Algebraic Expressions

Translating Verbal Phrases into Algebraic Expressions

In math, we often use words to describe number relationships. We can also write those same ideas using algebraic expressions. Learning how to change words into math symbols helps you solve problems more easily.

An algebraic expression is a math phrase that includes numbers, operation symbols, and sometimes a variable. A variable is a letter that stands for an unknown number.

For example, if we do not know how many stickers Ava has, we can use the variable \(s\). If Ava gets 4 more stickers, we can write that as \(s + 4\).

Why is this important? Many word problems in math use phrases like “more than,” “less than,” or “twice as much.” If you can translate these phrases into algebraic expressions, you can understand and solve problems more confidently.

Main idea: Read the words carefully, find the operation being described, and replace the unknown amount with a variable.

Here are the four basic operations you will often see in word phrases:

• Addition: plus, added to, increased by, more than, the sum of
• Subtraction: minus, decreased by, less than, fewer than, the difference of
• Multiplication: times, multiplied by, of, the product of, twice, triple
• Division: divided by, the quotient of, per, split into equal groups

Step-by-step strategy

1. Choose a variable for the unknown number.
2. Look for clue words that tell you which operation to use.
3. Write the expression in the correct order.
4. Check that the expression matches the meaning of the words.

Important order warning: Some phrases do not go in the same order as the words are spoken.

For example, “5 less than a number” does not mean \(5 - n\). It means you start with the number and subtract 5:

$$n - 5$$

The words “less than” and “more than” can be tricky, so read them carefully.

Helpful phrase patterns

• “a number plus 7” \(\rightarrow n + 7\)
• “8 more than a number” \(\rightarrow n + 8\)
• “a number minus 3” \(\rightarrow n - 3\)
• “6 less than a number” \(\rightarrow n - 6\)
• “4 times a number” \(\rightarrow 4n\)
• “the product of 5 and a number” \(\rightarrow 5n\)
• “a number divided by 2” \(\rightarrow \frac{n}{2}\)
• “the quotient of a number and 3” \(\rightarrow \frac{n}{3}\)

Notice that in multiplication, we usually write \(4n\) instead of \(4 \times n\). Both mean the same thing, but \(4n\) is the usual algebra form.

Worked Example 1: A simple addition phrase

Translate: 9 more than a number

Step 1: Let the unknown number be \(x\).

Step 2: The words “more than” tell us to add.

Step 3: Add 9 to the number.

$$x + 9$$

Answer: The algebraic expression is \(x + 9\).

Worked Example 2: A subtraction phrase with tricky order

Translate: 12 less than a number

Step 1: Let the unknown number be \(m\).

Step 2: The words “less than” tell us to subtract.

Step 3: Start with the number, then subtract 12.

$$m - 12$$

Answer: The algebraic expression is \(m - 12\).

Be careful: \(12 - m\) would mean something different. That would mean 12 minus the number.

Worked Example 3: A multiplication phrase

Translate: the product of 7 and a number

Step 1: Let the unknown number be \(p\).

Step 2: The words “product of” mean multiplication.

Step 3: Multiply 7 by the number.

$$7p$$

Answer: The algebraic expression is \(7p\).

Worked Example 4: A two-step phrase

Translate: 3 times a number, decreased by 5

Step 1: Let the unknown number be \(y\).

Step 2: “3 times a number” becomes \(3y\).

Step 3: “decreased by 5” means subtract 5.

$$3y - 5$$

Answer: The algebraic expression is \(3y - 5\).

This example shows that some verbal phrases have more than one operation. Break the phrase into smaller parts and translate each part.

More common words and what they mean

• twice a number means \(2n\)
• triple a number means \(3n\)
• half of a number means \(\frac{n}{2}\)
• the sum of 8 and a number means \(8 + n\)
• the difference of a number and 4 means \(n - 4\)
• the quotient of a number and 6 means \(\frac{n}{6}\)

How to check your work

After you write an expression, ask yourself:

• Did I choose a variable for the unknown amount?
• Did I use the correct operation?
• Did I put the numbers and variable in the correct order?
• Does my expression match the words exactly?

Common mistakes to avoid

• Mixing up “less than” and “minus”
• Writing the numbers in the wrong order
• Forgetting that words like “twice” mean multiply by 2
• Using an equation instead of an expression

An expression does not include an equals sign. For example, \(x + 4\) is an expression. But \(x + 4 = 10\) is an equation.

