Question

How do you simplify \[\dfrac{\left( 3{{y}^{\dfrac{1}{4}}} \right)}{\left( 4{{x}^{\dfrac{-2}{3}}}{{y}^{\dfrac{3}{2}}}3{{y}^{\dfrac{1}{2}}} \right)}\]?

Answer 1

Hint: The given algebraic expression is to be simplified. We have the expression with two variables namely x and y. The expression first needs to solved so that powers of similar base gets added up or subtracted depending upon the algebraic properties, that is, \[\dfrac{{{x}^{a}}}{{{x}^{b}}}={{x}^{a-b}}\] and \[{{x}^{a}}.{{x}^{b}}={{x}^{a+b}}\]. After solving the expression, we simplify the expression and hence we get the simplified form of the expression.

Complete step by step solution:
According to the given question, we have an algebraic expression which has been simplified. We will be using some algebraic properties to solve and simplify the expression.
Let’s begin by writing the given expression we have,
\[\dfrac{\left( 3{{y}^{\dfrac{1}{4}}} \right)}{\left( 4{{x}^{\dfrac{-2}{3}}}{{y}^{\dfrac{3}{2}}}3{{y}^{\dfrac{1}{2}}} \right)}\]
We will first separate the constant terms and the variables, we have,
\[\Rightarrow \dfrac{3\left( {{y}^{\dfrac{1}{4}}} \right)}{4.3.\left( {{x}^{\dfrac{-2}{3}}}{{y}^{\dfrac{3}{2}}}{{y}^{\dfrac{1}{2}}} \right)}\]
Here, we can see that the expression has 3 both in the numerator and denominator and so is cancelled and we have,
\[\Rightarrow \dfrac{\left( {{y}^{\dfrac{1}{4}}} \right)}{4\left( {{x}^{\dfrac{-2}{3}}}{{y}^{\dfrac{3}{2}}}{{y}^{\dfrac{1}{2}}} \right)}\]
We will first solve the y terms in the expression. In the denominator, we have two y terms. As we know that when bases are the same the powers get added up. We get,
\[\Rightarrow \dfrac{\left( {{y}^{\dfrac{1}{4}}} \right)}{4\left( {{x}^{\dfrac{-2}{3}}}{{y}^{\dfrac{3}{2}}}^{+\dfrac{1}{2}} \right)}\]
This is according to the fact that, \[{{x}^{a}}.{{x}^{b}}={{x}^{a+b}}\].
Now, we will solve and proceed and we get,
\[\Rightarrow \dfrac{\left( {{y}^{\dfrac{1}{4}}} \right)}{4\left( {{x}^{\dfrac{-2}{3}}}{{y}^{\dfrac{4}{2}}} \right)}\]
\[\Rightarrow \dfrac{\left( {{y}^{\dfrac{1}{4}}} \right)}{4\left( {{x}^{\dfrac{-2}{3}}}{{y}^{2}} \right)}\]
We know that, \[\dfrac{{{x}^{a}}}{{{x}^{b}}}={{x}^{a-b}}\]. Applying this formula in the expression we get,
\[\Rightarrow \dfrac{\left( {{y}^{\dfrac{1}{4}-2}} \right)}{4\left( {{x}^{\dfrac{-2}{3}}} \right)}\]
We will proceed to solve, we get,
\[\Rightarrow \dfrac{\left( {{y}^{\dfrac{1-8}{4}}} \right)}{4\left( {{x}^{\dfrac{-2}{3}}} \right)}\]
\[\Rightarrow \dfrac{\left( {{y}^{\dfrac{-7}{4}}} \right)}{4\left( {{x}^{\dfrac{-2}{3}}} \right)}\]
We also know that, \[{{x}^{-1}}=\dfrac{1}{x}\] and \[\dfrac{1}{{{x}^{-1}}}=x\]. Applying these properties in the above expression we get the new simplified expression as:
\[\Rightarrow \dfrac{\left( {{x}^{\dfrac{2}{3}}} \right)}{4\left( {{y}^{\dfrac{7}{4}}} \right)}\]
That is,
\[\Rightarrow \dfrac{{{x}^{\dfrac{2}{3}}}}{4{{y}^{\dfrac{7}{4}}}}\]
Therefore, the simplified form of the expression is \[\dfrac{{{x}^{\dfrac{2}{3}}}}{4{{y}^{\dfrac{7}{4}}}}\].

Note:
In the above expression, we did not solve anything in particular for x term as there was only x-term. We dealt with it at the end by just changing its position as it had a negative sign in the power. The expression was quite big, so there are chances to write the wrong values while solving so it is advised to solve it in steps.