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Simplifying Expressions

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.


Few examples of algebraic expressions are as follows: 

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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Algebraic Expressions

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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 $10a^3b^3c$

Rational expression is defined as fraction. The numerator and denominator of the fraction is polynomial. Rational expressions are simplified by finding out the common factors present in both the numerator and denominator. After finding out the common factor we cancel it out and simplify the expression. We need to remember few of the most common factorization formula using polynomial function. They are as follows:
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) $a^3 + b^3 + c^3 -3abc$ = $(a + b + c)(a^2 + b^2 + c^2 – ab – bc – ca)$

Radical Expressions

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Radical expressions are defined as expressions which contains the radical symbol √. The value or variable under this symbol is called the radicand. The different radical symbols are as follows: 
The radical symbols are represented as:
  • 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 $\sqrt[n]{}$

Simplifying Radical Expressions with Exponents


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

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

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

= $(2x^2 - 2x)\ \sqrt{2}$ 

= $2x(x - 1) \sqrt{2}$ 

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

Simplifying Radical Expressions Fractions


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 $2 \sqrt{5}$ = $10 \sqrt{5}$.

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

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

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

Variable Expressions

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A variable expression 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:

Example: Simplify $3x^2 – 8x^3 + x(2x + 4x^2)$

Solution:
Step 1: Remove the parenthesis by using the distributive property.

$3x^2$– $8x^3 $+ x(2x + 4$x^2$)

$3x^2$–$8x^3 $+ $2x^2 $+ 4$x^3$

Step 2:
Combine like terms. The like terms in the expression are $3x^2$and $2x^2$  & $-8x^3$and $4x^3$. These two pair of like 

terms are added in order to simplify

$5x^{2} $ – $4x^{3}$

Step 3:
Thus ,the simplified variable expression is  $5x^{2} $– $4x^{3}$

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

Expressions with Square Roots

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The process of obtaining the 'square root' of a number is termed as 'solving' or 'simplification'. The number is broken down to its factors to obtain the number which satisfies the condition of being the square root of the number.

Let us simplify $\sqrt{125x^5y^2z^3}$

Given $\sqrt{125x^5y^2z^3}$
Step 1:
Case 1:
Factor 125 = 5 * 5 * 5 = $5^{3}$

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

Case 2:
The exponent of $x^5$ is odd, then $x^5$ = $x^{4}$ $\times$ x

Case 3:
The exponent of $y^{2}$ is even, the square root of exponent is 1.

Case 4:
The exponent of $z^3$ is odd, then $z^{3}$ = $z^{2}$ $\times$ z

Step 2:

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

[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 $\sqrt{5xz}$.

Logarithmic Expressions

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Logarithmic expressions is defined as expressions which contains logarithm of variables and constants connected to each other with the help of arithmetic operations. Simplification of logarithmic expressions needs good understanding of logarithmic rules. Few of the basic logarithmic rules are as follows:
  • log m + log n = log(mn)
  • log m - log n = log($\frac{m}{n}$)
  • log mn = n log m

Example: log 3000 – log 3 

Solution:
log 3000 – log 3

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

Absolute Value Expressions

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Absolute value expressions are defined as expressions which contain absolute value of a number or a variable. The symbol to denote absolute value is | |. The value residing inside the bars is known as the argument. It is actually a measure of the distance of that argument from the 0 mark on a number line. Absolute value mean only positive value. Simplifying absolute value expressions is begun by separating it into two separate equations and then solving it separately. 

The absolute function can be defined as:

|x| = $\left\{\begin{matrix}
x, & x \geq 0\\
-x, & x < 0
\end{matrix}\right.$

Let us take up the following example:

Example: |x + 4| = 18

Solution: 
Starting with separating it into two equations

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

Complex Expressions

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A Complex expression is a rational expression whose numerator and denominator contain one or more rational expressions.

Simplifying Expressions

Simplifying Complex Radical Expressions


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.

Simplifying Expressions with Exponents

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To simplify expressions with exponents, first apply power to power rules. Then, combine like terms and arrange the terms from highest exponent to lowest exponent.

Power Rules:

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)3 - (x2)4(5x5)

Given (2x)3 - (x2)4(5x5)

(2x)3 - (x2)4(5x5) = 2$\times$  x3 - x$\times$ 5 $\times$ x5

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

= 8 x3 - 5 x8 - 5 [$x^m$ * $x^n$ = $x^{m+n}$]

= 8 x3 - 5x3

= (8 - 5) x3

= 3 x3

Simplifying Expressions with Rational Exponents


Let us simplify $\frac{12x^2z^5}{24x^3z^2}$

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

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

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

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

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

Simplifying Expressions with Negative Exponents


Let us simplify 10(x -9)2y(x-5y-1)2

Given 10(x-9)2y(x-5y-1)2

10(x -9)2y(x -5y -1)2 = 10(x -18) y (x -10 y -2) Using power law, $(x^m)^n$ = $x^{mn}$

= 10 x -18 + (-10) y1 + (-2) [$x^m$ * $x^n$ = $x^{m+n}$]

= 10 x -18 - 10 y1 - 2

= 10 x -28 y - 1

Examples

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Question 1:  Simplify $\frac{ (\frac {3x^2 – 6x }{ 6x^2 – 24)} } {\frac{(3x^2 + x }{ x + 2} )}$

Solution: 
Consider the equation

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

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

Question 4: Simplify the logarithm expression: 
           
                      $log_2 18$ – $log_ 36$ + $log_2 128$

Solution: Step 1: We first find out the prime factors of each of the log values given

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

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

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

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

Step 2: Applying the logarithmic rule of addition and subtraction log m + log n = log mn and log m – log n = log ($\frac{m} { n}$), we get

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

Step 3: Cancelling out the common factors $2^{2}$ $\times$ $3^{2}$ , we get

$log_2  2^{6}$

= 6 ( because log_2 (2) = 1)


Question 5: Simplify the exponential function
                 
                               $\frac{2 (3 a^4 b^-2 c^-2) ^{-2}}{3 a^2 b^2 c^2}$

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

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!!

Question 6: Simplify the rational expression: 

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

Solution: Step 1: To simplify a rational expression we need to find out the common factors present in both the numerator and 
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)

Step 2: After completing finding out the factors of each of the polynomial present in the given problem, we try to find out the 
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)

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

Step 3: Thus the simplified form is $\frac{(x – 10) (x + 2) }{(x + 3)(x – 1)}$

Solved Examples

Question 1: Simplify $\frac{\frac{1}{x} + \frac{3}{2x}}{\frac{x}{x+2} + 1}$
Solution:
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

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

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

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

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

Question 2: Evaluate 7x3 - 4x2 + 6x - 9 for x = -4.
Solution:
Given 7x3 - 4x2 + 6x - 9

Put x = -4 in given expression

=> 7x3 - 4x2 + 6x - 9 = 7(-4)3 - 4(-4)2 + 6(-4) - 9

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

= -448 - 64 - 32

= -544

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