Algebraic Expressions in Math: Definition, Example and Equation

Algebraic Expression is a mathematical expression that is made of numbers, and variables connected with any arithmetical operation between them. Algebraic forms are used to define unknown conditions in real life or situations that include unknown variables.

An algebraic expression is made up of terms, there can be one or more than one term present in the expression. Suppose we have to find the age of Arum if the age of his sister is twice the age of Arun and the sum of their age is 24 years.

This situation can be easily explained using the Algebraic Expressions let the age of Arun be x then the age of his sister be 2x and the sum of their ages is x + 2x = 24 this is an algebraic expression.

Some other examples of algebraic expressions are,

• 5x + 4y
• 11x – 12
• 2y - 13, etc.

Here the above three expressions are algebraic expressions they have variables x, and y and constant terms -12, and -13.

Variables, Constants, Terms, and Coefficients in Algebraic Expression

The image showing the constant and variables of an Algebraic Expression is shown below.

• Variables: Variables are the unknown values that are present in the algebraic expression. For instance, 4y + 5z has y and z as variables.
• Coefficients: Coefficients are the fixed values (real numbers) attached to the variables. They are multiplied by the variables. For example, in 5x² + 3 the coefficient of x² is 5. 
• Term: A Term can be a constant, a variable, or a combination of both. Each term is separated by either addition or subtraction. For example, 3x + 5, 3x, and 5 are the two terms of the algebraic expression 3x + 5.
• Constant: In an algebraic expression, constants are fixed values that do not change. They are numerical values without any variables attached.
• Arithemtic Operation: These are basic mathematical operations that involve numbers and are used to perform calculations. For example, in 2x2 - 3, "-" is subtraction which is an arithmetic operation.

Types of Algebraic Expressions

There are various types of algebraic expressions based on the number of terms in the algebraic expression. The three main types of algebraic expressions are

• Monomial
• Binomial
• Trinomial
• Polynomial

Simplify Algebraic Expressions

Simplifying algebraic expressions is easy and very basic. First, understand what are like and unlike terms. Like terms have the same sign and unlike terms have opposite signs.

To simplify the given algebraic expression, first, find out the terms having the same power. Then, if the terms are like terms, add them; if they are unlike terms, find the difference between the terms.

The most simplified form of an algebraic expression is one where no same power terms are not repeated. 

For instance, let's simplify 4x⁵ + 3x³ - 8x² + 67 - 4x² + 6x³,

Same powers that are repeated are cubic and square, upon combining them together, the expression becomes, 4x⁵ + (3x³ + 6x³) - (8x² - 4x²) + 67.

Now, simplifying the expression, the final answer obtained is 4x⁵ + 9x³ - 12x² + 67.

This term does not have any terms repeated that have the same power.

Addition of Algebraic Expressions

When an addition operation is performed on two algebraic expressions, like terms are added with like terms only, i.e., coefficients of the like terms are added.

Example: Add (25x + 34y + 14z) and (9x − 16y + 6z + 17).

Solution:

(25x + 34y + 14z) + (9x − 16y + 6z + 17)

By writing like terms together, we get
= (25x + 9x) + (34y − 16y) + (14z + 6z) + 17

By adding like terms, we get
= 34x + 18y + 20z + 17.

Hence, (25x + 34y + 14z) + (9x − 16y + 6z + 17) = 34x + 18y + 20z + 17.

Read More: Addition of Algebraic Expressions.

Subtraction of Algebraic Expressions

To subtract an algebraic expression from another, we have to add the additive inverse of the second expression to the first expression.

Example: Subtract (3b² − 5b) from (5b² + 6b + 8) .

Solution:

(5b² + 6b + 8) − (3b² − 5b)

= (5b² + 6b + 8) - 3b² + 5b
= (5b² − 3b²) + (6b + 5b) + 8 
= 2b² + 11b + 8

Read More: Subtraction of Algebraic Expression.

Multiplication of Algebraic Expressions

When a multiplication operation is performed on two algebraic expressions, we have to multiply every term of the first expression with every term of the second expression and then combine all the products.

Example: Multiply (3x + 2y) with (4x + 6y − 8z)

Solution:

(3x + 2y)(4x + 6y − 8z) 

= 3x(4x) + 3x(6y) − 3x(8z) + 2y(4x) + 2y(6y) − 2y(8z)
= 12x² + 18xy − 24xz + 8xy + 12y² − 16yz
= 12x² + 12y² + 26xy − 16yz − 24xz

Read More: Multiplication of Algebraic Expression

Division of Algebraic Expressions

When we have to divide an algebraic expression from another, we can factorize both the numerator and the denominator, then cancel all the possible terms, and simplify the rest, or we can use the long division method when we cannot factorize the algebraic expressions.

Example: Solve: (x² + 5x + 6)/(x + 2)

Solution:

= (x² + 5x + 6)/(x + 2)

After factorizing (x² + 5x + 6) = (x + 2) (x + 3)

= [(x + 2) (x + 3)]/(x + 2)
= (x + 3)

Read More: Division of Algebraic Expressions

Examples Problems on Algebraic Expressions

Example 1: Find out the constant from the following algebraic expressions,

• x³ + 4x² - 6
• 9 + y⁵

Solution: 

Constants are the terms that do not have any variable attached to them, therefore, in the first case, -6 is the constant, and in the second case, 9 is the constant.

Example 2: Find out the number of terms present in the following expressions,

• 4x² + 7x - 8
• 5y⁷ - 12

Solution:

Terms are separated by each other either by addition or subtraction sign. Therefore, in the first case, there are 3 terms and in the second case, there are 2 terms.

Example 3: Simplify the algebraic term, z⁵ + z³ - y⁶ + 7z⁵ - 8y⁶ + 34 + 10z³

Solution:

In the expression, there are terms with the same power and same variable that are repeated, first bring them together,

(z⁵ + 7z⁵) + (z³ + 10z³) - (y⁶ - 8y⁶) + 34.
= 8z⁵ + 11z³ - 9y⁶ + 34. (simplifying like terms)

Example 4: Add (13x² + 11), ( – 25x² + 26x + 42) and (-33x – 29).

Solution:

Let F = (13x² + 11) + ( – 25x² + 26x + 42) + (-33x – 29)
⇒ F = 13x² – 25x² + 26x – 33x + 11 + 42 – 29
⇒ F = -12x² – 7x + 24

Hence, (13x² + 11) + ( – 25x² + 26x + 42) + (-33x – 29) = -12x² – 7x + 24.

Example 5: Solve (5x + 4y + 6z)² + (3y – 7x)².

Solution:

Given, Let F = (5x + 4y + 6z)² + (3y – 7x)²

From algebraic formulae, we have

• (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
• (a – b)² = a² – 2ab + b²

Thus, F = (5x)² + (4y)² + (6z)² + 2(5x)(4y) + 2(4y)(6z) + 2(6z)(5x) + [(3y)² – 2(3y)(7x) + (7x)²]
⇒ F= 25x² + 16y² + 36z² + 40xy + 48yz + 60zx +9y2 – 42xy + 49x²
⇒ F = 74x² + 25y² + 36z² – 2xy + 48yz + 60zx

Hence, (5x + 4y + 6z)² + (3y – 7x)² = 74x² + 25y² + 36z² – 2xy + 48yz + 60zx.