplying a variable. A constant is a number by itself. Terms are the parts of the expression separated by plus or minus signs.

If you can break an expression into terms and then look for variables, coefficients, and constants, you can understand the anatomy of almost any algebraic expression.

Translating Real-World Constraints into Algebra

In math, a constraint is a rule or condition that must be true.

In real life, we hear constraints in sentences like:

• "You can spend at most $20."
• "The total number of players is 15."
• "The length is 3 more than the width."
• "You must read at least 30 pages."

Algebra helps us turn these words into math symbols. This is called translating real-world constraints into algebra.

When we translate words into algebra, we often use:

• Variables to stand for unknown numbers
• Expressions to show relationships
• Equations to show that two amounts are equal
• Inequalities to show limits like greater than, less than, at least, or at most

Learning this skill helps you understand word problems and model real situations with math.

Step 1: Choose a variable

A variable is a letter that stands for a number we do not know yet.

For example:

• Let \(x\) = number of notebooks
• Let \(m\) = number of miles walked
• Let \(w\) = width of a rectangle

Always decide what your variable means before writing an equation or inequality.

Step 2: Look for clue words

Word problems often contain clue words that tell you which operation or symbol to use.

• Sum, total, more than usually mean addition
• Difference, less than, fewer than usually mean subtraction
• Times, product, twice, of usually mean multiplication
• Per, each, quotient usually mean division

Some clue words tell you about a limit:

• At least means greater than or equal to: \(\ge\)
• At most means less than or equal to: \(\le\)
• More than means \(>\)
• Less than means \(<\)
• Exactly means equal to: \(=\)

Important: Be careful with phrases like less than. The order can matter.

For example:

• "5 less than a number" means \(x - 5\)
• "A number is less than 5" means \(x < 5\)

Step 3: Decide whether you need an expression, equation, or inequality

Use an expression when you are only describing a quantity.

Example: "Three more than a number" becomes \(x + 3\).

Use an equation when two amounts are equal.

Example: "The total cost is 18 dollars" becomes something like \(3x = 18\).

Use an inequality when there is a limit or boundary.

Example: "You can spend at most 18 dollars" becomes something like \(3x \le 18\).

Main idea 1: Translating totals and combinations

Many real-world problems combine two or more parts into one total.

Words like total, combined, and altogether often lead to addition.

If one amount depends on another, use a variable and write each part carefully.

Worked Example 1: Total number of tickets

An event sold child tickets and adult tickets. Let \(c\) be the number of child tickets sold. The number of adult tickets sold was 12 more than the number of child tickets. Write an expression for the total number of tickets sold.

Step A: Define the variable.

Let \(c\) = number of child tickets.

Step B: Write the other amount.

Adult tickets = \(c + 12\).

Step C: Add to find the total.

Simplify:

Answer: The total number of tickets sold is \(2c + 12\).

Main idea 2: Translating rate situations

A rate compares two different kinds of quantities, such as dollars per hour, miles per hour, or cost per item.

In many rate problems:

• Cost = price per item \(\times\) number of items
• Distance = speed \(\times\) time
• Pay = hourly rate \(\times\) number of hours

If there is a starting fee, add it after multiplying.

Worked Example 2: Budget with a rate

A bike rental shop charges \(\$6\) per hour. Mia has no more than \(\$24\) to spend. Let \(h\) be the number of hours she can rent the bike. Write an inequality.

Step A: Define the variable.

Let \(h\) = number of hours.

Step B: Write the cost expression.

Since the bike costs \(\$6\) per hour, the total cost is \(6h\).

Step C: Translate the constraint.

"No more than \(\$24\)" means at most \(24\), so use \(\le\).

Answer: The inequality is \(6h \le 24\).

This inequality shows that the cost must stay at or below 24 dollars.

Main idea 3: Translating geometric constraints

Geometry problems often describe lengths, widths, perimeters, and areas using words.

You can translate these descriptions into algebra by writing each measurement with a variable.

Common geometry facts for 7th Grade include:

• Perimeter of a rectangle: $$P = 2l + 2w$$
• Area of a rectangle: $$A = lw$$

If one side depends on another, write that relationship first.

Worked Example 3: Rectangle perimeter

The length of a rectangle is 4 units more than its width. The perimeter is 28 units. Write an equation.

Step A: Choose a variable.

Let \(w\) = width.

Step B: Write the length.