Practice your thinking

Try translating these on your own:

• 5 more than a number
• the quotient of a number and 8
• twice a number plus 6
• 10 less than 4 times a number

Possible answers:

• \(n + 5\)
• \(\frac{n}{8}\)
• \(2n + 6\)
• \(4n - 10\)

Summary

Translating verbal phrases into algebraic expressions means changing math words into symbols and variables. First, choose a variable for the unknown number. Then look for clue words like “sum,” “difference,” “product,” or “quotient” to decide which operation to use.

Always pay close attention to word order, especially in phrases like “less than” and “more than.” With practice, you will get better at turning word phrases into algebraic expressions quickly and correctly.

Identifying Parts of an Expression

In algebra, an expression is a math phrase made of numbers, variables, and operation signs. Expressions do not have an equals sign. For example, \(3x + 5\) is an expression.

When you understand the parts of an expression, it becomes much easier to read algebra, solve problems, and explain your thinking. In this lesson, you will learn how to identify terms, coefficients, variables, and constants.

1. What is an expression?

An expression can include:

• numbers, like \(4\), \(12\), or \(100\)
• variables, like \(x\), \(y\), or \(n\)
• operation signs, like \(+\), \(-\), \(\times\), or division

Here are some examples of expressions:

• \(7 + 2\)
• \(4x\)
• \(3y - 8\)
• \(2a + 5b + 9\)

2. Terms

Terms are the parts of an expression that are separated by addition or subtraction signs.

For example, in the expression

$$4x + 7 - 2y$$

the terms are:

• \(4x\)
• \(7\)
• \(-2y\)

Notice that the subtraction sign belongs with the term after it. So in \(4x + 7 - 2y\), the last term is \(-2y\), not just \(2y\).

A term can be:

• a number only, like \(6\)
• a variable only, like \(x\)
• a number multiplied by a variable, like \(5x\)
• more than one variable multiplied together, like \(ab\)

3. Variables

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

In these expressions, the variables are:

• In \(6x\), the variable is \(x\).
• In \(3a + 9\), the variable is \(a\).
• In \(2m + 4n\), the variables are \(m\) and \(n\).

Variables help us write rules and patterns without knowing the exact number yet.

4. Coefficients

A coefficient is the number multiplied by a variable.

In the term \(5x\), the coefficient is \(5\) because \(5\) is multiplied by \(x\).

More examples:

• In \(9y\), the coefficient is \(9\).
• In \(-3n\), the coefficient is \(-3\).
• In \(\frac{1}{2}p\), the coefficient is \(\frac{1}{2}\).

Sometimes a variable appears by itself, like \(x\). That really means \(1x\), so the coefficient is 1.

Also, if you see \(-z\), that means \(-1z\), so the coefficient is -1.

5. Constants

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

In the expression

$$8x + 4$$

the constant is \(4\).

In

$$3a - 10$$

the constant is \(-10\).

Remember: constants are just numbers. They stay the same unless the whole expression changes.

6. How to identify the parts of an expression

When you look at an expression, you can follow these steps:

1. Find the terms by splitting at addition and subtraction signs.
2. Look for any variables in each term.
3. Find the coefficient of each variable term.
4. Find any constant term, which has no variable.

For example, look at

$$6x - 3 + 2y$$

• Terms: \(6x\), \(-3\), \(2y\)
• Variables: \(x\), \(y\)
• Coefficients: \(6\) and \(2\)
• Constant: \(-3\)

Worked Example 1

Identify the parts of the expression \(7m + 12\).

Step 1: Find the terms.

• \(7m\)
• \(12\)

Step 2: Find the variable.

• The variable is \(m\).

Step 3: Find the coefficient.

• In \(7m\), the coefficient is \(7\).

Step 4: Find the constant.

• The constant is \(12\).

Answer:

• Terms: \(7m\), \(12\)
• Variable: \(m\)
• Coefficient: \(7\)
• Constant: \(12\)

Worked Example 2

Identify the parts of the expression \(4x - 9\).

Step 1: Terms

• \(4x\)
• \(-9\)

Step 2: Variable

• The variable is \(x\).

Step 3: Coefficient

• The coefficient of \(x\) is \(4\).

Step 4: Constant

• The constant is \(-9\).

Answer:

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

Worked Example 3

Identify the parts of the expression \(3a + 5b + 11\).

Step 1: Terms

• \(3a\)
• \(5b\)
• \(11\)

Step 2: Variables

• The variables are \(a\) and \(b\).

Step 3: Coefficients

• The coefficient of \(a\) is \(3\).
• The coefficient of \(b\) is \(5\).

Step 4: Constant

• The constant is \(11\).

