Question

How do you simplify \[\dfrac{{{x}^{2}}}{x{{y}^{-4}}}.\dfrac{2{{x}^{-3}}{{y}^{4}}}{3x{{y}^{-1}}}?\]

Answer 1

Hint: We use cancellation of similar terms to simplify \[\dfrac{{{x}^{2}}}{x{{y}^{-4}}}.\dfrac{2{{x}^{-3}}{{y}^{4}}}{3x{{y}^{-1}}}.\] After cutting off the similar terms up to the same power from the numerator and the denominator, the rest of the terms will remain as they are. Remember \[{{x}^{-n}}=\dfrac{1}{{{x}^{n}}}.\] Apply this to make the fraction only of positive exponents. Also remember \[{{\left( \dfrac{1}{x} \right)}^{n}}=\dfrac{1}{{{x}^{n}}}.\] Furthermore, \[{{\left( \dfrac{1}{{{x}^{n}}} \right)}^{m}}=\dfrac{1}{{{\left( {{x}^{n}} \right)}^{m}}}=\dfrac{1}{{{x}^{nm}}}.\] Recall that we have learnt \[{{x}^{n}}.{{x}^{m}}={{x}^{n+m}}\] and \[{{x}^{-n}}.{{x}^{-m}}={{x}^{-n-m}}.\]

Complete step by step solution:
Consider \[\dfrac{{{x}^{2}}}{x{{y}^{-4}}}.\dfrac{2{{x}^{-3}}{{y}^{4}}}{3x{{y}^{-1}}}.\]
We use the cancellation of terms to simplify the given algebraic expression.
Let us do it as follows:
\[\Rightarrow \dfrac{{{x}^{2}}}{x{{y}^{-4}}}.\dfrac{2{{x}^{-3}}{{y}^{4}}}{3x{{y}^{-1}}}=\dfrac{2{{x}^{2}}{{x}^{-3}}{{y}^{4}}}{3{{x}^{2}}{{y}^{-5}}}\]
Here, we have used the identities \[{{x}^{n}}.{{x}^{m}}={{x}^{n+m}}\] and \[{{x}^{-n}}.{{x}^{-m}}={{x}^{-n-m}}.\]
We are going to cut off \[{{x}^{2}}\] from both the numerator and the denominator in the coming step,
\[\Rightarrow \dfrac{{{x}^{2}}}{x{{y}^{-4}}}.\dfrac{2{{x}^{-3}}{{y}^{4}}}{3x{{y}^{-1}}}=\dfrac{2{{x}^{-3}}{{y}^{4}}}{3{{y}^{-5}}}\]
We are going to apply the identity \[\dfrac{{{x}^{n}}}{{{x}^{m}}}={{x}^{n-m}},\] that is \[\dfrac{{{x}^{n}}}{{{x}^{-m}}}={{x}^{n+m}}.\]
Thus, we get
\[\Rightarrow \dfrac{{{x}^{2}}}{x{{y}^{-4}}}.\dfrac{2{{x}^{-3}}{{y}^{4}}}{3x{{y}^{-1}}}=\dfrac{2{{x}^{-3}}{{y}^{4+5}}}{3}\]
We will get,
\[\Rightarrow \dfrac{{{x}^{2}}}{x{{y}^{-4}}}.\dfrac{2{{x}^{-3}}{{y}^{4}}}{3x{{y}^{-1}}}=\dfrac{2{{x}^{-3}}{{y}^{9}}}{3}\]
We are going to use the identity \[{{x}^{-n}}=\dfrac{1}{{{x}^{n}}}.\]
This is same as the identity \[{{x}^{-n}}=\dfrac{1}{{{x}^{n}}}.\]
This way we obtain, \[\dfrac{{{x}^{2}}}{x{{y}^{-4}}}.\dfrac{2{{x}^{-3}}{{y}^{4}}}{3x{{y}^{-1}}}=\dfrac{2{{y}^{9}}}{3{{x}^{3}}}\]

Therefore, the simplified form of the given algebraic expression \[\dfrac{{{x}^{2}}}{x{{y}^{-4}}}.\dfrac{2{{x}^{-3}}{{y}^{4}}}{3x{{y}^{-1}}}\] is \[\dfrac{2{{y}^{9}}}{3{{x}^{3}}}.\]

Note: We can always take the terms with negative exponents from numerator to denominator or from denominator to numerator in order to make them terms with positive exponents. We can site this identity \[{{x}^{-n}}=\dfrac{1}{{{x}^{n}}}\] as an example for the fact that is written here.
As an alternative method we can use the following procedure to simplify the given algebraic expression:
Since the expression \[\dfrac{{{x}^{2}}}{x{{y}^{-4}}}.\dfrac{2{{x}^{-3}}{{y}^{4}}}{3x{{y}^{-1}}}\] is a product of two fractions, we can multiply the terms in the numerators and also the terms in the denominators directly as follows,
\[\Rightarrow \dfrac{{{x}^{2}}}{x{{y}^{-4}}}.\dfrac{2{{x}^{-3}}{{y}^{4}}}{3x{{y}^{-1}}}=\dfrac{2{{x}^{2}}.{{x}^{-3}}.{{y}^{4}}}{3x.x.{{y}^{-4}}.{{y}^{-1}}}\]
We are using the identities given above,
\[\Rightarrow \dfrac{{{x}^{2}}}{x{{y}^{-4}}}.\dfrac{2{{x}^{-3}}{{y}^{4}}}{3x{{y}^{-1}}}=\dfrac{2{{x}^{2-3}}.{{y}^{4}}}{3{{x}^{2}}.{{y}^{-5}}}\]
Hence, \[\dfrac{{{x}^{2}}}{x{{y}^{-4}}}.\dfrac{2{{x}^{-3}}{{y}^{4}}}{3x{{y}^{-1}}}=\dfrac{2{{x}^{-1}}.{{y}^{4+5}}}{3{{x}^{2}}}=\dfrac{2{{y}^{4+5}}}{3{{x}^{2+1}}}=\dfrac{2{{y}^{9}}}{3{{x}^{3}}}.\]