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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.

x, & x \geq 0\\
-x, & x < 0
\end{matrix}\right.$Let us take up the following example: Example: |x + 4| = 18 Solution: Starting with separating it into two equations x + 4 = 18 x + 4 – 4 = 18 – 4 x = 14 And, x + 4 = -18 x + 4 – 4 = -18 – 4 x = -22 Thus, the two values of variable x which satisfies the given absolute value expression are x = 14 and x = -22 ## Complex Expressions Back to Top A Complex expression is a rational expression whose numerator and denominator contain one or more rational expressions. ### Simplifying Complex Radical Expressions Use the product and quotient rules for simplifying radicals: Let m and n be real numbers, variables or algebraic expressions. If nth root of both are real, then, •$\sqrt[n]{mn}$=$\sqrt[n]{m}\times\sqrt[n]{n}$•$\sqrt[n]{\frac{m}{n}}$=$\frac{\sqrt[n]{m}}{\sqrt[n]{n}}$, n$\neq$0. ## Simplifying Expressions with Exponents Back to Top To simplify expressions with exponents, first apply power to power rules. Then, combine like terms and arrange the terms from highest exponent to lowest exponent. Power Rules: If m and n be real numbers then,$x^m\timesx^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}$] = 10 x -18 - 10 y1 - 2 = 10 x -28 y - 1 ## Examples Back to Top Question 1: Simplify$\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\times3^2$36 =$2 \times  2 \times  3 \times  3$=$2^{2}\times3^2$128 =$2 \times  2 \times  2 \times  2 \times  2 \times  2 \times  2$=$2^7log_2 (2 \times  3^{2})$–$log_2 (2^{2} \times  3^{2})$+$log_2 (2^{7})$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$2^{2}\times3^{2}$, 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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