Algebraic expressions are defined as expressions which contain terms composed of variables, integers, constants, exponents. The terms are combined using algebraic operations which are addition, subtraction, multiplication, division and exponentiation. We can simplify algebraic expressions by following the simple rule of PEMDAS which stands for Parenthesis, Exponents, Multiplication, Division, Addition and Subtraction.

So, initially we remove the parenthesis by multiplying factors, then use the exponent rule to solve terms containing exponents, carry out multiplication and division followed by combing like terms by addition and subtraction. On the other hand, equation is defined as an expression with equal sign.

So, initially we remove the parenthesis by multiplying factors, then use the exponent rule to solve terms containing exponents, carry out multiplication and division followed by combing like terms by addition and subtraction. On the other hand, equation is defined as an expression with equal sign.

1. $2(x + 4) – 3(x – 5) + 5y$

2. $8x + 6y + x^3y+ 1$

3. $8x^3 + y^3 + 6x^2y + 3xy^2$

4. $(a + b) * (a^2 – 3ab^2 + 4y^2)$

There may be different types of expressions depending upon the types of variables used. Some important of them are: algebraic expression, rational expression, logarithmic expression, exponential expression, polynomial expression, radicals, trigonometrical expression etc.

By simplifying an algebraic expression or any mathematical expression we mean removing the complexity of the expression by writing it in a compact or most efficient manner such that the value of the algebraic expression remains unaltered. Simplification mainly involves collecting like terms. Like terms are those terms in an expression which have the same variables with same exponents but the coefficient may or may not be the same like 3xy^2, -xy^2$ and 8xy^2. Combining of like terms is done using the algebraic operations addition and subtraction. Parenthesis or terms present inside the bracket must always be solved first before we collect the like terms. Students always find difficulty in simplifying problems. But, it's actual purpose is to break down problems into steps and then, solving it. There are different ways of simplifying and different methods followed by different tutors for simplifying expressions. Get the help from our expert online tutors and be an expert yourself.

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The term exponent is defined as number of times a constant or a variable is multiplied. The symbol to denote an exponent is ‘^’. It is expressed in the form x^a, where x is the base and a is the power the base is raised to. It mean the variable x need to be multiplied with itself a times. There are certain rules of exponent which need to be followed while solving an expression containing exponents. The rules are as follows:

Product rule having same base: $a^m * a^n $= $a^(m + n)$

Product rule having same exponent but different base: $x^m * y^m$ = $(x * y)^m$

Quotient rule having same base: $\frac {a^n} {a^m}$ = $a^{(n – m)}$

Quotient rule having same exponent but different base: $\frac {a^m} {b^m}$ = $ (\frac{a}{b})^{m} $

Power rule: $(a^n) ^ m $= $a^nm$

Negative exponent:$ x^{-a} $= $\frac {1} {x^a}$

Zero rule:$ a^0 $ = 1

**For example:**

Simplify $(2a^2b)$ $\times$ $(5ab^2c)$

**Solution:**** Step 1:** We open the parenthesis and write the like variables and constants side by side

2 $\times$ 5 $\times$ $a^2$ $\times$ a $\times$ b $\times$ $b^2$ $\times$ c

**Step 2:** Applying the product rule with same base we add the exponents having same base

$10\ \times\ a^{(2 + 1)}\ \times\ b^{(1 + 2)}\ \times\ c$

**Step 3:** Solving the expression now, we get the answer $10a^3b^3c$

1) $a^2 + 2ab + b^2$ = $(a + b)^2$

2) $a^2 – 2ab + b^2 $ = $(a – b)^2$

3) $a^2 – b^2$ = $(a + b)(a- b)$

4) $a^3 + b^3$ = $(a + b)(a^2 – ab + b^2)$

5) $a^3 – b^3$ = $(a – b)(a^2 + ab + b^2)$

6) $a^3 + b^3 + 3a^2b + 3ab^2$ = $(a + b)^3$

7) $a^3 – b^3 – 3a^2b + 3ab^2$ = $(a – b)^3$

8) $a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$ = $(a + b + c)^2$

9) $a^3 + b^3 + c^3 -3abc$ = $(a + b + c)(a^2 + b^2 + c^2 – ab – bc – ca)$

^{}

- Square root is written as radical symbol, $\sqrt{}$.
- Cube root is written as $\sqrt[3]{}$
- Fourth root is written as $\sqrt[4]{}$
- Fifth root is written as $\sqrt[5]{}$
- nth root is written as $\sqrt[n]{}$