Length = \(w + 4\).

Step C: Use the perimeter formula.

Substitute the expressions:

Answer: The equation is $$2(w + 4) + 2w = 28$$

This equation represents the geometric constraint in algebra form.

Main idea 4: Translating “at least” and “at most” situations

Many real-world constraints are not exact totals. Instead, they set a minimum or maximum.

These are modeled with inequalities.

• "At least 10" means $$x \ge 10$$
• "At most 10" means $$x \le 10$$
• "More than 10" means $$x > 10$$
• "Less than 10" means $$x < 10$$

Worked Example 4: Reading goal

Jordan wants to read at least 150 pages this week. He has already read 42 pages. Let \(p\) be the number of pages he still needs to read. Write an inequality.

Step A: Define the variable.

Let \(p\) = pages Jordan still needs to read.

Step B: Write the total pages read by the end of the week.

He already read 42 pages, and he will read \(p\) more pages, so the total is \(42 + p\).

Step C: Translate "at least 150".

"At least" means \(\ge\).

Answer: The inequality is \(42 + p \ge 150\).

How to check your translation

After writing algebra from words, check your work by asking:

• Did I define my variable clearly?
• Did I use the correct operation: add, subtract, multiply, or divide?
• Did I choose equation or inequality correctly?
• Do words like "at least" or "at most" match my symbol?
• Does my algebra make sense in the real situation?

Common mistakes to avoid

• Forgetting to define the variable. Always say what the letter means.
• Using the wrong inequality symbol. Remember: at least \(\ge\), at most \(\le\).
• Mixing up word order. "3 less than a number" is \(x - 3\), not \(3 - x\).
• Leaving out part of the total. If there is a starting amount and an added amount, include both.
• Confusing an expression with an equation. An expression has no equal sign. An equation does.

Try thinking through these quick translations

1. "A number increased by 9"

   Expression: \(x + 9\)

2. "The total cost for \(n\) movie tickets at \(\$8\) each is no more than \(\$40\)"

   Inequality: \(8n \le 40\)

3. "The length of a garden is 5 feet longer than its width"

   If width is \(w\), then length is \(w + 5\)

4. "A runner jogs 3 miles per day for \(d\) days"

   Expression: \(3d\)

Summary

Translating real-world constraints into algebra means turning words into math symbols.

Start by choosing a variable. Then look for clue words that tell you what operations or symbols to use. Decide whether the situation needs an expression, an equation, or an inequality.

Remember that real-world constraints often include totals, rates, and limits. With practice, you can model these situations clearly and correctly in algebra.

Evaluating Algebraic Expressions means finding the value of an expression when you know what the variables stand for.

For example, in the expression \(x + 5\), if \(x = 3\), then you replace \(x\) with \(3\). The expression becomes \(3 + 5\), which equals \(8\).

This is an important algebra skill because variables can stand for different numbers. Once you know the value of the variable, you can calculate the value of the whole expression.

What is a variable?

A variable is a letter that stands for a number. Common variables are \(x\), \(y\), and \(n\).

• In \(x + 4\), the variable is \(x\).
• In \(3m - 2\), the variable is \(m\).
• In \(a + b\), the variables are \(a\) and \(b\).

What is an algebraic expression?

An algebraic expression is a math phrase that includes numbers, variables, and operations such as addition, subtraction, multiplication, and division.

• \(x + 7\)
• \(4n\)
• \(2a - 3b\)
• \(\frac{m}{5} + 1\)

How to evaluate an algebraic expression

To evaluate an expression, follow these steps:

1. Substitute the given value for each variable.
2. Use parentheses around substituted numbers, especially if a number is negative.
3. Simplify using the order of operations.

Order of operations tells us the correct order to solve a math expression:

1. Parentheses
2. Exponents if there are any
3. Multiplication and division from left to right
4. Addition and subtraction from left to right

Even if there are no exponents, it is still important to multiply or divide before adding or subtracting.

Important note about multiplication

In algebra, writing a number right next to a variable means multiplication.

• \(3x\) means \(3 \times x\)
• \(ab\) means \(a \times b\)
• \(2(y + 4)\) means \(2 \times (y + 4)\)

Worked Example 1: One variable, one operation

Evaluate \(x + 9\) when \(x = 4\).

Step 1: Substitute.

Step 2: Simplify.

