Answer
$y^{1/2}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Use the laws of exponents to simplify the given expression, $
\dfrac{y^{5/3}\cdot y^{-2}}{y^{-5/6}}
.$
$\bf{\text{Solution Details:}}$
Using the Product Rule of the laws of exponents which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{y^{\frac{5}{3}+(-2)}}{y^{-5/6}}
\\\\=
\dfrac{y^{\frac{5}{3}-2}}{y^{-5/6}}
.\end{array}
To simplify the expression $
\dfrac{5}{3}-2
,$ change the expressions to similar fractions (same denominator) by using the $LCD$. The $LCD$ of the denominators $
3
$ and $
1
$ is $
3
$ since it is the lowest number that can be exactly divided by the denominators. Multiplying the terms by an expression equal to $1$ that will make the denominator equal to the $LCD$ results to
\begin{array}{l}\require{cancel}
\dfrac{5}{3}-2\cdot\dfrac{3}{3}
\\\\=
\dfrac{5}{3}-\dfrac{6}{3}
\\\\=
\dfrac{5-6}{3}
\\\\=
\dfrac{-1}{3}
\\\\=
-\dfrac{1}{3}
.\end{array}
The expression, $
\dfrac{y^{\frac{5}{3}-2}}{y^{-5/6}}
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{y^{-\frac{1}{3}}}{y^{-5/6}}
.\end{array}
Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to
\begin{array}{l}\require{cancel}
y^{-\frac{1}{3}-\left(-\frac{5}{6} \right)}
\\\\=
y^{-\frac{1}{3}+\frac{5}{6}}
.\end{array}
To simplify the expression $
-\dfrac{1}{3}+\dfrac{5}{6}
,$ change the expressions to similar fractions (same denominator) by using the $LCD$. The $LCD$ of the denominators $
3
$ and $
6
$ is $
6
$ since it is the lowest number that can be exactly divided by the denominators. Multiplying the terms by an expression equal to $1$ that will make the denominator equal to the $LCD$ results to
\begin{array}{l}\require{cancel}
-\dfrac{1}{3}\cdot\dfrac{2}{2}+\dfrac{5}{6}
\\\\=
-\dfrac{2}{6}+\dfrac{5}{6}
\\\\=
\dfrac{-2+5}{6}
\\\\=
\dfrac{3}{6}
\\\\=
\dfrac{1}{2}
.\end{array}
The expression, $
y^{-\frac{1}{3}+\frac{5}{6}}
,$ is equivalent to
\begin{array}{l}\require{cancel}
y^{\frac{1}{2}}
\\\\=
y^{1/2}
.\end{array}