Answer:

• Terms: \(3a\), \(5b\), \(11\)
• Variables: \(a\), \(b\)
• Coefficients: \(3\), \(5\)
• Constant: \(11\)

Worked Example 4

Identify the parts of the expression \(-x + 6y - 14\).

This example is a little trickier because one variable has no number written in front of it.

Step 1: Terms

• \(-x\)
• \(6y\)
• \(-14\)

Step 2: Variables

• The variables are \(x\) and \(y\).

Step 3: Coefficients

• In \(-x\), the coefficient is \(-1\).
• In \(6y\), the coefficient is \(6\).

Step 4: Constant

• The constant is \(-14\).

Answer:

• Terms: \(-x\), \(6y\), \(-14\)
• Variables: \(x\), \(y\)
• Coefficients: \(-1\), \(6\)
• Constant: \(-14\)

Common mistakes to avoid

• Forgetting the sign of a term. In \(5x - 2\), the constant is \(-2\), not \(2\).
• Thinking every number is a coefficient. A coefficient must be attached to a variable. In \(4x + 7\), \(4\) is a coefficient, but \(7\) is a constant.
• Forgetting that a lone variable has a coefficient of 1. In \(x + 8\), the coefficient of \(x\) is \(1\).
• Forgetting that \(-x\) means \(-1x\).

Quick check

Try identifying the parts of these expressions on your own:

• \(9p + 4\)
• \(2m - 7n + 3\)
• \(-k - 15\)

You should ask yourself:

• What are the terms?
• Which letters are the variables?
• What numbers are the coefficients?
• Is there a constant?

Summary

An expression is a math phrase without an equals sign. The terms are the parts separated by addition or subtraction.

A variable is a letter that stands for a number. A coefficient is the number multiplied by a variable. A constant is a number by itself with no variable.

If you remember to keep negative signs with their terms, you will be able to identify the parts of an expression correctly and confidently.

Evaluating Algebraic Expressions via Substitution

In math, a variable is a letter that stands for a number. An algebraic expression is a math phrase with numbers, variables, and operations like addition, subtraction, multiplication, or division.

For example, in the expression \(x + 5\), the variable is \(x\). If we know the value of \(x\), we can find the value of the whole expression.

This process is called evaluating an expression. When we substitute, we replace the variable with its given value and then solve.

Why is substitution useful?

• It helps us find the value of an expression.
• It shows how letters can represent different numbers.
• It connects arithmetic you already know with algebra.

Step-by-step method

1. Read the expression carefully.
2. Replace each variable with the number given.
3. Use parentheses around the substituted number if it helps you stay organized.
4. Solve using the order of operations.

Order of operations reminder

• Do what is inside parentheses first.
• Then do multiplication and division.
• Then do addition and subtraction.
• Work from left to right when operations have the same level.

Important note about multiplication

In algebra, writing a number next to a variable means multiplication. For example:

• \(3x\) means \(3 \times x\)
• \(ab\) means \(a \times b\)

This is very important when substituting. If \(x = 4\), then \(3x = 3 \times 4 = 12\), not 34.

Worked Example 1: One variable, one operation

Evaluate \(n + 7\) when \(n = 5\).

Step 1: Substitute 5 for \(n\).

$$n + 7 = 5 + 7$$

Step 2: Add.

$$5 + 7 = 12$$

Answer: The value of the expression is \(12\).

Worked Example 2: Multiplication with a variable

Evaluate \(4m\) when \(m = 3\).

Step 1: Replace \(m\) with 3.

$$4m = 4(3)$$

Step 2: Multiply.

$$4 \times 3 = 12$$

Answer: The value of the expression is \(12\).

Worked Example 3: Two operations

Evaluate \(2x + 6\) when \(x = 4\).

Step 1: Substitute 4 for \(x\).

$$2x + 6 = 2(4) + 6$$

Step 2: Use order of operations. Multiply first.

$$2(4) + 6 = 8 + 6$$

Step 3: Add.

$$8 + 6 = 14$$

Answer: The value of the expression is \(14\).

Worked Example 4: More than one variable

Evaluate \(a + 3b\) when \(a = 2\) and \(b = 5\).

Step 1: Substitute each value into the expression.

$$a + 3b = 2 + 3(5)$$

Step 2: Multiply first.

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

Step 3: Add.

$$2 + 15 = 17$$

Answer: The value of the expression is \(17\).

What if the variable is negative?

Sometimes the value you substitute is a negative number. Be careful and use parentheses.

For example, evaluate \(y + 8\) when \(y = -3\).

Substitute carefully:

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

Now add:

$$(-3) + 8 = 5$$

So the value is \(5\).

