Properties and Equivalent Expressions | Grade 6 Math | LevelKite

You already know how to evaluate an expression: you replace the variable with a number and then compute. In this lesson you will see that two expressions can look different and still give the same value every time. You will also learn the rules that explain why.

In this lesson, each variable, such as $x$ , stands for a number, and it can be any number.

A property of an operation is a rule that is true for every number. Here is one you have used for years: $3 + 5 = 8$ and $5 + 3 = 8$ . Switching the order of two numbers you add never changes the result. This works for every pair of numbers, so it is a property.

The table below lists the properties of addition and multiplication. A few words will help you read it. Addends are the numbers you add, and the result is the sum. Factors are the numbers you multiply, and the result is the product. The letters $a$ , $b$ , and $c$ stand for any numbers.

Properties of addition and multiplication

Property	What it says	Number example	For any numbers a, b, c
Commutative property of addition	Changing the order of the addends does not change the sum.	$3 + 5 = 5 + 3$ , because both sides equal $8$	$a + b = b + a$
Commutative property of multiplication	Changing the order of the factors does not change the product.	$3 \cdot 5 = 5 \cdot 3$ , because both sides equal $15$	$a \cdot b = b \cdot a$
Associative property of addition	Changing which addends are grouped in parentheses does not change the sum.	$(2 + 3) + 4 = 5 + 4 = 9$ and $2 + (3 + 4) = 2 + 7 = 9$	$(a + b) + c = a + (b + c)$
Associative property of multiplication	Changing which factors are grouped in parentheses does not change the product.	$(2 \cdot 3) \cdot 4 = 6 \cdot 4 = 24$ and $2 \cdot (3 \cdot 4) = 2 \cdot 12 = 24$	$(a \cdot b) \cdot c = a \cdot (b \cdot c)$
Identity property of addition	Adding $0$ to a number leaves the number unchanged.	$7 + 0 = 7$	$a + 0 = a$
Identity property of multiplication	Multiplying a number by $1$ leaves the number unchanged.	$7 \cdot 1 = 7$	$a \cdot 1 = a$
Distributive property	A number times a sum equals that number times each addend, with the products added.	$2(3 + 4) = 2 \cdot 7 = 14$ and $2 \cdot 3 + 2 \cdot 4 = 6 + 8 = 14$	$a(b + c) = a \cdot b + a \cdot c$

Worked example

Name the property

Each equation is true for every number $x$ . Name the property that each one shows: (a) $x + 9 = 9 + x$ ; (b) $(4 \cdot x) \cdot 5 = 4 \cdot (x \cdot 5)$ ; (c) $x \cdot 1 = x$ ; (d) $3(x + 5) = 3 \cdot x + 3 \cdot 5$ .

In equation (a), both sides add the same two addends, x and 9. Only their order is switched. Changing the order of addends is the commutative property of addition.
$x + 9 = 9 + x$
In equation (b), the factors 4, x, and 5 stay in the same order. Only the parentheses move, so the grouping changes. Changing the grouping of factors is the associative property of multiplication.
$(4 \cdot x) \cdot 5 = 4 \cdot (x \cdot 5)$
In equation (c), x is multiplied by 1 and the result is still x. Multiplying by 1 leaves a number unchanged, so this is the identity property of multiplication.
$x \cdot 1 = x$
In equation (d), the left side multiplies 3 by a sum. The right side multiplies 3 by each addend and then adds the two products. That is the distributive property.
$3(x + 5) = 3 \cdot x + 3 \cdot 5$

Answer

Equation (a) shows the commutative property of addition, (b) shows the associative property of multiplication, (c) shows the identity property of multiplication, and (d) shows the distributive property.

Now compare three expressions: $x + 2$ , $2 + x$ , and $2x$ . Remember that $2x$ means $2 \cdot x$ . Here $x$ can be any number, and the table shows five of them. As you read the table, look for two things. Which expressions match in every row? Which expression matches the others in only one row?

Values of three expressions

Value of x	x + 2	2 + x	2x
$0$	$0 + 2 = 2$	$2 + 0 = 2$	$2 \cdot 0 = 0$
$1$	$1 + 2 = 3$	$2 + 1 = 3$	$2 \cdot 1 = 2$
$2$	$2 + 2 = 4$	$2 + 2 = 4$	$2 \cdot 2 = 4$
$3$	$3 + 2 = 5$	$2 + 3 = 5$	$2 \cdot 3 = 6$
$10$	$10 + 2 = 12$	$2 + 10 = 12$	$2 \cdot 10 = 20$

The expressions $x + 2$ and $2 + x$ match in every row. That is not luck. The commutative property of addition says that $x + 2 = 2 + x$ for every number $x$ , including all the numbers that are not in the table. Expressions like these are called equivalent.

The expression $2x$ is different. It matches $x + 2$ when $x = 2$ , because both equal $4$ . But when $x = 3$ , $x + 2 = 5$ and $2x = 6$ . So $2x$ and $x + 2$ are not equivalent, even though they agree at one value.

Key concept

Equivalent expressions

Two expressions are equivalent when they have the same value for every value the variable is allowed to take. In this lesson, that means every number.

Each property of addition and multiplication is true for every number. So rewriting an expression with a property always gives an equivalent expression.
Matching at one value, or even at several values, does not prove that two expressions are equivalent, because you cannot test every number.
A value where two expressions differ is called a counterexample. One counterexample proves that the expressions are not equivalent.

Worked example

Decide which expression is equivalent

Which expression is equivalent to $2 \cdot (x \cdot 3)$ : $x + 5$ or $6x$ ? Here $x$ can be any number.

Start by testing a value. Replace x with the same number, 1, in all three expressions. Each one gives 6. One matching value does not prove that expressions are equivalent, so test another value.
$2 \cdot (1 \cdot 3) = 2 \cdot 3 = 6 \qquad 1 + 5 = 6 \qquad 6 \cdot 1 = 6$
Replace x with 2 in all three expressions. The original expression gives 12, but x + 5 gives 7. This is a counterexample, so x + 5 is not equivalent to the original expression.
$2 \cdot (2 \cdot 3) = 2 \cdot 6 = 12 \qquad 2 + 5 = 7 \qquad 6 \cdot 2 = 12$
The expression 6x matched both times. But you cannot test every number, so matches alone are not proof. Use properties instead. The commutative property of multiplication lets you switch the order of the factors x and 3.
$2 \cdot (x \cdot 3) = 2 \cdot (3 \cdot x)$
The associative property of multiplication lets you group the 2 with the 3, and 2 times 3 is 6. Both properties are true for every number, so the original expression equals 6x for every value of x.
$2 \cdot (3 \cdot x) = (2 \cdot 3) \cdot x = 6 \cdot x = 6x$

Answer

The expression $6x$ is equivalent to $2 \cdot (x \cdot 3)$ , because the properties show that the two are equal for every value of $x$ . The expression $x + 5$ is not equivalent, because it gives a different value when $x = 2$ .

Now return to the distributive property from the properties table. You need it for an expression such as $3(x + 4)$ . You cannot add $x + 4$ first, because you do not know the value of $x$ . The distributive property lets you multiply anyway.

Here is why it works. The expression $3(x + 4)$ means $3$ groups of $(x + 4)$ :

$(x + 4) + (x + 4) + (x + 4)$

The commutative and associative properties of addition let you reorder and regroup the addends:

$(x + x + x) + (4 + 4 + 4)$

Three copies of $x$ make $3 \cdot x$ , and three copies of $4$ make $3 \cdot 4$ . So $3(x + 4) = 3 \cdot x + 3 \cdot 4 = 3x + 12$ . The $3$ multiplies each term inside the parentheses. The terms of an expression are the parts that are added or subtracted.

The same pattern holds for any numbers $a$ , $b$ , and $c$ . That is the distributive property:

a(b + c) = a \cdot b + a \cdot c

Worked example

Multiply a number by a sum

Use the distributive property to multiply $5(x + 3)$ . Here $x$ can be any number.

Find the parts, so you know what to multiply. The factor outside the parentheses is 5. The terms inside the parentheses are x and 3.
$5(x + 3)$
Multiply the outside factor by each term inside, and keep the addition sign between the two products. The distributive property is true for every number, so the new expression is equivalent to the old one.
$5(x + 3) = 5 \cdot x + 5 \cdot 3$
Simplify each product. 5 times x is written 5x, and 5 times 3 is 15.
$5 \cdot x + 5 \cdot 3 = 5x + 15$

Answer

The product is $5x + 15$ . The expressions $5(x + 3)$ and $5x + 15$ are equivalent, so they have the same value for every value of $x$ .

The distributive property also works when the terms inside the parentheses are subtracted. For any numbers $a$ , $b$ , and $c$ , the rule is $a(b - c) = a \cdot b - a \cdot c$ . For example, $2(10 - 3) = 2 \cdot 7 = 14$ , and $2 \cdot 10 - 2 \cdot 3 = 20 - 6 = 14$ . This makes sense, because $2$ groups of $(10 - 3)$ have two copies of $10$ with two copies of $3$ taken away. So you still multiply the outside factor by each term. Just keep the subtraction sign between the two products.

Worked example

Multiply when the terms are subtracted

Use the distributive property to multiply $4(2y - 7)$ . Here $y$ can be any number.

Multiply the outside factor, 4, by each term inside. Keep the subtraction sign between the two products.
$4(2y - 7) = 4 \cdot 2y - 4 \cdot 7$
Simplify the first product. 2y means 2 times y, so the associative property of multiplication lets you group the 4 with the 2.
$4 \cdot (2 \cdot y) = (4 \cdot 2) \cdot y = 8y$
Simplify the second product by multiplying 4 and 7.
$4 \cdot 7 = 28$
Replace each product with its simpler form. Each replacement keeps the value the same, and the subtraction sign stays between them.
$4(2y - 7) = 8y - 28$
Check with one value. Replace y with 5 in both expressions. Both give 12. This check can catch an arithmetic slip, but one matching value is not proof. The properties are what make the two expressions equal for every value of y.
$4(2 \cdot 5 - 7) = 4(10 - 7) = 4 \cdot 3 = 12 \qquad 8 \cdot 5 - 28 = 40 - 28 = 12$

Answer

The product is $8y - 28$ . The expressions $4(2y - 7)$ and $8y - 28$ are equivalent, so they have the same value for every value of $y$ .

Properties of addition and multiplication are true for every number. So when you rewrite an expression with a property, you get an equivalent expression: one that has the same value for every value the variable is allowed to take. The distributive property is one of these rewrites. It tells you to multiply the outside factor by each term inside the parentheses.

Remember how to judge two expressions. One counterexample proves that they are not equivalent. Matching at one value, or at a few values, does not prove that they are equivalent, so use the properties.

Next, you will combine like terms and factor expressions using the distributive property.