Answer: \(13\)

Worked Example 2: One variable, multiplication and subtraction

Evaluate \(5n - 6\) when \(n = 3\).

Step 1: Substitute.

Step 2: Multiply first.

Now the expression is:

Step 3: Subtract.

Answer: \(9\)

Worked Example 3: Two variables

Evaluate \(2a + 3b\) when \(a = 4\) and \(b = 2\).

Step 1: Substitute both values.

Step 2: Multiply.

Now the expression is:

Step 3: Add.

Answer: \(14\)

Worked Example 4: Using parentheses and order of operations

Evaluate \(3(x + 2)\) when \(x = 5\).

Step 1: Substitute.

Step 2: Solve inside the parentheses first.

Now the expression is:

Step 3: Multiply.

Answer: \(21\)

Be careful with negative numbers

If a variable equals a negative number, always put the number in parentheses when you substitute.

For example, evaluate \(4 - y\) when \(y = -2\).

Substitute carefully:

Subtracting a negative is the same as adding:

If you forget the parentheses, it is easy to make a mistake.

Common mistakes to avoid

• Forgetting to substitute every variable. If an expression has two variables, replace both.
• Ignoring order of operations. Multiply and divide before adding and subtracting.
• Forgetting that a number next to a variable means multiply. For example, \(4x\) means \(4 \times x\), not \(4 + x\).
• Leaving out parentheses for negative numbers. Write \((-3)\), not just \(-3\), when substituting into an expression.

Quick check

Try these on your own:

• Evaluate \(m + 8\) when \(m = 7\).
• Evaluate \(6p\) when \(p = 4\).
• Evaluate \(2x + y\) when \(x = 3\) and \(y = 5\).
• Evaluate \(2(n - 1)\) when \(n = 6\).

Answers

• \(7 + 8 = 15\)
• \(6(4) = 24\)
• \(2(3) + 5 = 6 + 5 = 11\)
• \(2(6 - 1) = 2(5) = 10\)

Summary

To evaluate an algebraic expression, replace each variable with its given value and then simplify. Use parentheses when substituting, especially for negative numbers. Finally, follow the order of operations carefully to find the correct value.

Properties of Real Numbers help us understand how numbers behave when we add, subtract, multiply, and divide. These properties are important in algebra because they let us rewrite expressions in easier ways without changing their value.

When you start working with variables like \(x\) and \(y\), these properties become very useful. They help you reorder numbers, group numbers, and recognize what does or does not change an expression.

In this lesson, you will learn the commutative property, associative property, identity property, zero property, and inverse property. You will also see how to use them in numerical and algebraic expressions.

1. Commutative Property

The word commutative means that numbers can change places, and the answer stays the same.

This property works for addition and multiplication.

• Addition: \(a+b=b+a\)
• Multiplication: \(a\cdot b=b\cdot a\)

Examples:

• \(3+5=5+3\)
• \(4\cdot 7=7\cdot 4\)
• \(x+9=9+x\)
• \(2y= y\cdot 2\)

Important: The commutative property does not work for subtraction or division.

• \(8-3\neq 3-8\)
• \(12\div 4\neq 4\div 12\)

2. Associative Property

The word associative means that numbers can be grouped in different ways, and the answer stays the same.

This property also works for addition and multiplication.

• Addition: \((a+b)+c=a+(b+c)\)
• Multiplication: \((a\cdot b)\cdot c=a\cdot (b\cdot c)\)

Examples:

• \((2+6)+4=2+(6+4)\)
• \((3\cdot 5)\cdot 2=3\cdot (5\cdot 2)\)
• \((x+3)+7=x+(3+7)\)

Important: The associative property does not work for subtraction or division.

For example,

Since \(4\neq 8\), subtraction is not associative.

3. Identity Property

An identity number is a number that keeps another number the same.

There are two identity properties:

• Additive Identity: Adding zero does not change a number.
• Multiplicative Identity: Multiplying by one does not change a number.

Formulas:

• \(a+0=a\)
• \(a\cdot 1=a\)

Examples:

• \(12+0=12\)
• \(x+0=x\)
• \(9\cdot 1=9\)
• \(1\cdot y=y\)

4. Zero Property

The zero property of multiplication tells us that any number multiplied by zero equals zero.

Formula:

Examples:

• \(7\cdot 0=0\)
• \(0\cdot 15=0\)
• \(x\cdot 0=0\)

This is different from the additive identity. With addition, adding zero keeps the number the same. With multiplication, multiplying by zero makes the result zero.

