How do you simplify the expression $\left( \dfrac{2{{x}^{7}}{{y}^{8}}}{6x{{y}^{9}}} \right)$?
Answer
563.4k+ views
Hint: In the question, we have been given a rational expression which has to be simplified. For simplifying it, we need to cancel out the highest common factor from the numerator and the denominator. For this, we have to factorize the numerator and the denominator. Then we have to use the properties of the exponents which are given by $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$ and ${{a}^{-m}}=\dfrac{1}{{{a}^{m}}}$ to finally obtain the simplified expression.
Complete step-by-step solution:
Let us consider the expression given in the above question by writing it as
$\Rightarrow E=\dfrac{2{{x}^{7}}{{y}^{8}}}{6x{{y}^{9}}}$
We can see that the terms in the numerator and the denominator are already written in the factored form apart from the number $6$ in the denominator. It can be factored as $2\cdot 3$ so that we can write the above expression as
$\Rightarrow E=\dfrac{2{{x}^{7}}{{y}^{8}}}{2\cdot 3x{{y}^{9}}}$
Cancelling $2$ from the numerator and the denominator we get
$\begin{align}
& \Rightarrow E=\dfrac{{{x}^{7}}{{y}^{8}}}{3x{{y}^{9}}} \\
& \Rightarrow E=\dfrac{1}{3}\left( \dfrac{{{x}^{7}}{{y}^{8}}}{x{{y}^{9}}} \right) \\
\end{align}$
Writing the x terms and the y terms together, we get
$\Rightarrow E=\dfrac{1}{3}\left( \dfrac{{{x}^{7}}}{x} \right)\left( \dfrac{{{y}^{8}}}{{{y}^{9}}} \right)$
From the properties of the exponents, we know that $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$. Applying this property on the above expression, we get
\[\begin{align}
& \Rightarrow E=\dfrac{1}{3}{{x}^{7-1}}{{y}^{8-9}} \\
& \Rightarrow E=\dfrac{1}{3}{{x}^{6}}{{y}^{-1}} \\
\end{align}\]
From the negative rule of the exponent, we have ${{a}^{-m}}=\dfrac{1}{{{a}^{m}}}$ which means that we can write the above expression as
$\begin{align}
& \Rightarrow E=\dfrac{1}{3}{{x}^{6}}\left( \dfrac{1}{{{y}^{1}}} \right) \\
& \Rightarrow E=\dfrac{{{x}^{6}}}{3y} \\
\end{align}$
Hence, the expression given in the above question has been finally simplified as $\dfrac{{{x}^{6}}}{3y}$.
Note: We need to be comfortable with all the important rules of exponents in order to be able to solve these types of questions. Make sure that the final expression does not contain any negative power neither in the numerator nor in the denominator.
Complete step-by-step solution:
Let us consider the expression given in the above question by writing it as
$\Rightarrow E=\dfrac{2{{x}^{7}}{{y}^{8}}}{6x{{y}^{9}}}$
We can see that the terms in the numerator and the denominator are already written in the factored form apart from the number $6$ in the denominator. It can be factored as $2\cdot 3$ so that we can write the above expression as
$\Rightarrow E=\dfrac{2{{x}^{7}}{{y}^{8}}}{2\cdot 3x{{y}^{9}}}$
Cancelling $2$ from the numerator and the denominator we get
$\begin{align}
& \Rightarrow E=\dfrac{{{x}^{7}}{{y}^{8}}}{3x{{y}^{9}}} \\
& \Rightarrow E=\dfrac{1}{3}\left( \dfrac{{{x}^{7}}{{y}^{8}}}{x{{y}^{9}}} \right) \\
\end{align}$
Writing the x terms and the y terms together, we get
$\Rightarrow E=\dfrac{1}{3}\left( \dfrac{{{x}^{7}}}{x} \right)\left( \dfrac{{{y}^{8}}}{{{y}^{9}}} \right)$
From the properties of the exponents, we know that $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$. Applying this property on the above expression, we get
\[\begin{align}
& \Rightarrow E=\dfrac{1}{3}{{x}^{7-1}}{{y}^{8-9}} \\
& \Rightarrow E=\dfrac{1}{3}{{x}^{6}}{{y}^{-1}} \\
\end{align}\]
From the negative rule of the exponent, we have ${{a}^{-m}}=\dfrac{1}{{{a}^{m}}}$ which means that we can write the above expression as
$\begin{align}
& \Rightarrow E=\dfrac{1}{3}{{x}^{6}}\left( \dfrac{1}{{{y}^{1}}} \right) \\
& \Rightarrow E=\dfrac{{{x}^{6}}}{3y} \\
\end{align}$
Hence, the expression given in the above question has been finally simplified as $\dfrac{{{x}^{6}}}{3y}$.
Note: We need to be comfortable with all the important rules of exponents in order to be able to solve these types of questions. Make sure that the final expression does not contain any negative power neither in the numerator nor in the denominator.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference Between Plant Cell and Animal Cell

How many hours before the closure of election must class 9 social science CBSE

What is pollution? How many types of pollution? Define it

What is the Full Form of ICSE / ISC ?

Name 10 Living and Non living things class 9 biology CBSE


