How To Simplify Rational Expressions? - Examples With Detailed Solutions

How to simplify rational expressions? Grade 11 examples are presented along with detailed solutions and full explanations. More questions with answers at the bottom of the page are also included.

 How to simplify rational expressions? How to simplify rational expressions? The rules of addition, subtraction, multiplication and division of rational expressions are used to simplify complex expressions. Rational expressions with the same denominator are added or subtracted as follows: $\dfrac{A}{B} \pm \dfrac{C}{B} = \dfrac{A \pm C}{B}$ Rational expressions are multplied as follows: $\dfrac{A}{B} \cdot \dfrac{C}{D} = \dfrac{A \cdot C}{B \cdot D}$ We divide two rational expressions by multiplying the first rational expression by the reciprocal of the second rational expression as follows: $\dfrac{A}{B} \div \dfrac{C}{D} = \dfrac{A}{B} \cdot \dfrac{D}{C} = \dfrac{A \cdot D}{B \cdot C} \;\; \text{or} \;\; \dfrac{ \dfrac{A}{B} }{ \dfrac{C}{D}} = \dfrac{A}{B} \cdot \dfrac{D}{C} = \dfrac{A \cdot D}{B \cdot C}$ If you have difficulties in simplifying rational expressions, review the tutorials in How to Add, Subtract and Simplify Rational Expressions and How to multiply, divide and simplify rational expressions and then start the present tutorial. The examples with detailed solutions and explanations in this tutorials will help you overcome any difficulties in simplifying rational expressions on the condition that you understand every step involved in solving these questions and also spend more time practicing if needed. I will present the examples below starting with fractions first and then with rational expressions, with more challenging questions as you walk through the tutorial. You need to understand each step! Example 1: Simplify: $\dfrac{\dfrac{2}{3} + 5}{4} +\dfrac{1}{2}$. Solution: We first simplify the numerator $\dfrac{2}{3} + 5$ of the fraction $\dfrac{\dfrac{2}{3} + 5}{4}$. Convert to common denominator. $\dfrac{\dfrac{2}{3} + 5}{4} +\dfrac{1}{2} = \dfrac{\dfrac{2}{3} + 5 \cdot \dfrac{3}{3}}{4} +\dfrac{1}{2} = \dfrac{\dfrac{17}{3}}{4} +\dfrac{1}{2}$ Change 4 to a fraction 4 / 1 and simplify $\dfrac{\dfrac{17}{3}}{4}$ by dividing the fractions which comes to a multiplication by the reciprocal. $= \dfrac{\dfrac{17}{3}}{\dfrac{4}{1}} +\dfrac{1}{2} = \dfrac{17}{3} \cdot \dfrac{1}{4} +\dfrac{1}{2}$ Multiply the fractions $\dfrac{17}{3} \cdot \dfrac{1}{4}$. $= \dfrac{17 \cdot 1}{3 \cdot 4}+\dfrac{1}{2} = \dfrac{17}{12} +\dfrac{1}{2}$ Convert to common denominator and add. $= \dfrac{17}{12} +\dfrac{1}{2} \cdot \dfrac{6}{6} = \dfrac{17+6}{12} = \dfrac{23}{12}$ Example 2: Simplify: $\dfrac{\dfrac{x+1}{x-2}+\dfrac{x}{x+1}}{\dfrac{1}{x+1}}$. Solution: We first simplify the numerator $\dfrac{x+1}{x-2}+\dfrac{x}{x+1}$ by converting to the same denominator and applying the rule of the sum of two rational expressions whose lowest common denominator is (x - 2)(x + 1). $\dfrac{\dfrac{x+1}{x-2}+\dfrac{x}{x+1}}{\dfrac{1}{x+1}} = \dfrac{\dfrac{x+1}{x-2} \cdot \dfrac{x+1}{x+1} +\dfrac{x}{x+1} \cdot \dfrac{x-2}{x-2} }{\dfrac{1}{x+1}} = \dfrac{ \dfrac{(x+1)^2+x(x-2)}{(x+1)(x-2)} }{\dfrac{1}{x+1}}$ We now apply the rule of division of rational expressions by multiplying by the reciprocal of $\dfrac{x+1}{1}$. $= \dfrac{((x+1)^2+x(x-2))}{(x+1)(x-2)} \cdot \dfrac{x+1}{1}$ Simplify by cancelling common factors. $= \dfrac{((x+1)^2+x(x-2))}{{\colorcancel{red}{(x+1)}}(x-2)} \cdot \dfrac{\colorcancel{red}{(x+1)}}{1} = \dfrac{((x+1)^2+x(x-2))}{(x-2)}$ Expand and simplify. $= \dfrac{2x^2+1}{x-2} \;\; \text{for} \;\; x \ne -1$ Example 3: Simplify: Simplify: $\dfrac{\dfrac{x-1}{3x+2}+3}{\dfrac{x+1}{6x+4}} - 2$. Solution: We simplify $\dfrac{x-1}{3x+2}+3$ by first converting to the same denominator $\dfrac{\dfrac{x-1}{3x+2}+3}{\dfrac{x+1}{6x+4}} - 2 = \dfrac{\dfrac{x-1}{3x+2}+3 \cdot \dfrac{3x+2}{3x+2}}{\dfrac{x+1}{6x+4}} - 2$ then applying the rule of the sum of two rational expressions. $= \dfrac{\dfrac{x - 1 +3 \cdot (3x+2)}{3x+2}}{\dfrac{x+1}{6x+4}} - 2$ Expand and simplify $x - 1 +3 \cdot (3x+2)$ . $= \dfrac{\dfrac{10x + 5}{3x+2}}{\dfrac{x+1}{6x+4}} - 2$ We use the rule of division of rational expressions which reduces to a multiplication by the reciprocal on: $\dfrac{\dfrac{10x + 5}{3x+2}}{\dfrac{x+1}{6x+4}} = \dfrac{10x + 5}{3x+2} \cdot \dfrac{6x+4}{x+1}$. $= \dfrac{10x + 5}{3x+2} \cdot \dfrac{6x+4}{x+1} - 2$ Factor $= \dfrac{5(2x + 1)}{3x+2} \cdot \dfrac {2(3x+2)}{x+1} - 2 =$ and simplify . $= \dfrac{5(2x + 1)}{{\colorcancel{red}{3x+2}}} \cdot \dfrac {2{\colorcancel{red}{(3x+2)}}}{x+1} - 2 = \dfrac{10 (2x+1)}{x+1}-2$ Convert to the same denominator, apply the rule of addition of rational expressions and simplify. $\dfrac{10(2x+1)}{x+1}-2 \cdot \dfrac{x+1}{x+1} = \dfrac{10(2x+1) - 2(x+1)}{x+1} = \dfrac{2(9x+4)}{x+1} \;\; \text{for} \;\; x \ne -2/3$ Example 4: Simplify: $\dfrac{ \dfrac{2-x}{x+2}-\dfrac{4}{x+2}}{\dfrac{1}{x+3}-\dfrac{4}{x+3}}$ Solution: Apply the rule of subtraction of rational expressions to $\dfrac{2-x}{x+2}-\dfrac{4}{x+2}$ and $\dfrac{1}{x+3}-\dfrac{4}{x+3}$ $\dfrac{ \dfrac{2-x}{x+2}-\dfrac{4}{x+2}}{\dfrac{1}{x+3}-\dfrac{4}{x+3}} = \dfrac{ \dfrac{2-x -4}{x+2}}{\dfrac{1-4}{x+3}}$ then simplify. $= \dfrac{ \dfrac{-x - 2}{x+2}}{\dfrac{-3}{x+3}}$ Apply the rule