Common mistakes to avoid

• Forgetting to substitute: Make sure every variable is replaced with its value.
• Ignoring multiplication: \(5x\) means \(5 \times x\), not 5 next to x.
• Skipping order of operations: In \(2x + 3\), multiply first, then add.
• Mixing up variable values: If \(a = 2\) and \(b = 6\), do not switch them.
• Not using parentheses for negative numbers: Write \((-4)\), not just \(-4\), when substituting into expressions.

Try thinking through these

• If \(p = 6\), then \(p - 2 = 6 - 2 = 4\).
• If \(k = 7\), then \(3k = 3 \times 7 = 21\).
• If \(r = 2\), then \(r + 4r = 2 + 4(2) = 2 + 8 = 10\).

How to check your work

• Did you replace every variable?
• Did you multiply before adding or subtracting?
• Did you use the correct number for each variable?
• Does your final answer make sense?

Summary

To evaluate an algebraic expression, replace each variable with its given value. Then solve the expression using the order of operations. Remember that a number next to a variable means multiplication, and use parentheses when substituting negative numbers.

Applying the Distributive Property means multiplying a number outside parentheses by each term inside the parentheses.

This helps us remove grouping symbols and rewrite expressions in a simpler form.

For example, in the expression \(3(x+4)\), the \(3\) must be multiplied by both \(x\) and \(4\).

We can write it like this:

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

This rule works with addition and subtraction inside the parentheses.

The distributive property says:

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

and

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

Think of it as: multiply outside by every term inside.

Here are the main steps:

1. Look at the factor outside the parentheses.
2. Multiply that factor by the first term inside.
3. Multiply that factor by the second term inside.
4. Keep the correct operation sign: plus or minus.
5. Write the new expression without parentheses.

Important idea: a term is a part of an expression separated by a plus or minus sign.

In \((x+5)\), the terms are \(x\) and \(5\). In \((y-2)\), the terms are \(y\) and \(2\).

When there is no symbol between a number and parentheses, it means multiplication.

So \(4(a+3)\) means \(4\times(a+3)\).

Worked Example 1

Simplify:

$$ 2(x+5) $$

Multiply \(2\) by each term inside the parentheses:

$$ 2(x+5)=2\cdot x+2\cdot 5 $$

Simplify each product:

$$ 2x+10 $$

Answer: \(2x+10\)

Worked Example 2

Simplify:

$$ 6(m-3) $$

Multiply \(6\) by both terms inside the parentheses:

$$ 6(m-3)=6\cdot m-6\cdot 3 $$

Now simplify:

$$ 6m-18 $$

Answer: \(6m-18\)

Notice that the subtraction sign stays in the expression. We multiply the outside factor by each term, but we keep the minus sign between the terms.

Worked Example 3

Simplify:

$$ 5(2p+7) $$

The terms inside the parentheses are \(2p\) and \(7\).

Distribute the \(5\):

$$ 5(2p+7)=5\cdot 2p+5\cdot 7 $$

Multiply:

$$ 10p+35 $$

Answer: \(10p+35\)

Worked Example 4

Simplify:

$$ 3(4x-2) $$

Multiply \(3\) by \(4x\), and multiply \(3\) by \(-2\):

$$ 3(4x-2)=3\cdot 4x-3\cdot 2 $$

Simplify:

$$ 12x-6 $$

Answer: \(12x-6\)

Common Mistakes to Avoid

• Forgetting to multiply every term: In \(4(a+2)\), do not write \(4a+2\). The correct answer is \(4a+8\).
• Dropping the minus sign: In \(7(n-1)\), the correct result is \(7n-7\), not \(7n+7\).
• Only multiplying the number term: In \(2(x+6)\), you must multiply both \(x\) and \(6\).

A quick check is to ask yourself: “Did I multiply the outside number by every term inside the parentheses?”

You can also use the distributive property with numbers only.

For example:

$$ 4(10+3)=4\cdot 10+4\cdot 3=40+12=52 $$

This is the same idea. The only difference is that algebraic expressions may include variables like \(x\), \(m\), or \(p\).

Practice Ideas

• \(2(a+4)\)
• \(9(x-2)\)
• \(3(5y+1)\)
• \(8(2n-6)\)

Try to expand each one by multiplying the number outside the parentheses by each term inside.

Summary

The distributive property helps you remove parentheses in an expression.

To use it, multiply the factor outside the parentheses by every term inside. Then write the new expression without parentheses.

Remember:

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

and

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

If you multiply carefully and keep the correct plus or minus sign, you can simplify expressions correctly.

Factoring Algebraic Expressions means rewriting an expression as a product. In 6th Grade, the main idea is to factor out the greatest common factor (GCF) from all the terms.