5. Inverse Property

An inverse “undoes” a number.

There are two common inverse ideas:

• Additive Inverse: A number plus its opposite equals zero.
• Multiplicative Inverse: A number times its reciprocal equals one.

Formulas:

• \(a+(-a)=0\)
• \(a\cdot \frac{1}{a}=1\), for \(a\neq 0\)

Examples:

• \(5+(-5)=0\)
• \(-8+8=0\)
• \(4\cdot \frac{1}{4}=1\)
• \(\frac{3}{5}\cdot \frac{5}{3}=1\)

In 7th Grade, you will often use the additive inverse when combining integers, such as \(9+(-9)=0\).

Why These Properties Matter in Algebra

In algebra, expressions can look complicated. Properties help you rewrite expressions in smarter ways.

For example, in \(5+x\), you can switch the order to write \(x+5\) using the commutative property of addition. This can make expressions easier to read or compare.

You can also regroup terms to make mental math easier. For example, in \((8+2)+7\), it is helpful to regroup as \(8+(2+7)\) so you can quickly add \(2+7=9\).

Worked Example 1: Using the Commutative Property

Rewrite \(6+a\) with the variable first.

Step 1: Notice that this is addition.

Step 2: Use the commutative property to switch the order.

Answer: \(a+6\)

Worked Example 2: Using the Associative Property

Rewrite \((4+9)+1\) in a different grouping, then find the sum.

Step 1: Use the associative property of addition.

Step 2: Add inside the parentheses.

Answer: \(14\)

Worked Example 3: Identifying a Property

What property is shown by this statement?

Step 1: Multiplying by \(1\) keeps the value the same.

Step 2: This matches the multiplicative identity property.

Answer: Identity Property of Multiplication

Worked Example 4: Combining Properties

Simplify and name the properties used:

Step 1: Regroup using the associative property.

Step 2: Add inside the parentheses.

So,

Step 3: Recognize that a number and its opposite make zero.

Answer: The expression simplifies to \(0\). This uses the associative property and the inverse property.

How to Tell the Properties Apart

• Commutative: changes the order
• Associative: changes the grouping
• Identity: keeps the number the same
• Zero Property: multiplying by zero gives zero
• Inverse: combines a number with its opposite or reciprocal to make \(0\) or \(1\)

Common Mistakes to Avoid

• Do not use the commutative property with subtraction or division.
• Do not use the associative property with subtraction or division.
• Remember that \(a+0=a\), but \(a\cdot 0=0\).
• Do not confuse identity with inverse. Identity keeps a number the same, but inverse combines with a number to make \(0\) or \(1\).

Quick Check

1. Name the property: \(9+m=m+9\)
2. Name the property: \((2\cdot 5)\cdot x=2\cdot (5\cdot x)\)
3. Name the property: \(y+0=y\)
4. Name the property: \(8\cdot 0=0\)
5. Name the property: \(11+(-11)=0\)

Answers:

1. Commutative Property of Addition
2. Associative Property of Multiplication
3. Identity Property of Addition
4. Zero Property of Multiplication
5. Inverse Property of Addition

Summary

The properties of real numbers help you work with numbers and variables more easily. The commutative property changes order, the associative property changes grouping, the identity property keeps a number the same, the zero property tells what happens when multiplying by zero, and the inverse property shows how numbers can undo each other.

When solving algebra problems, these properties help you rewrite expressions in ways that are simpler and easier to understand. Learning to recognize them will make you stronger in both arithmetic and algebra.

Combining Like Terms is a way to simplify an algebraic expression. When we simplify, we rewrite the expression so it is easier to read and work with, but it still means the same thing.

Before learning how to combine like terms, it helps to understand what a term is. A term is a number, a variable, or a number multiplied by a variable. In the expression \(3x + 5 - 2x + 7\), the terms are \(3x\), \(5\), \(-2x\), and \(7\).

A like term has the same variable part. That means the same letters raised to the same powers. For example, \(4x\) and \(-9x\) are like terms because they both have \(x\). Also, \(3y^2\) and \(10y^2\) are like terms because they both have \(y^2\).

Terms that do not have the same variable part are unlike terms. For example, \(5x\) and \(5y\) are not like terms. Also, \(2x\) and \(2x^2\) are not like terms because \(x\) and \(x^2\) are different.

