
How do you simplify \[\dfrac{{{x}^{3}}{{y}^{7}}}{{{x}^{2}}{{y}^{3}}}?\]
Answer
549.9k+ views
Hint: We use cancellation of similar terms to simplify \[\dfrac{{{x}^{3}}{{y}^{7}}}{{{x}^{2}}{{y}^{3}}}.\] That is, we cancel the same number of the same terms from the numerator and the denominator. In an algebraic fraction, like arithmetic fraction, we consider the power of the variables in both the numerator and the denominator. 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.
Complete step by step solution:
Consider the algebraic fraction \[\dfrac{{{x}^{3}}{{y}^{7}}}{{{x}^{2}}{{y}^{3}}}.\]
We have to cut off the similar terms from the numerator and the denominator.
Consider the variable \[x,\] the power of \[x\] in the numerator is \[3\] and the power of \[x\] in the denominator is \[2.\]
Therefore, the cancellation is as follows:
\[\Rightarrow \dfrac{{{x}^{3}}{{y}^{7}}}{{{x}^{2}}{{y}^{3}}}=\dfrac{x.x.x.{{y}^{7}}}{x.x.{{y}^{3}}}\]
We cut off \[x.x\] from both the numerator and the denominator, only one \[x\] will remain in the numerator and no \[x\] will be there in the denominator.
Now we get,
\[\Rightarrow \dfrac{{{x}^{3}}{{y}^{7}}}{{{x}^{2}}{{y}^{3}}}=\dfrac{x{{y}^{7}}}{{{y}^{3}}}\]
Now we consider the variable \[y,\] the power of \[y\] in the numerator is \[7\] and the power of \[y\] in the denominator is \[3.\]
The cancellation is as follows:
\[\Rightarrow \dfrac{{{x}^{3}}{{y}^{7}}}{{{x}^{2}}{{y}^{3}}}=\dfrac{x.y.y.y.y.y.y.y}{y.y.y}\]
Cut off \[y.y.y\] from both the numerator and the denominator, only \[{{y}^{4}}\] will be remained in the numerator and there will be no \[y\] in the denominator.
So,
\[\Rightarrow \dfrac{{{x}^{3}}{{y}^{7}}}{{{x}^{2}}{{y}^{3}}}=\dfrac{x{{y}^{4}}}{1}\]
Hence the simplified form of the given algebraic fraction \[\dfrac{{{x}^{3}}{{y}^{7}}}{{{x}^{2}}{{y}^{3}}}=x{{y}^{4}}.\]
Note: We cancel the terms only if it's multiplied with \[1\] or any other terms.
We are not allowed to do cancellation of terms if the terms in the numerator or denominator are added to or subtracted from other terms (same variable or different variables) in the numerator or the denominator.
Complete step by step solution:
Consider the algebraic fraction \[\dfrac{{{x}^{3}}{{y}^{7}}}{{{x}^{2}}{{y}^{3}}}.\]
We have to cut off the similar terms from the numerator and the denominator.
Consider the variable \[x,\] the power of \[x\] in the numerator is \[3\] and the power of \[x\] in the denominator is \[2.\]
Therefore, the cancellation is as follows:
\[\Rightarrow \dfrac{{{x}^{3}}{{y}^{7}}}{{{x}^{2}}{{y}^{3}}}=\dfrac{x.x.x.{{y}^{7}}}{x.x.{{y}^{3}}}\]
We cut off \[x.x\] from both the numerator and the denominator, only one \[x\] will remain in the numerator and no \[x\] will be there in the denominator.
Now we get,
\[\Rightarrow \dfrac{{{x}^{3}}{{y}^{7}}}{{{x}^{2}}{{y}^{3}}}=\dfrac{x{{y}^{7}}}{{{y}^{3}}}\]
Now we consider the variable \[y,\] the power of \[y\] in the numerator is \[7\] and the power of \[y\] in the denominator is \[3.\]
The cancellation is as follows:
\[\Rightarrow \dfrac{{{x}^{3}}{{y}^{7}}}{{{x}^{2}}{{y}^{3}}}=\dfrac{x.y.y.y.y.y.y.y}{y.y.y}\]
Cut off \[y.y.y\] from both the numerator and the denominator, only \[{{y}^{4}}\] will be remained in the numerator and there will be no \[y\] in the denominator.
So,
\[\Rightarrow \dfrac{{{x}^{3}}{{y}^{7}}}{{{x}^{2}}{{y}^{3}}}=\dfrac{x{{y}^{4}}}{1}\]
Hence the simplified form of the given algebraic fraction \[\dfrac{{{x}^{3}}{{y}^{7}}}{{{x}^{2}}{{y}^{3}}}=x{{y}^{4}}.\]
Note: We cancel the terms only if it's multiplied with \[1\] or any other terms.
We are not allowed to do cancellation of terms if the terms in the numerator or denominator are added to or subtracted from other terms (same variable or different variables) in the numerator or the denominator.
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