This is like doing the reverse of the distributive property. For example, if we know that

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

then factoring goes backward:

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

In this lesson, you will learn how to spot the greatest common factor and use it to factor expressions correctly.

Important words:

• Term: a part of an expression separated by plus or minus signs
• Factor: a number or variable multiplied by another number or variable
• Greatest Common Factor (GCF): the biggest factor that all terms share
• Factoring: rewriting an expression as a product by taking out a common factor

Let’s start by remembering the distributive property:

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

Factoring is the reverse:

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

So when you factor, you look at all the terms and ask, “What do they all have in common?”

How to factor by taking out the GCF

1. Find the GCF of the numbers in the terms.
2. Check whether the terms also share a variable, like \(x\).
3. Take that common factor and place it outside parentheses.
4. Write what is left from each term inside the parentheses.
5. Check your answer by distributing to see if you get the original expression back.

Worked Example 1: Numbers only

Factor:

$$12+18$$

Step 1: Find the greatest common factor of 12 and 18.

The factors of 12 are 1, 2, 3, 4, 6, 12.

The factors of 18 are 1, 2, 3, 6, 9, 18.

The greatest common factor is \(6\).

Step 2: Write 6 outside parentheses.

$$12+18=6(?+?)$$

Step 3: Ask, “What is left?”

• \(12 \div 6=2\)
• \(18 \div 6=3\)

So the factored form is:

$$12+18=6(2+3)$$

Check by distributing:

$$6(2+3)=6\cdot2+6\cdot3=12+18$$

It matches, so the factoring is correct.

Worked Example 2: A variable in both terms

Factor:

$$8x+12x$$

Step 1: Find the GCF of the numbers 8 and 12.

The GCF is \(4\).

Step 2: Both terms also have \(x\), so \(x\) is part of the common factor.

So the greatest common factor is \(4x\).

Step 3: Put \(4x\) outside parentheses.

$$8x+12x=4x(?+?)$$

Step 4: Divide each term by \(4x\).

• \(8x \div 4x=2\)
• \(12x \div 4x=3\)

So:

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

Check:

$$4x(2+3)=4x\cdot2+4x\cdot3=8x+12x$$

Worked Example 3: A more typical algebra expression

Factor:

$$15x+20$$

Step 1: Find the GCF of 15 and 20.

The GCF is \(5\).

Step 2: Does each term have \(x\)?

No. Only \(15x\) has \(x\), so you cannot factor out \(x\).

That means the GCF is just \(5\).

Step 3: Put \(5\) outside parentheses.

$$15x+20=5(?+?)$$

Step 4: Divide each term by 5.

• \(15x \div 5=3x\)
• \(20 \div 5=4\)

So the factored form is:

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

Check:

$$5(3x+4)=5\cdot3x+5\cdot4=15x+20$$

Worked Example 4: Three terms

Factor:

$$6x+9+12x$$

Step 1: Find the GCF of 6, 9, and 12.

The GCF is \(3\).

Step 2: Does every term have \(x\)?

No, because \(9\) does not have \(x\).

So the greatest common factor is \(3\).

Step 3: Put 3 outside parentheses.

$$6x+9+12x=3(?+?+?)$$

Step 4: Divide each term by 3.

• \(6x \div 3=2x\)
• \(9 \div 3=3\)
• \(12x \div 3=4x\)

So:

$$6x+9+12x=3(2x+3+4x)$$

Check:

$$3(2x+3+4x)=6x+9+12x$$

How to find the GCF quickly

• Look for the biggest number that divides all the coefficients.
• If every term has the same variable, include it in the GCF.
• If even one term does not have that variable, do not include the variable.

For example:

• In \(10x+15x\), the GCF is \(5x\).
• In \(10x+15\), the GCF is \(5\), not \(5x\).
• In \(14y+21y+7y\), the GCF is \(7y\).

Common mistakes to avoid

• Taking out a factor that is not common to every term.
  Example: In \(12x+18\), you cannot factor out \(x\) because 18 does not have \(x\).
• Not using the greatest common factor.
  Example: \(8x+12\) could be written as \(2(4x+6)\), but it is better to factor out the greatest common factor: \(4(2x+3)\).
• Forgetting to divide every term.
  Every term must be divided by the factor you took out.
• Not checking with distribution.
  Always multiply back to make sure your factored expression matches the original one.

Practice thinking

When you see an expression, ask yourself these questions:

1. What number do all the terms share?
2. Do all the terms share the same variable?
3. What is left in each term after I divide by the GCF?
4. Does distributing bring me back to the starting expression?