Think of like terms as objects of the same type. You can add 3 apples and 2 apples to get 5 apples. But you cannot combine 3 apples and 2 oranges into 5 of the same thing. In algebra, \(3x + 2x = 5x\), but \(3x + 2y\) cannot be combined.

When you combine like terms, you only add or subtract the coefficients. The coefficient is the number in front of the variable. In \(6x\), the coefficient is 6. In \(-3y\), the coefficient is \(-3\).

Here is the basic idea:

• \(2x + 5x = 7x\)
• \(9a - 4a = 5a\)
• \(3m + 2n\) cannot be combined because \(m\) and \(n\) are different variables.

Constants are also like terms with each other. A constant is a number with no variable. For example, \(4\) and \(-9\) are like terms, so they can be combined.

For example:

Now let’s look at a step-by-step method for combining like terms.

1. Find the terms. Look at the parts separated by plus or minus signs.
2. Group like terms. Put terms with the same variable part together.
3. Add or subtract the coefficients.
4. Write the simplified expression.

Worked Example 1

Simplify \(4x + 3x\).

Both terms have \(x\), so they are like terms. Add the coefficients:

So,

Worked Example 2

Simplify \(6y - 2y + 5\).

The terms are \(6y\), \(-2y\), and \(5\).

\(6y\) and \(-2y\) are like terms, so combine them:

The constant \(5\) stays the same.

So the simplified expression is:

Worked Example 3

Simplify \(3a + 7 - 5a + 2\).

First, group like terms:

• Variable terms: \(3a\) and \(-5a\)
• Constants: \(7\) and \(2\)

Now combine each group:

So the simplified expression is:

Worked Example 4

Simplify \(2x + 4 + 5x - 3 + x\).

First, identify the like terms.

• \(2x\), \(5x\), and \(x\) are like terms.
• \(4\) and \(-3\) are like terms.

Remember that \(x\) means \(1x\).

Now combine the \(x\)-terms:

Combine the constants:

So the simplified expression is:

Here are some important reminders:

• Only combine like terms. The variable part must match exactly.
• Keep the variable part the same. For example, \(3x + 2x = 5x\), not \(5x^2\).
• Watch the signs. A minus sign belongs to the term after it. For example, \(7k - 10k = -3k\).
• Remember invisible 1s. If you see \(x\), it means \(1x\).

Let’s look at what cannot be combined:

• \(2x + 3y\) cannot be combined because \(x\) and \(y\) are different.
• \(4x + 6x^2\) cannot be combined because \(x\) and \(x^2\) are different.
• \(5ab + 2a\) cannot be combined because \(ab\) and \(a\) are different variable parts.

A good way to check yourself is to ask: Do these terms have exactly the same variable part? If yes, combine them. If not, leave them separate.

You can also use combining like terms to rewrite expressions in a cleaner order. Usually, variable terms are written first, followed by constants. For example, instead of \(5 + 3x\), we often write \(3x + 5\).

Let’s try a few quick practice ideas:

• \(7m + 2m = 9m\)
• \(10p - p = 9p\)
• \(4 + 8 - 1 = 11\)
• \(3x + 2 + 4x + 6 = 7x + 8\)

Summary

Combining like terms means adding or subtracting terms that have the same variable part. You combine the coefficients and keep the variable part unchanged. Constants can be combined with other constants. If the variable parts do not match exactly, the terms cannot be combined.

Distributive Property Expansion is a way to multiply one term by everything inside parentheses.

This is an important algebra skill because it helps us rewrite expressions in a simpler form. When we expand an expression, we remove the parentheses by multiplying correctly.

The distributive property says:

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

This means the number or variable outside the parentheses must be multiplied by each term inside the parentheses.

If there is subtraction inside the parentheses, the rule is very similar:

$$a(b-c)=ab-ac$$

So the outside term still goes to every term inside.

Main idea: Multiply the outside factor by every term inside the parentheses.

• Multiply coefficients.
• Multiply variables when needed.
• Keep track of positive and negative signs.
• Then write the new expression without parentheses.

Here is a simple example of the pattern:

$$3(x+4)=3\cdot x+3\cdot 4=3x+12$$

The 3 is distributed to both the 52 and the 4.

It can help to draw arrows from the outside term to each term inside the parentheses. This reminds you not to miss any term.

Why this works: If you think of parentheses as a grouped amount, then multiplying by the group means multiplying every part of the group.

