Imagine you are playing a building game, like Minecraft or Roblox. You can build a house using different sets of blocks, but in the end, the house still takes up the same space and looks the same from the outside. That is what we do in algebra when we write equivalent expressions—they may look different, but they have the same value for every number you put in. 🎯
In this lesson, you will learn how to use the properties of operations (like the distributive property) to change expressions into different but equal forms. This is a key skill in algebra and will help you with equations, word problems, and even real-life situations like shopping and splitting costs with friends.
The main idea: use properties of operations to rewrite expressions, such as turning \(3(2 + x)\) into \(6 + 3x\), or turning \(y + y + y\) into \(3y\).
The idea of “same value, different look” is shown clearly in [Figure 1], where blocks are grouped in different ways but represent the same total.

Equivalent expressions are expressions that have the same value for every value of the variable, even if they look different.
Examples:
To check if two expressions are equivalent, you can plug in the same value for the variable.
Example check: Are \(y + y + y\) and \(3y\) equivalent?
They give the same result, so they are equivalent.
We mainly use three properties to generate equivalent expressions:
You have seen these with numbers in earlier grades. Now we extend them to algebraic expressions that include variables like \(x\) and \(y\).
The commutative property tells us we can change the order when we add or multiply, and the value stays the same.
Examples with numbers:
Examples with variables:
These are all equivalent expressions because only the order changed.
The associative property tells us we can change the grouping when adding or multiplying, and the value stays the same.
Examples with numbers:
Examples with variables:
We are not changing the numbers or variables, just how they are grouped.
The distributive property connects multiplication and addition (or subtraction). It says:
\[a(b + c) = ab + ac\]
We “distribute” the outside factor \(a\) to each term inside the parentheses.
Examples with numbers:
Examples with variables:
We can also use it in reverse to factor, which means putting something back into parentheses:
\[ab + ac = a(b + c)\]
For example, \(24x + 18y = 6(4x + 3y)\) because \(6 \cdot 4x = 24x\) and \(6 \cdot 3y = 18y\).
The steps of distributing and factoring are shown clearly in [Figure 2], using arrows from the outside factor to each term inside the parentheses.

Now we put these properties to work to create equivalent expressions.
Goal: Remove parentheses by multiplying.
Example A: \(3(2 + x)\)
Step-by-step solution
So \(3(2 + x)\) and \(6 + 3x\) are equivalent expressions.
Example B: \(4(x + 5)\)
So \(4(x + 5) = 4x + 20\).
Example C: \(2(3y + 4)\)
Goal: Put an expression back into parentheses with a common factor outside.
Example D: \(24x + 18y\)
Step-by-step solution
So \(24x + 18y\) and \(6(4x + 3y)\) are equivalent.
Example E: \(12x + 8\)
We say terms are like terms if they have the same variable raised to the same power.
The property behind combining like terms is actually the distributive property plus the commutative and associative properties.
Example F: \(y + y + y\)
Step-by-step solution
So \(y + y + y\) is equivalent to \(3y\).
This process is shown visually in [Figure 3], where three \(y\)-tiles are grouped and then replaced by a single block labeled \(3y\).

Here are three fully worked examples to practice seeing how properties create equivalent expressions.
Problem: Rewrite \(5(2x + 3)\) as an equivalent expression without parentheses.
Solution:
Answer: \(5(2x + 3) = 10x + 15\).
Problem: Write \(14x + 21\) as a product using the distributive property.
Solution:
Answer: \(14x + 21 = 7(2x + 3)\).
Problem: Simplify \(2y + 5 + 3y\) to an equivalent expression with as few terms as possible.
Solution:
Answer: \(2y + 5 + 3y = 5y + 5\).
These properties are not just for math class. They show up in everyday situations.
Suppose you buy 3 T-shirts, each costing \(x\) dollars, and each has an extra decoration that costs 2 dollars. The total cost is:
\[3(x + 2)\]
This means “3 groups of (shirt price + decoration).” Using the distributive property:
\[3(x + 2) = 3x + 6\]
Now you see the same situation as “total shirt cost” plus “total decoration cost.” Both expressions are equivalent and describe the same purchase. 💡
Imagine you have 4 bags, and each bag has \(2 + y\) cookies (2 chocolate cookies and \(y\) sugar cookies). The total number of cookies is:
\[4(2 + y)\]
Using the distributive property:
\[4(2 + y) = 8 + 4y\]
This tells you there are 8 chocolate cookies and \(4y\) sugar cookies in total.
Say you play 5 rounds of a game. In each round you earn 10 coins plus \(p\) bonus points. The total reward is:
\[5(10 + p)\]
Using the distributive property:
\[5(10 + p) = 50 + 5p\]
So you always get 50 coins plus \(5p\) bonus points overall. Whether you think in groups \(5(10 + p)\) or as a total \(50 + 5p\), it is the same amount.
Using properties of operations to generate equivalent expressions helps you: