How do you simplify (15x^4y^2)/(40x^3y^3) and what are the ecluded values fot he variables?

Mar 28, 2017

See the entire solution process below:

Explanation:

First, rewrite the expression as shown below and simplify the constants:

$\left(\frac{15}{40}\right) \left({x}^{4} / {x}^{3}\right) \left({y}^{2} / {y}^{3}\right) = \frac{5 \times 3}{5 \times 8} \left({x}^{4} / {x}^{3}\right) \left({y}^{2} / {y}^{3}\right) =$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} \times 3}{\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} \times 8} \left({x}^{4} / {x}^{3}\right) \left({y}^{2} / {y}^{3}\right) = \frac{3}{8} \left({x}^{4} / {x}^{3}\right) \left({y}^{2} / {y}^{3}\right)$

Now, use these two rules of exponents to simplify the $x$ and $y$ terms:

${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} - \textcolor{b l u e}{b}}$ and ${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = \frac{1}{x} ^ \left(\textcolor{b l u e}{b} - \textcolor{red}{a}\right)$

$\frac{3}{8} \left({x}^{\textcolor{red}{4}} / {x}^{\textcolor{b l u e}{3}}\right) \left({y}^{\textcolor{red}{2}} / {y}^{\textcolor{b l u e}{3}}\right) = \frac{3}{8} \left({x}^{\textcolor{red}{4} - \textcolor{b l u e}{3}}\right) \left(\frac{1}{x} ^ \left(\textcolor{b l u e}{3} - \textcolor{red}{2}\right)\right) = \frac{3}{8} \left({x}^{1}\right) \left(\frac{1}{y} ^ 1\right) =$

$\frac{3 {x}^{1}}{9 {y}^{1}}$

Now, we can use this rule of exponents to complete the simplification:

${a}^{\textcolor{red}{1}} = a$

$\frac{3 {x}^{\textcolor{red}{1}}}{9 {y}^{\textcolor{red}{1}}} = \frac{3 x}{8 y}$

Because we cannot divide by $0$, from the original expression $40 {x}^{3} {y}^{3} \ne 0$, therefore $x \ne 0$ and $y \ne 0$