For example, if you have 2 groups of 5 d + 3 d, then you really have:

$$2(d+3)=d+3+d+3=2d+6$$

This matches the distributive property.

Worked Example 1: Positive whole number outside parentheses

Expand:

$$4(x+5)$$

Step 1: Multiply 4 by each term inside the parentheses.

$$4\cdot x+4\cdot 5$$

Step 2: Simplify each product.

$$4x+20$$

Answer: $$4(x+5)=4x+20$$

Worked Example 2: Subtraction inside parentheses

Expand:

$$6(y-2)$$

Step 1: Distribute 6 to both terms.

$$6\cdot y+6\cdot (-2)$$

Step 2: Multiply carefully.

$$6y-12$$

Answer: $$6(y-2)=6y-12$$

Notice that subtracting 2 means the second term is negative. That is why the result is 56y-1252, not 56y+1252.

Worked Example 3: Negative coefficient outside parentheses

Expand:

$$-3(2m+7)$$

Step 1: Multiply 5-352 by each term.

$$(-3)\cdot 2m+(-3)\cdot 7$$

Step 2: Multiply.

$$-6m-21$$

Answer: $$-3(2m+7)=-6m-21$$

When the outside number is negative, each product may change sign. Be extra careful with signs.

Worked Example 4: Fractional coefficient

Expand:

$$\frac{1}{2}(8n-6)$$

Step 1: Distribute \(\frac{1}{2}\) to each term.

$$\frac{1}{2}\cdot 8n+\frac{1}{2}\cdot (-6)$$

Step 2: Multiply.

$$4n-3$$

Answer: $$\frac{1}{2}(8n-6)=4n-3$$

Fractions work the same way as whole numbers. The outside factor still multiplies every term inside.

How to multiply variables

If the outside factor and inside term both have variables, multiply the numbers and then the variables.

Example:

$$2x(3x+4)$$

Distribute \(2x\):

$$2x\cdot 3x+2x\cdot 4$$

Now simplify:

$$6x^2+8x$$

You may see this type of problem as you practice more. The same distributive idea still works.

Common mistakes to avoid

• Forgetting a term: Every term inside the parentheses must be multiplied.
• Sign mistakes: A negative outside the parentheses changes signs when multiplied.
• Only multiplying the first term: For example, \(5(a+2)\neq 5a+2\). The correct expansion is \(5a+10\).
• Rushing with fractions: Multiply the fraction by each term carefully.

Helpful sign reminders

• Positive times positive = positive
• Positive times negative = negative
• Negative times positive = negative
• Negative times negative = positive

Quick check examples

Try to think through these:

• $$2(a+9)=2a+18$$
• $$5(k-1)=5k-5$$
• $$-4(p-3)=-4p+12$$
• $$\frac{3}{4}(8x+12)=6x+9$$

Steps you can follow every time

1. Look at the factor outside the parentheses.
2. Multiply it by the first term inside.
3. Multiply it by the second term inside.
4. If there are more terms, keep going until every term is used.
5. Simplify the products.
6. Check your signs.

Summary

The distributive property helps you expand expressions by multiplying the outside factor by every term inside the parentheses.

Remember the patterns:

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

$$a(b-c)=ab-ac$$

If the outside factor is negative or a fraction, the rule stays the same. The most important thing is to multiply every term and watch the signs carefully.

Factoring Linear Expressions means rewriting an expression as a product. In 7th grade, this usually means finding the Greatest Common Factor (GCF) of all the terms and pulling it out.

Factoring is the reverse of the distributive property. For example, if you know that

$$3(x+2)=3x+6,$$

then factoring goes backward:

$$3x+6=3(x+2).$$

In this lesson, you will learn how to factor linear expressions by finding the GCF and using the distributive property in reverse.

Why do we factor?

• It helps us rewrite expressions in a simpler, more organized form.
• It shows the parts that are being multiplied.
• It helps later when solving equations and working with algebra.

Step 1: Find the greatest common factor.

The GCF is the largest factor that every term has in common.

For numbers, look for the biggest number that divides each term. For example, in \(8x\) and \(12\), the GCF of 8 and 12 is 4.

If the terms have the same variable, that variable can also be part of the GCF. For example, in \(6x\) and \(9x\), both terms have \(x\), so the GCF is \(3x\).

Step 2: Divide each term by the GCF.