Let’s try a few quick factorizations

• \(9x+27=9(x+3)\)
• \(4x+16=4(x+4)\)
• \(7x+14x=7x(1+2)\)
• \(18+24x=6(3+4x)\)

Summary

Factoring algebraic expressions means rewriting an expression as a product by taking out the greatest common factor. This is the reverse of the distributive property.

To factor, find what all the terms have in common, take that common factor outside parentheses, and write what is left inside. Then check your work by distributing.

Combining Like Terms helps us make algebra expressions simpler and easier to understand.

When we combine like terms, we add or subtract parts of an expression that are the same kind of term.

For example, in the expression \(3x + 2x\), both terms have \(x\). Since they are the same kind of term, we can combine them:

$$3x + 2x = 5x$$

This lesson will show you how to spot like terms, how to combine them, and how to avoid common mistakes.

What is a term?

A term is one part of an expression. Terms are usually separated by plus signs or minus signs.

In the expression \(4x + 7 - 2x + 3\), the terms are:

• \(4x\)
• \(7\)
• \(-2x\)
• \(3\)

What are like terms?

Like terms are terms that have the same variable part.

Examples of like terms:

• \(5x\) and \(2x\)
• \(7y\) and \(-3y\)
• \(4a\) and \(10a\)
• \(8\) and \(-1\) because both are constants

Examples of terms that are not like terms:

• \(3x\) and \(3y\)
• \(5a\) and \(5b\)
• \(2x\) and \(2\)

Even if the numbers match, the variable part must also match for the terms to be like terms.

The coefficient

The coefficient is the number in front of the variable.

• In \(6x\), the coefficient is \(6\).
• In \(-4y\), the coefficient is \(-4\).
• In \(x\), the coefficient is \(1\), because \(x = 1x\).

When we combine like terms, we only add or subtract the coefficients. The variable stays the same.

For example:

$$8x - 3x = 5x$$

We subtracted the coefficients: \(8 - 3 = 5\), and kept the \(x\).

How to combine like terms

1. Find the terms that match.
2. Add or subtract their coefficients.
3. Keep the same variable part.
4. Combine constants too, if there are any.

Worked Example 1

Simplify \(2x + 5x\).

Both terms have \(x\), so they are like terms.

Add the coefficients:

$$2 + 5 = 7$$

Keep the variable:

$$2x + 5x = 7x$$

Answer: \(7x\)

Worked Example 2

Simplify \(9y - 4y\).

Both terms have \(y\), so they are like terms.

Subtract the coefficients:

$$9 - 4 = 5$$

Keep the variable:

$$9y - 4y = 5y$$

Answer: \(5y\)

Worked Example 3

Simplify \(3x + 7 + 4x - 2\).

First, find the like terms:

• \(3x\) and \(4x\)
• \(7\) and \(-2\)

Combine the \(x\)-terms:

$$3x + 4x = 7x$$

Combine the constants:

$$7 - 2 = 5$$

Put them together:

$$3x + 7 + 4x - 2 = 7x + 5$$

Answer: \(7x + 5\)

Worked Example 4

Simplify \(6a - 2 + 3a + 8 - a\).

Find the like terms:

• \(6a\), \(3a\), and \(-a\)
• \(-2\) and \(8\)

Remember that \(-a\) means \(-1a\).

Combine the variable terms:

$$6a + 3a - a = 6a + 3a - 1a = 8a$$

Combine the constants:

$$-2 + 8 = 6$$

Put them together:

$$6a - 2 + 3a + 8 - a = 8a + 6$$

Answer: \(8a + 6\)

Important ideas to remember

• You can only combine like terms.
• The variable part must match exactly.
• Add or subtract the coefficients.
• Do not change the variable when combining like terms.
• Constants can be combined with other constants.

Common mistakes

• Mistake: \(2x + 3 = 5x\)
  This is wrong because \(2x\) and \(3\) are not like terms.
• Mistake: \(4x + 2y = 6xy\)
  This is wrong because \(x\)-terms and \(y\)-terms are not like terms.
• Mistake: forgetting that \(-x = -1x\).

Quick check

Try these on your own:

• \(5m + 2m\)
• \(10p - 6p\)
• \(4x + 3 + 2x + 1\)
• \(7b - b + 5 - 9\)

Answers:

• \(7m\)
• \(4p\)
• \(6x + 4\)
• \(6b - 4\)

Summary

Combining like terms means simplifying an expression by putting together terms with the same variable part.

You add or subtract the coefficients and keep the variable the same. Constants can also be combined with other constants.

When you look carefully for matching terms, algebra becomes much easier to work with.

Evaluating Expressions with Grouping Symbols means finding the value of a number sentence that has operation signs and symbols like parentheses.

Grouping symbols tell us which part to work on first. The most common grouping symbols in 4th grade are parentheses: \((\ )\).

When we evaluate an expression, we do not guess. We follow the steps in the correct order so the answer is correct every time.

Important idea: An expression is a math phrase with numbers and operation signs, such as \(6 + (4 \times 3)\). It does not have an equals sign until we show its value.

Grouping symbols help organize the math. They show that some numbers belong together, like a team. We solve what is inside the grouping symbols before working on the rest.

Order to follow:

• Step 1: Solve what is inside parentheses first.
• Step 2: Multiply or divide.
• Step 3: Add or subtract.

This order matters. If we do the operations in the wrong order, we can get the wrong answer.

Look at these two expressions:

• \(2 + 3 \times 4\)
• \((2 + 3) \times 4\)

They look similar, but the grouping symbols change what we do first.

For \(2 + 3 \times 4\), we multiply first:

$$2 + 3 \times 4 = 2 + 12 = 14$$

For \((2 + 3) \times 4\), we add inside the parentheses first:

$$ (2 + 3) \times 4 = 5 \times 4 = 20 $$

So grouping symbols can change the value of an expression.

Worked Example 1

Evaluate \(7 + (5 - 2)\).

Step 1: Solve inside the parentheses.

$$5 - 2 = 3$$

Now the expression becomes:

$$7 + 3$$

Step 2: Add.

$$7 + 3 = 10$$

Answer: \(10\)

Worked Example 2

Evaluate \(6 \times (4 + 1)\).

Step 1: Solve inside the parentheses.

$$4 + 1 = 5$$

Now the expression becomes:

$$6 \times 5$$

Step 2: Multiply.

$$6 \times 5 = 30$$

Answer: \(30\)

Worked Example 3

Evaluate \(18 - (3 \times 4)\).

Step 1: Solve inside the parentheses first.

$$3 \times 4 = 12$$

Now the expression becomes:

$$18 - 12$$

Step 2: Subtract.

$$18 - 12 = 6$$

Answer: \(6\)

Worked Example 4

Evaluate \((8 + 4) \div 3\).

Step 1: Solve inside the parentheses.

$$8 + 4 = 12$$

Now the expression becomes:

$$12 \div 3$$

Step 2: Divide.

$$12 \div 3 = 4$$

Answer: \(4\)

How to check your work

• Did you solve inside the parentheses first?
• After that, did you multiply or divide before adding or subtracting?
• Did you rewrite the expression carefully after each step?
• Did you make sure each new expression matches the step before it?

Common mistakes to avoid

• Ignoring the parentheses. Always do that part first.
• Doing operations left to right without checking the order. Remember: parentheses first.
• Changing more than one thing at a time. It helps to solve one step, then rewrite the expression.

Here is an example of a common mistake:

Expression: \(9 - (2 + 3)\)

Wrong way: subtract first and do \(9 - 2 = 7\), then \(7 + 3 = 10\)

This is wrong because the parentheses were not solved first.

Right way:

$$9 - (2 + 3) = 9 - 5 = 4$$

Helpful strategy: Cover the outside part with your finger and look only at what is inside the parentheses. Solve that part first. Then bring back the rest of the expression.

Let’s compare one more pair

• \(12 - 4 \div 2\)
• \((12 - 4) \div 2\)

First expression:

$$12 - 4 \div 2 = 12 - 2 = 10$$

Second expression:

$$ (12 - 4) \div 2 = 8 \div 2 = 4 $$

The answers are different because the grouping symbols changed what we solved first.

Summary

To evaluate expressions with grouping symbols, always start with what is inside the parentheses. Then multiply or divide. Last, add or subtract.

If you follow the steps carefully and rewrite the expression after each step, you can solve expressions correctly and confidently.

Lesson: Proving Expression Equivalence

In algebra, two expressions are equivalent if they always have the same value, no matter what value you choose for the variable.

For example, the expressions \(x + x\) and \(2x\) are equivalent because both mean “two groups of \(x\).” If \(x=3\), both expressions equal \(6\). If \(x=10\), both expressions equal \(20\). They match every time.

Learning how to prove expressions are equivalent helps you check your work, rewrite expressions in simpler forms, and understand how algebra works.

What does “prove equivalent” mean?

To prove two expressions are equivalent, you need to show they are equal for all values of the variable, not just one example.

There are two main ways 6th graders can check equivalence:

• Substitute the same value into both expressions and compare the results.
• Simplify the expressions using arithmetic and properties until they look the same or clearly have the same value.

Important: Testing just one value can give you a clue, but it does not always fully prove equivalence. It is better to test several values or simplify the expressions.

Method 1: Substitute values

Substituting means replacing the variable with a number.

Suppose you want to compare \(3n + 2n\) and \(5n\).

If \(n=4\):

$$3(4) + 2(4) = 12 + 8 = 20$$ $$5(4)=20$$

They match for \(n=4\).

If \(n=7\):

$$3(7) + 2(7) = 21 + 14 = 35$$ $$5(7)=35$$

They match again. This suggests the expressions are equivalent.

Substitution is a great way to test equivalence. But the strongest way is to also use simplifying.

Method 2: Simplify the expressions

Simplifying means combining like terms or using what you know about operations.

For example:

$$3n + 2n = 5n$$

Since \(3n\) and \(2n\) are like terms, they combine to make \(5n\). So the expressions are equivalent.

Combining like terms

Like terms have the same variable part. For example:

• \(4x\) and \(7x\) are like terms.
• \(3y\) and \(y\) are like terms.
• \(5a\) and \(2b\) are not like terms because the variables are different.

You can combine like terms by adding or subtracting their number parts.

Examples:

• \(4x + 7x = 11x\)
• \(9y - 2y = 7y\)
• \(6a + a = 7a\)

Using the distributive property

The distributive property helps you rewrite expressions. It says:

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

For example:

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

So \(3(x+4)\) and \(3x+12\) are equivalent expressions.

You can also work backward:

$$4m+8 = 4(m+2)$$

Both forms mean the same thing.

Worked Example 1: A simple like-terms example

Are \(6x + 2x\) and \(8x\) equivalent?

Step 1: Simplify.

$$6x + 2x = 8x$$

Yes, they are equivalent.

Check with substitution: Let \(x=5\).

$$6(5)+2(5)=30+10=40$$ $$8(5)=40$$

Both expressions equal \(40\), so the check works too.

Worked Example 2: A subtraction example

Are \(9y - 4y\) and \(5y\) equivalent?

Step 1: Combine like terms.

$$9y - 4y = 5y$$

Yes, the expressions are equivalent.

Check with substitution: Let \(y=3\).

$$9(3)-4(3)=27-12=15$$ $$5(3)=15$$

The values match.

Worked Example 3: Using the distributive property

Are \(2(a+5)\) and \(2a+10\) equivalent?

Step 1: Distribute the 2.

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

Yes, they are equivalent.

Check with substitution: Let \(a=4\).

$$2(4+5)=2(9)=18$$ $$2(4)+10=8+10=18$$

Both expressions equal \(18\).

Worked Example 4: Expressions that are not equivalent

Are \(3x + 4\) and \(7x\) equivalent?

Let’s test a value. If \(x=2\):

$$3(2)+4=6+4=10$$ $$7(2)=14$$

The values are different, so the expressions are not equivalent.

We can also see why by looking at the terms. In \(3x+4\), the \(+4\) is a constant, but in \(7x\), everything depends on \(x\). These expressions do not mean the same thing.

How to decide if two expressions are equivalent

1. Look for like terms you can combine.
2. Use the distributive property if there are parentheses.
3. Simplify each expression as much as you can.
4. See whether the simplified forms match.
5. If needed, substitute the same number into both expressions to check.

Common mistakes to avoid

• Only checking one value: One matching value does not always prove equivalence.
• Combining unlike terms: You cannot combine \(3x+4\) into \(7x\).
• Forgetting to distribute to every term: In \(3(x+2)\), the 3 must multiply both \(x\) and 2.
• Mixing up terms and factors: In \(2x+2\), the 2 is not multiplying the whole expression unless you write parentheses: \(2(x+1)\).

Try to think about meaning

Expressions are equivalent if they describe the same amount in different ways.

For example, imagine you have 4 bags with \(m\) marbles in each bag, and then 8 extra marbles. That can be written as:

$$4m+8$$

If you group the marbles differently, you can also write it as:

$$4(m+2)$$

Both expressions describe the same total, so they are equivalent.

Quick practice ideas

• Compare \(5n+n\) and \(6n\).
• Compare \(4(x+3)\) and \(4x+12\).
• Compare \(7y+2\) and \(9y\).
• Compare \(10a-3a\) and \(7a\).

Summary

To prove expression equivalence, you show that two expressions always have the same value. You can do this by simplifying with like terms and the distributive property, and by substituting the same value into both expressions to check your thinking.

If two expressions simplify to the same form, they are equivalent. If they give different values for the same variable, they are not equivalent.