of division of rational expressions and factor $= \dfrac{-x - 2}{x+2} \cdot \dfrac{x+3}{-3} = \dfrac{-(x + 2)}{x+2} \cdot \dfrac{x+3}{-3}$ and simplify $= \dfrac{-{\colorcancel{red}{(x+2)}}}{{\colorcancel{red}{x+2}}} \cdot \dfrac{x+3}{-3} = \dfrac{x+3}{3} \;\; \text{for} \;\; x \ne -2$ Example 5: Simplify: $\dfrac{ \dfrac{2-x}{x+2} \cdot \dfrac{4}{x+3}}{\dfrac{1}{x+3}-\dfrac{4}{x+3}}$ Solution: Apply the rule of multiplication to $\dfrac{2-x}{x+2} \cdot \dfrac{4}{x+3}$ and the rule of subtraction to $\dfrac{1}{x+3}-\dfrac{4}{x+3}$. $\dfrac{ \dfrac{2-x}{x+2} \cdot \dfrac{4}{x+3}}{\dfrac{1}{x+3}-\dfrac{4}{x+3}} = \dfrac{ \dfrac{4(2-x)}{(x+2)(x+3)}}{\dfrac{1 - 4}{x+3}}$ Apply the rule of division: $= \dfrac{4(2-x)}{(x+2)(x+3)} \cdot \dfrac{x+3}{-3}$ Simplify. $= \dfrac{4(2-x)}{(x+2){\colorcancel{red}{(x+3)}}} \cdot \dfrac{{\colorcancel{red}{(x+3)}}}{-3} = -\dfrac{4(2-x)}{3(x+2)} \;\; \text{for} \;\; x \ne -3$ Example 6: Simplify: $\dfrac{1}{1+\dfrac{1}{x+\dfrac{1}{x+1}}}$. Solution: We first convert the terms in $x + \dfrac{1}{x+1}$ to the same denominator. $\dfrac{1}{1+\dfrac{1}{x+\dfrac{1}{x+1}}} = \dfrac{1}{1+\dfrac{1}{x \cdot \dfrac{x+1}{x+1}+\dfrac{1}{x+1}}}$ We now add $x \cdot \dfrac{x+1}{x+1}+\dfrac{1}{x+1}$ and simplify. $= \dfrac{1}{1+\dfrac{1}{\dfrac{x^2+x+1}{x+1}}}$ Note that $\dfrac{1}{\dfrac{x^2+x+1}{x+1}}$ is a reciprocal is equal to $\dfrac{x+1}{x^2+x+1}$. Hence $= \dfrac{1}{1+ \dfrac{x+1}{x^2+x+1}}$ Convert the terms in $1+ \dfrac{x+1}{x^2+x+1}$ to the same denominator. $= \dfrac{1}{1 \cdot \dfrac{x^2+x+1}{x^2+x+1}+ \dfrac{x+1}{x^2+x+1}}$ Simplify and apply the rule of addition to $\dfrac{x^2+x+1}{x^2+x+1}+ \dfrac{x+1}{x^2+x+1}$. $= \dfrac{1}{ \dfrac{x^2+x+1}{x^2+x+1}+ \dfrac{x+1}{x^2+x+1}} = \dfrac{1}{ \dfrac{x^2+x+1+x+1}{x^2+x+1}} = \dfrac{1}{ \dfrac{x^2+2x+2}{x^2+x+1}}$ The last expression is the reciprocal of a rational expression which may be written as. $= \dfrac{x^2+x+1}{x^2+2x+2}$ More Questions: Simplify the following expressions - Answers at the bottom of the page. a) $\dfrac{\dfrac{2}{5} + 7}{\dfrac{4}{3}} +\dfrac{1}{3}$ b) $\dfrac{\dfrac{x-2}{x+3}+\dfrac{x}{x+3}}{\dfrac{1}{2x+6}}$ c) $\dfrac{\dfrac{x-1}{3x-12}+3}{\dfrac{x+1}{x-4}} - \dfrac{2}{3}$ d) $\dfrac{ \dfrac{2-x}{x+1}-\dfrac{4}{x+1}}{\dfrac{1}{x-5}-\dfrac{4}{x-5}}$ e) $\dfrac{ \dfrac{2-x}{x+2} - \dfrac{4}{x+3}}{\dfrac{1}{x+3} \cdot \dfrac{4}{x+3}}$ f) $\dfrac{\dfrac{1}{2}}{1+\dfrac{1}{\dfrac{1}{x-1}-2}}$ Answers to the above questions a) $\dfrac{353}{60}$ b) $4(x-1)$ c) $\dfrac{8x-39}{3(x+1)}$ d)$\dfrac{ (x+2 ) (x-5)}{3(x+1)}$ e) $\dfrac{(-x^2-5x-2)(x+3)}{4(x+2)}$ f)$\dfrac{-2x+3}{2(-x+2)}$

Updated: 20 January 2017 (A Dendane)