After you find the GCF, divide each term by it. The answers go inside parentheses.

For example, to factor \(10x+15\):

• The GCF of 10 and 15 is 5.
• Divide each term by 5: \(10x \div 5=2x\) and \(15 \div 5=3\).
• So, \(10x+15=5(2x+3)\).

Step 3: Check by distributing.

A great way to check your factoring is to distribute and see if you get the original expression back.

For example:

$$5(2x+3)=10x+15$$

Since it matches the original expression, the factoring is correct.

Important idea: Every term must be divided by the GCF.

If an expression has two or three terms, the factor you pull out must be common to all of them.

For example, in \(12x+18-6\), the numbers 12, 18, and 6 all share a GCF of 6. So 6 can be factored out of every term.

Worked Example 1

Factor:

$$8x+12$$

Step A: Find the GCF of 8 and 12. It is 4.

Step B: Divide each term by 4.

• \(8x \div 4=2x\)
• \(12 \div 4=3\)

Answer:

$$8x+12=4(2x+3)$$

Check:

$$4(2x+3)=8x+12$$

Worked Example 2

Factor:

$$6x+9x$$

Both terms have a number factor and a variable factor in common.

Step A: The GCF of 6 and 9 is 3. Both terms also have \(x\). So the GCF is \(3x\).

Step B: Divide each term by \(3x\).

• \(6x \div 3x=2\)
• \(9x \div 3x=3\)

Answer:

$$6x+9x=3x(2+3)$$

This is a correct factored form because distributing gives the original expression back.

Check:

$$3x(2+3)=6x+9x$$

Worked Example 3

Factor:

$$15x+20$$

Step A: The GCF of 15 and 20 is 5.

Step B: Divide each term by 5.

• \(15x \div 5=3x\)
• \(20 \div 5=4\)

Answer:

$$15x+20=5(3x+4)$$

Check:

$$5(3x+4)=15x+20$$

Worked Example 4

Factor:

$$12x+18-6$$

This expression has three terms, so we need a factor common to all three.

Step A: The GCF of 12, 18, and 6 is 6.

Step B: Divide each term by 6.

• \(12x \div 6=2x\)
• \(18 \div 6=3\)
• \(-6 \div 6=-1\)

Answer:

$$12x+18-6=6(2x+3-1)$$

Check:

$$6(2x+3-1)=12x+18-6$$

How to know if an expression is fully factored

An expression is fully factored when there is no greater common factor left inside and outside the parentheses.

For example, \(12x+16\) can be factored as \(2(6x+8)\), but it is not fully factored because 6 and 8 still share a factor of 2.

The best factored form is:

$$12x+16=4(3x+4)$$

Common mistakes to avoid

• Not using the greatest common factor. Always look for the largest one.
• Forgetting a term. Every term must be divided by the factor you pull out.
• Dropping the sign. If a term is negative, keep the negative sign when dividing.
• Not checking your work. Distribute to make sure your factored form matches the original expression.

Helpful strategy

1. Circle the terms in the expression.
2. Find the biggest factor all terms share.
3. Write that factor outside parentheses.
4. Divide each term by the factor and write the results inside parentheses.
5. Check by distributing.

Try thinking through these on your own:

• \(14x+21=7(2x+3)\)
• \(9x+3=3(3x+1)\)
• \(20x+10-5=5(4x+2-1)\)

Summary

Factoring linear expressions means using the distributive property backward. You find the greatest common factor of all the terms, pull it out, and place what is left inside parentheses.

Always make sure every term is divided by the GCF, and check your answer by distributing. With practice, factoring becomes a quick way to rewrite expressions clearly and correctly.

Proving Expression Equivalence means deciding whether two algebraic expressions always have the same value.

Two expressions are equivalent if they match for every value of the variable. If they only match sometimes, then they are not equivalent.

For example, the expressions \(2(x+3)\) and \(2x+6\) are equivalent because no matter what number you use for \(x\), both expressions give the same result.

This idea is important because in algebra, expressions can look different but still mean the same thing. Your job is to learn how to prove when that is true and when it is not.

There are two main ways to check expression equivalence:

• Simplify or rewrite the expressions using algebra rules.
• Test values for the variable. This can help disprove equivalence if you find a value that gives different results.

Important: Testing values can show that two expressions are not equivalent if you find even one counterexample. But testing a few values does not fully prove that two expressions are equivalent for all values. To prove equivalence, it is best to use algebra rules.

