Simplifying algebraic expressions means to rewrite the given algebraic expression in a compact form. This is done mainly by adding or subtracting all the terms that can be added or subtracted, removing brackets and so on.

Each term in an algebraic expression is separated by a mathematical sign. The term without any variable is called as a constant. The numerical factor of a variable term is called its numerical coefficient.

Only like terms can be added or subtracted. Like terms are terms with the same variables raised to the same power. For example, 2x + 8x is an algebraic expression with two terms. These two terms are like terms as they have the same variable x and the power of the variable is one in both the terms. Hence, this expression can be simplified by adding up the like terms.

2x + 8x = 10x. This is a single term.

Now consider 3(x + 3y). This is also an algebraic expression which can be simplified by using Distributive Law.

**Distributive Law:** The sum of any two addends multiplied by a number is equal to the sum of the product of each addend and the number. For any three variables x, y and z, we have,

x(y+z) = xy + xz

Using the distributive law, we have, 3(x + 3y) = 3x + 9y. The bracket has been removed.

This cannot be simplified further as the two terms are not like terms. One variable is x and the other variable is y. The variables are not same.

We cannot simplify an expression like 3s + $s^{2}$. The two terms have the same variables but their powers are not the same. Hence, they are unlike terms.

In order to simplify an algebraic expression, we must be able to distinguish between like and unlike terms in an expression.

### Properties used to Simplify Algebraic Expression

Given below are some of the properties that are used in simplifying an algebraic expression:

**Commutative Property of Addition and Multiplication**

a + b = b + a

ab = ba

**Associative Property of Addition and Multiplication**

a + (b + c) = (a + b) + c

a(bc) = (ab)c

**Additive Identity Property**

a + 0 = 0 + a = a

Zero is the additive identity element

**Distributive Law**

a(b + c) = ab + ac

**Multiplicative Identity Property**

a x 1 = 1 x a = a

One is the Multiplicative identity element

**Multiplication by 0**

The product of any number and zero is always zero.

a x 0 = 0 x a = 0

**Additive Inverse Property**

a + (-a) = (-a) + a = 0

For every real number 'a', there exists an inverse -a such that the sum of the number a and its inverse -a is zero.

**Multiplicative Inverse Property**

For any real number 'a' except zero, there is another number $\frac{1}{a}$ (the reciprocal of a) such that a x ($\frac{1}{a}$) = 1

$a$ and $\frac{1}{a}$ are the Multiplicative Inverses of each other.

**Division by 1**

The quotient of any number divided by 1 is the number itself.

a ÷ 1 = a

**Division by 0**

Division by 0 is undefined. But, when 0 is divided by any number, the quotient is 0.

$\frac{a}{0}$ = undefined

$\frac{0}{a}$ = 0

With the help of the properties listed above and the steps listed below, we can simplify an algebraic expression efficiently.

**Step 1:** Simplify the expressions within brackets

**Step 2:** Simplify the exponents

**Step 3:** Complete the multiplication and division operations as they occur from left to right.

**Step 3:** Complete the addition and subtraction operations for like terms as they appear from left to right.### Examples on Simplifying Algebraic Expressions

Given below are some solved examples on how to simplify algebraic expressions.

**Example 1:**

Simplify the following

$[\frac{1}{x-1} - \frac{1}{x-3}] / [\frac{1}{x-3} - \frac{1}{x-5}]$

**Solution: **

Start by simplifying the numerator and the denominator separately.

$\frac{1}{x-1} - \frac{1}{x-3} = \frac{(x-3) - (x-1)}{(x-1)(x-3)}$

$= \frac{-2}{(x-1)(x-3)}$

$\frac{1}{x-3} - \frac{1}{x-5} = \frac{(x-5) - (x-3)}{(x-3)(x-5)}$

$= \frac{-2}{(x-3)(x-5)}$ =

We have the expression reduced to

$\frac{-2}{(x-1)(x-3)} / \frac{-2}{(x-3)(x-5)}$

Using the rule for division of algebraic fractions,

a/b ÷ c/d = ad/bc

$\frac{-2}{(x-1)(x-3)} / \frac{-2}{(x-3)(x-5)} = \frac{-2}{(x-1)(x-3)} \times \frac{(x-3)(x-5)}{-2}$

$= \frac{(-2)(x-3)(x-5)}{(x-1)(x-3)(-2)}$

$= \frac{(x-5)}{(x-1)}$

Hence the given expression is simplified to $\frac{(x-5)}{(x-1)}$

**Example 2:**

Simplify $2(3x^{2} + 6x - 1) + 3(5x + 1) - 5x^{2}$

**Solution: **

Rewrite the expression by using Distributive Property

$2(3x^{2} + 6x - 1) = 2(3x^{2}) + 2(6x) - 2(1)$

$ = 6x^{2} + 12x - 2$

$3(5x + 1) = 3(5x) + 3(1)$

$= 15x + 3$

The given expression is now reduced to,

$2(3x^{2} + 6x - 1) + 3(5x + 1) - 5x^{2} = 6x^{2} + 12x - 2 + 15x + 3 - 5x^{2}$

Grouping the like terms and constants together,

$6x^{2} + 12x - 2 + 15x + 3 - 5x^{2} = (6x^{2} - 5x^{2}) + (12x + 15x) + ( -2 + 3)$

Perform the operations that are specified within each of the brackets

$(6x^{2} - 5x^{2}) = x^{2}$

$(12x + 15x) = 27x$

$( -2 + 3) = 1$

Rewrite the expression

$(6x^{2} - 5x^{2}) + (12x + 15x) + ( -2 + 3) = x^{2} + 27x + 1$

There are no like terms. And so, this is the simplest form.

$2(3x^{2} + 6x - 1) + 3(5x + 1) - 5x^{2} = x^{2} + 27x + 1$