How do you simplify the expression $\left( \dfrac{2{{x}^{7}}{{y}^{8}}}{6x{{y}^{9}}} \right)$?
Answer
559.8k+ 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.
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