Useful algebra rules for proving equivalence:

• Distributive property: \(a(b+c)=ab+ac\)
• Combine like terms: \(3x+2x=5x\)
• Commutative property: order can change in addition or multiplication, such as \(x+5=5+x\)
• Associative property: grouping can change in addition or multiplication, such as \((x+2)+3=x+(2+3)\)

When working with expressions, always look for ways to rewrite them into a simpler form. If both expressions simplify to the same result, then they are equivalent.

Step-by-step method

1. Look at both expressions carefully.
2. Simplify each one by distributing, combining like terms, or rearranging terms.
3. Compare the simplified forms.
4. If the simplified forms are the same, the expressions are equivalent.
5. If they are different, try a test value. If one value gives different answers, they are not equivalent.

Worked Example 1: A basic equivalence

Are \(3(x+4)\) and \(3x+12\) equivalent?

Step 1: Simplify the first expression.

Use the distributive property:

$$3(x+4)=3x+12$$

Step 2: Compare.

The second expression is already \(3x+12\).

So both expressions are the same after simplifying.

Conclusion: \(3(x+4)\) and \(3x+12\) are equivalent.

Worked Example 2: Combining like terms

Are \(5x+2+x\) and \(6x+2\) equivalent?

Step 1: Combine like terms in the first expression.

$$5x+2+x = 5x+x+2 = 6x+2$$

Step 2: Compare.

The second expression is \(6x+2\).

Conclusion: \(5x+2+x\) and \(6x+2\) are equivalent.

Worked Example 3: Expressions that are not equivalent

Are \(2(x+5)\) and \(2x+5\) equivalent?

Step 1: Simplify the first expression.

$$2(x+5)=2x+10$$

Step 2: Compare.

We get \(2x+10\), but the second expression is \(2x+5\).

These are not the same.

Step 3: Use a test value to confirm.

Let \(x=1\).

• First expression: \(2(1+5)=2(6)=12\)
• Second expression: \(2(1)+5=7\)

The values are different.

Conclusion: \(2(x+5)\) and \(2x+5\) are not equivalent.

This test value is called a counterexample. A counterexample is one value that shows two expressions are not always equal.

Worked Example 4: More than one step

Are \(4(y+2)-3y\) and \(y+8\) equivalent?

Step 1: Distribute.

$$4(y+2)-3y = 4y+8-3y$$

Step 2: Combine like terms.

$$4y+8-3y = y+8$$

Step 3: Compare.

The second expression is \(y+8\).

Conclusion: \(4(y+2)-3y\) and \(y+8\) are equivalent.

How to disprove equivalence

Sometimes two expressions look similar, but they are not the same. If simplifying does not make them match, you can disprove equivalence by finding a counterexample.

For example, compare \(x+x\) and \(x^2\).

If \(x=2\):

• \(x+x = 2+2 = 4\)
• \(x^2 = 2^2 = 4\)

These match for \(x=2\), but that does not prove they are equivalent.

Try \(x=3\):

• \(x+x = 3+3 = 6\)
• \(x^2 = 3^2 = 9\)

Now they are different.

Conclusion: \(x+x\) and \(x^2\) are not equivalent.

This is a good reminder: one matching value is not enough. Equivalent expressions must match for all values.

Common mistakes to avoid

• Forgetting to distribute to every term. For example, \(3(x+2)\) is \(3x+6\), not \(3x+2\).
• Combining unlike terms. For example, \(3x+2\) cannot become \(5x\).
• Thinking one test value proves equivalence. A few test values can suggest a pattern, but algebraic rewriting is the best proof.
• Mixing up \(x+x\) and \(x^2\). These are different expressions.

Quick practice ideas

Try deciding whether each pair is equivalent:

• \(2a+3a\) and \(5a\)
• \(5(m+1)\) and \(5m+1\)
• \(7+n+2\) and \(n+9\)
• \(3(k+4)-k\) and \(2k+12\)

As you work, simplify both sides and compare. If they do not match, test a value to find a counterexample.

Summary

To prove two expressions are equivalent, rewrite them using algebra rules like the distributive property and combining like terms. If both expressions simplify to the same form, they are equivalent.

To disprove equivalence, find a counterexample: one value of the variable that makes the expressions give different answers. Remember, equivalent expressions must be equal for every value, not just one or two.