Let us simplify $2 \sqrt{2x^4}\ -\ x \sqrt{8}$

Given $2 \sqrt{2x^4}\ -\ x \sqrt{8}$

$2 \sqrt{2x^4}\ -\ x \sqrt{8}$ = $2x^2 \sqrt{2}\ -\ 2x \sqrt{2}$

= $(2x^2 - 2x)\ \sqrt{2}$

= $2x(x - 1) \sqrt{2}$

$\Rightarrow\ 2 \sqrt{2x^4}\ -\ x\sqrt{8}$ = $2x(x - 1) \sqrt{2}$

Let us simplify $\frac{x}{\sqrt{125}}$ + $\frac{3x}{\sqrt{20}}$

Given $\frac{x}{\sqrt{125}}$ + $\frac{3x}{\sqrt{20}}$

$\frac{x}{\sqrt{125}}$ + $\frac{3x}{\sqrt{20}}$ = $\frac{x}{\sqrt{5 * 5 * 5}}$ + $\frac{3x}{\sqrt{2 \times 2 \times 5}}$

= $\frac{x}{5 \sqrt{ 5}}$ + $\frac{3x}{2 \sqrt{5}}$

Denominators are different, so take LCM of $5 \sqrt{ 5}$ and $2 \sqrt{5}$ = $10 \sqrt{5}$.

= $\frac{2x + 15x}{10 \sqrt{5}}$

= $\frac{17x}{10 \sqrt{5}}$

$\Rightarrow$ $\frac{x}{\sqrt{125}}$ $+$ $\frac{3x}{\sqrt{20}}$ = $\frac{17x}{10 \sqrt{5}}$.

A **variable expressio****n** is defined as an expression which contains one or more variables with constants. These variables are connected to each other with the help of at least a single arithmetic operation (either + or – or * or /). Let us take up he following example:

Solution:

$3x^2$– $8x^3 $+ x(2x + 4$x^2$)

$3x^2$–$8x^3 $+ $2x^2 $+ 4$x^3$

Step 2:

terms are added in order to simplify

$5x^{2} $ – $4x^{3}$

Step 3:

Solve by Combining Like Terms

Algebraic expression can be solved by using distributive property or by combining like terms. Combining like terms is a process used to simplify an expression using addition and subtraction of the coefficients of terms.

Let us solve 2x + 3 - 6x = -7x

Combine like terms

2x - 6x + 3 = -7x

-4x + 3 = -7x

-4x + 7x = -3

3x = -3

x = -1

Factor 125 = 5 * 5 * 5 = $5^{3}$

Exponent = 3 (odd), then $5^3$ = $5^{2}$ $\times$ 5

The exponent of $x^5$ is odd, then $x^5$ = $x^{4}$ $\times$ x

The exponent of $y^{2}$ is even, the square root of exponent is 1.

$\sqrt{125x^5y^2z^3}$ = $\sqrt{5^2 x^4 y^2 z^2 5xz}$

[By the commutative property of multiplication]

= 5 $\times$ $x^2$ y z $\sqrt{5xz}$

[By the product property of square roots ]

= 5 $x^2$ y z $\sqrt{5xz}$

=> $\sqrt{125x^5y^2z^3}$ = 5$x^2$yz $\sqrt{5xz}$.

- log m + log n = log(mn)
- log m - log n = log($\frac{m}{n}$)
- log m
^{n}= n log m

Solution:

We need to break the value of 3000 in terms of its simplest factors

log (3 $\times$ 2 $\times$ 2 $\times$ 2 $\times$ 5 $\times$ 5 $\times$ 5) – log 3

log (3 $\times$ 2^3 $\times$ $5^{3}$) – log 3

Using the property of logarithm addition log ab = log a + log b

log 3 + $log 2^3 $+ $log 5^{3} $– log 3

or log 3 + 3 log 2 + 3 log 5 – log 3

or 3 log 2 + 3 log 5

|x| = $\left\{\begin{matrix}

x, & x \geq 0\\

-x, & x < 0

\end{matrix}\right.$

Let us take up the following example:

Solution:

x + 4 = 18

x + 4 – 4 = 18 – 4

x = 14

And,

x + 4 = -18

x + 4 – 4 = -18 – 4

x = -22

Thus, the two values of variable x which satisfies the given absolute value expression are x = 14 and x = -22

Use the product and quotient rules for simplifying radicals:

Let m and n be real numbers, variables or algebraic expressions. If nth root of both are real, then,

- $\sqrt[n]{mn}$ = $\sqrt[n]{m}$ $\times$ $\sqrt[n]{n}$
- $\sqrt[n]{\frac{m}{n}}$ = $\frac{\sqrt[n]{m}}{\sqrt[n]{n}}$, n $\neq$ 0.

