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Last modified on December 18th, 2023

Equivalent algebraic expressions are algebraic expressions that are equal in value but look different. Thus, if we have two algebraic expressions and they yield the same value for every possible substitution of the variables, then they are considered equivalent.

First, let us consider two numeric expressions ${2\times 3+2}$ and ${3^{2}-1}$, equal to 8 and thus, they are equivalent.

Now, let us consider two algebraic expressions, x + 4 and x + 1 + 1 + 1 + 1.

x + 4 can also be written as x + 1 + 1 + 1 + 1, which is the second expression given.

Thus, the above two expressions are equivalent algebraic expressions.

To write and identify equivalent expressions, we follow the steps given below:

- Use the properties of operations to form equivalent expressions.
- Compare the expressions.
- Reach to our conclusion.

Now, let us determine whether the expressions 8b – 9a and -9a + 8b are equivalent.

Changing the order of terms of the expression 8b – 9a, we get

-9a + 8b, the second expression.

Thus, they are equivalent.

Let us identify whether 3a + 2b + 8a + 8 and 11a + 6b are equivalent.

Combining the like terms of the first expression, we get

(3a + 8a) + 2b + 8 = 11a + 2b + 8, not equals to 11a + 6b

They are not equivalent.

Thus, we simplify one expression using the properties of operations here, and if two expressions are identical, they are equivalent.

While identifying two equivalent expressions, we use the equivalence symbol (â‰¡) to show that they are identical.

Let us consider some more examples using some other important properties of operations.

**Addition and Subtraction**

Let us consider the expressions 2x + 5z and 5z +2x.

Since addition is commutative, changing the order of terms does not change the sum.

Thus, 2x + 5z â‰¡ 5z + 2x.

Now, let us identify two other expressions, 3p – 7q and -7q + 3p.

Since subtraction is not commutative, changing the order of terms changes the result.

Taking the first expression, we get

3p – 7q = 3p + (-7q) = -7q +3p

The given expressions are identical.

Thus, 3p – 7q â‰¡ -7q + 3p

**Multiplication**

Let the two expressions be ${3\left( x+7\right)}$ and ${\left( x+7\right) \times 3}$.

Since multiplication is commutative, changing the order of terms does not change the product.

Thus, ${3\left( x+7\right)}$ â‰¡ ${\left( x+7\right) \times 3}$

Let us consider another two expressions: ${4\left( x+8\right)}$ and 4x + 32

Using the distributive property, we get

${4\left( x+8\right)}$ = ${4\times x+4\times 8}$ = 4x + 32

Thus, ${4\left( x+8\right)}$ â‰¡ 4x + 32.

Here, we can substitute any value for a variable in algebraic expressions to determine whether they are equivalent.

Let us consider the expressions ${\dfrac{3\times m\times n}{m}}$ and n + n + n.

If we take m = 0, the first expression cannot be defined.

Thus, considering m = 2 and n = 4 (any non-zero value), we get,

${\dfrac{3\times 2\times 4}{2}}$ and 4 + 4 + 4

Both the results are 12

Thus, ${\dfrac{3\times m\times n}{m}}$ â‰¡ n + n + n, when ${m\neq 0}$.

The previous example can also be simplified as follows:

The first expression gives us ${\dfrac{3\times m\times n}{m}}$

Canceling all the common factors and keeping the remaining ones unchanged, we get

${3\times n}$ = 3n

The second expression gives n + n + n = 3n

Thus, ${\dfrac{3\times m\times n}{m}}$ â‰¡ n + n + n, when ${m\neq 0}$.

Now, let us consider two expressions with exponents, x(x – 5) and ${-5x+x^{2}}$, and check whether they are equivalent.

By using the properties of operations in the first expression, we get

${x\times x-x\times 5}$

= ${x^{2}-5x}$

Comparing the given expressions,

${x^{2}-5x}$ = ${-5x+x^{2}}$

Thus, they are equivalent.

We can easily find the value of the unknown coefficient in an expression given that it is equivalent to another given expression.

Let us consider two equivalent expressions 5(7x + 3) and kx + 15 (where, k is an unknown coefficient)

As 5(7x + 3) â‰¡ kx + 15, k must equal the coefficient of x.

Now, using the properties of operations, we get

${5\times 7x+15}$

= 35x + 15

Now, solving for the coefficient of x, we get k = 35

**Find an equivalent algebraic expression to the given expressions:****a) 3(2x + 1) – 8****b) 3(2x – 1) + 2(x + 1)**

Solution:

a) The given expression is 3(2x + 1) – 8

Solving the parentheses by distributive property, we get

${3\times 2x+3\times 1-8}$

= 6x + 3 -8

= 6x – 5

6x – 5 is an equivalent expression to 3(2x + 1) – 8.

b) The given expression is 3(2x – 1) + 2(x + 1)

Solving the parentheses by distributive property, we get

${3\times 2x-3\times 1+2\times x+2\times 1}$

= 6x – 3 +2x + 2

Combining the like terms, we get

(6x + 2x) + (-3 + 2)

= 8x – 1

8x – 1 is an equivalent expression to 3(2x – 1) + 2(x + 1).

Last modified on December 18th, 2023