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This is one of our many quizzes on Simplifying rational expressions. Click on this to access more such competitive quizzes on Rational numbers.

Simplifying rational expressions Quiz 2

mathematics

 Simplify $\dfrac{\dfrac{x+y}{x}}{\dfrac{1}{x}+\dfrac{1}{y}}$.

 To solve the above question, first take L.C.M of the denominator and then multiply it by the reciprocal of the denominator. After this, remove all the like terms from the equation to get your answer.

$ \text{We are given that: }\dfrac{\dfrac{x+y}{x}}{\dfrac{1}{x}+\dfrac{1}{y}} \\ \text{Now take L}\text{.C}\text{.M of the denominator}\text{.} \\ \dfrac{\dfrac{x+y}{x}}{\dfrac{x+y}{xy}} \\ \Rightarrow \dfrac{x+y}{x}\times \dfrac{xy}{x+y} \\ \text{Now remove the like terms from this equation}\text{.} \\ \text{You get: }y\text{.} \\ \text{The final answer is }y\text{.}$

 Solve $\dfrac{3{{x}^{3}}-4{{x}^{2}}y+x{{y}^{2}}}{49{{x}^{2}}y-{{y}^{3}}}$.

 To solve the above question, you have to first take out all the common multiples and then solve the quadratic equation. Divide the common brackets from the numerator and denominator to get the final answer.

 $ \text{We are given: }\dfrac{7{{x}^{3}}-4{{x}^{2}}y+x{{y}^{2}}}{9{{x}^{2}}y-{{y}^{3}}} \\ \text{first step:} \\ \dfrac{x\left( x-y \right)\left( 3x-y \right)}{y\left( 3x+y \right)\left( 3x-y \right)} \\ \text{second step: }\dfrac{x\left( x-y \right)}{y\left( 3x+y \right)} \\ \text{Final answer: }\dfrac{x\left( x-y \right)}{y\left( 3x+y \right)}$

 Simplify the following rational expression: $\dfrac{\dfrac{3}{x+1}}{\dfrac{x+4}{{{x}^{2}}+11x+10}}$.

 First you have to change the expression into product form. Then form the factors of the quadratic equation using the splitting method. After these steps you can cancel out the common terms from the numerator and denominator to get the final answer.

 $ \text{We are given: }\dfrac{\dfrac{3}{x+1}}{\dfrac{x+4}{{{x}^{2}}+11x+10}} \\ \text{First we will change this expression in a product}\text{.} \\ \text{we get, }\dfrac{3}{x+1}\times \dfrac{{{x}^{2}}+11x+10}{x+4} \\ \text{Now solve the quadratic equation in the above expression, to get:} \\ \Rightarrow \dfrac{3}{x+1}\times \dfrac{\left( x+1 \right)\left( x+10 \right)}{x+4} \\ \text{Now cancel the common terms} \\ \Rightarrow \dfrac{3\left( x+10 \right)}{x+4} \\ \text{Final answer: }\dfrac{3\left( x+10 \right)}{x+4}  $ 

 Solve the following expression: $\dfrac{x+10}{{{\left( 3x+8 \right)}^{3}}}+\dfrac{x}{{{\left( 3x+8 \right)}^{2}}}$.

 Start the question by taking the L.C.M to find the least common denominator after that multiply each term by an appropriate quantity. Then simply add the terms in the numerator to get the final answer.

 $\dfrac{x+10}{{{\left( 3x+8 \right)}^{3}}}+\dfrac{x}{{{\left( 3x+8 \right)}^{2}}} \\ \text{Now we have to take the L}\text{.C}\text{.M of the above expression}\text{.} \\ \text{Take the highest power on each term in the denominator}\text{. After this multiply each term by appropriate quantity}\text{.} \\ \text{We get:} \\ \dfrac{\left( x+10 \right)+x\left( 3x+8 \right)}{{{\left( 3x+8 \right)}^{3}}} \\ \Rightarrow \dfrac{x+10+3{{x}^{2}}+8x}{{{\left( 3x+8 \right)}^{3}}} \\ \Rightarrow \dfrac{3{{x}^{2}}+9x+10}{{{\left( 3x+8 \right)}^{3}}} \\ \text{So, final answer is }\dfrac{3{{x}^{2}}+9x+10}{{{\left( 3x+8 \right)}^{3}}}  $

 Solve the expression: $\dfrac{{{x}^{2}}-11x+24}{{{x}^{2}}-18x+80}\div \dfrac{{{x}^{2}}-9x+20}{{{x}^{2}}-15x+50}$.

  To solve the above question, you have to first change the form of the question from division to product form and then you have to make factors of the quadratic equations. Remove all the like terms to get the final answer.

 $ \text{We have: }\dfrac{{{x}^{2}}-11x+24}{{{x}^{2}}-18x+80}\div \dfrac{{{x}^{2}}-9x+20}{{{x}^{2}}-15x+50} \\ \text{Inverse the denominator to change into product form:} \\ \dfrac{{{x}^{2}}-11x+24}{{{x}^{2}}-18x+80}\times \dfrac{{{x}^{2}}-15x+50}{{{x}^{2}}-9x+20} \\ \text{Now, make factors of the above quadratic equations}\text{.} \\ \Rightarrow \dfrac{\left( x-8 \right)\left( x-3 \right)}{\left( x-8 \right)\left( x-10 \right)}\times \dfrac{\left( x-5 \right)\left( x-10 \right)}{\left( x-5 \right)\left( x-4 \right)} \\ \Rightarrow \dfrac{x-3}{x-4} $