If m and n be real numbers then,

$x^m$ $\times$ $x^n$ = $x^{m+n}$

$\frac{x^m}{x^n}$ = $x^{m-n}$

$(x^m)^n$ = $x^{mn}$

$x^{-m}$ = $\frac{1}{x^m}$

$(xy)^m = x^m $\times$ y^m$

Let us simplify (2x)

(2x)

[Using power law, $(x^m)^n$ = $x^{mn}$ and $(xy)^m = x^m * y^m$]

= 8

= 8

= (8 - 5)

= 3

Given $\frac{12x^2z^5}{24x^3z^2}$

$\frac{12x^2z^5}{24x^3z^2}$ = $\frac{x^{2 - 3} z^{5 - 2}}{2}$

(Cancel common terms and using law, $\frac{x^m}{x^n}$ = $x^{m-n}$ ]

= $\frac{1}{2}$ $x^{-1}$ $z^{3}$

= $\frac{1}{2x}$ $z^{3}$

= $\frac{z^3}{2x}$

=> $\frac{12x^2z^5}{24x^3z^2}$

Given 10(x

10(x

= 10 x

= 10 x

= 10 x

Solution:

$\frac{ (\frac {3x^2 – 6x }{ 6x^2 – 24)} } {\frac{(3x^2 + x }{ x + 2}) }$

We need to first factorize each of the numerators and denominators Starting with,

$3x^2$ + 6x

In this equation the common factor is 3x, if we take out the common facor 3x we get x – 2. So $3x^2$ + 6x = 3x(x-2)

The same procedure we will follow for next equation i.e., $6x^2$ + 24

In this the common factor is 6, if we take 6 out then we will remain with $x^2$ – 4.

We can factorize $x^2$ – 4 by applying the formula $a^2$ – $b^2$ =( a+b)(a-b)

We get (x+2)(x-2)

Hence, $6x^2$ + 24 = 6(x+2)(x-2)

Next, we need to factorize 3x2 + x, here x is the common factor. Hence we get 3x2 + x = x(3x+1)

Hence we got,

3x2 + 6x = 3x(x-2)

$6x^2$ + 24 = 6(x+2)(x-2)

$3x^2$ + x = x(3x+1)

Substitute all these factors in the equation we get

Cancelling the common terms we get $\frac{1} {2(3x+1)}$

Thus, the simplified form of the given rational function is $\frac{1}{2 (3x + 1)}$

$log_2 18$ – $log_ 36$ + $log_2 128$

18 = $2 \times 3 \times 3$ = $2$ $\times$ $3^2$

36 = $2 \times 2 \times 3 \times 3$ = $2^{2}$ $\times$ $3^2$

128 = $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$ = $2^7$

$log_2 (2 \times 3^{2})$ – $log_2 (2^{2} \times 3^{2})$+ $log_2 (2^{7})$

$\frac{log_ 2 (2 \times 3^2 \times 2^7)}{2^2 \times 3^2}$

$log_2 2^{6}$

= 6 ( because log_2 (2) = 1)

$\frac{2 (3 a^4 b^-2 c^-2) ^{-2}}{3 a^2 b^2 c^2}$

Applying the power rule in the numerator, we get

$\frac{2 (3^{-2} a^{-8} b^{4} c^{4}) }{3 a^{2} b^{2} c^{2}}$

Using the exponential quotient rule having same base $\frac {a^n }{ a^m}$ = $a^{(n – m)}$, we get

$2 (3^{-2-1} a^{-8-2} b^{4-2} c^{4-2}$

$2 (3^{-3} a^{-10} b^{2} c^{2})$

Changing the negative exponent to positive exponent by taking its reciprocal, we get

$\frac{2 b^2 c^2 }{3^3 a^{10}}$

$\frac{2 b^2 c^2 }{27 a^{10}}$ ..Answer!!

$\frac{\frac {(x^2 – 9x – 10} {x^2 + x - 6})} {\frac{(x^2 – 1}{ x^2 – 4})}$

denominator first.

Let us factorize the numerator x^2 – 9x – 10. The values of a, b and c are a = 1, b = -9 and c = -10.The product of a and c is

a $\times$ c = 1 $\times$ -10 = -10 and the pair of factors of ac which on adding we get the value of b is -10 and 1. So,

$x^2 – 9x – 10$

=$x^2 – 10x + x – 10$

Factoring out the common term from first two terms and next two terms, we get

= x(x – 10) + 1(x – 10)

= (x + 1)(x – 10)

The numerator of the second fraction is x^2 – 1. Using the difference of square formula, (a^2 – b^2) = (a + b)(a – b)

$x^2$– 1

= (x + 1)(x – 1)

Factorizing the denominator of the first fraction,$x^2$+ x – 6. The values of a, b and c are a = 1, b = 1 and c = -6. The product of

a and c is a $\times$ c = 1 $\times$ -6 = -6 and the pair of factors of ac which on adding we get the value of b is 3 and -2. So,

$x^2$+ x – 6

= $x^2 $+ 3x – 2x – 6

Factoring out the common term from first two terms and next two terms, we get

= x(x + 3) – 2(x + 3)

= (x – 2)(x + 3)

Factorizing the denominator of the second fraction, x^2 – 4. Using the difference of square formula, (a^2 – b^2) = (a + b)(a – b)

$x^2$– 4

= (x + 2)(x – 2)

common factors present both in the numerator and denominator.

$\frac {\frac{(x + 1)(x – 10) }{(x – 2)(x + 3)} }{ \frac{(x + 1)(x – 1) }{ (x + 2)(x – 2)}}$

Dividing by a fraction mean multiplying with its reciprocal

$\frac{(x + 1)(x – 10) }{(x – 2)(x + 3)} \ast \frac{(x + 2)(x – 2)}{(x + 1)(x – 1)}$

The common factors are (x + 1) and (x – 2)

$\frac{(x – 10)}{(x + 3)} \ast \frac{(x + 2)}{(x – 1)}$

Given $\frac{\frac{1}{x} + \frac{3}{2x}}{\frac{x}{x+2} + 1}$

**Step 1:**

Simplify numerator and denominator separately, write the numerator and denominator as a single rational expression.

=> $\frac{1}{x} + \frac{3}{2x}$ = $\frac{2 + 3}{2x}$ (LCM of x and 2x = 2x)

= $\frac{5}{2x}$

and $\frac{x}{x+2}$ + 1 = $\frac{x + 1(x + 2)}{x +2}$ (LCM of x + 2 and 1 = x + 2)

= $\frac{x + x + 2}{x + 2}$

= $\frac{2x + 2}{x + 2}$

Now, the numerator and denominator of the complex fraction are single rational expressions.

**Step 2:**

Multiply by the reciprocal of denominator

$\frac{5}{2x}$ * $\frac{x + 2}{2x + 2}$

$\frac{5(x + 2)}{2x(2x + 2)}$

$\frac{5x +10}{4x^2 + 4x}$

=> $\frac{\frac{1}{x} + \frac{3}{2x}}{\frac{x}{x+2} + 1}$ = $\frac{5x +10}{4x^2 + 4x}$

Simplify numerator and denominator separately, write the numerator and denominator as a single rational expression.

=> $\frac{1}{x} + \frac{3}{2x}$ = $\frac{2 + 3}{2x}$ (LCM of x and 2x = 2x)

= $\frac{5}{2x}$

and $\frac{x}{x+2}$ + 1 = $\frac{x + 1(x + 2)}{x +2}$ (LCM of x + 2 and 1 = x + 2)

= $\frac{x + x + 2}{x + 2}$

= $\frac{2x + 2}{x + 2}$

Now, the numerator and denominator of the complex fraction are single rational expressions.

Multiply by the reciprocal of denominator

$\frac{5}{2x}$ * $\frac{x + 2}{2x + 2}$

$\frac{5(x + 2)}{2x(2x + 2)}$

$\frac{5x +10}{4x^2 + 4x}$

=> $\frac{\frac{1}{x} + \frac{3}{2x}}{\frac{x}{x+2} + 1}$ = $\frac{5x +10}{4x^2 + 4x}$

Given 7x^{3} - 4x^{2} + 6x - 9

Put x = -4 in given expression

=> 7x^{3} - 4x^{2} + 6x - 9 = 7(-4)^{3} - 4(-4)^{2} + 6(-4) - 9

= 7 * - 64 - 4 * 16 - 24 - 9

= -448 - 64 - 32

= -544

Put x = -4 in given expression

=> 7x

= 7 * - 64 - 4 * 16 - 24 - 9

= -448 - 64 - 32

= -544

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