Question

Simplify the expression. The simplified expression should have no negative exponents. $$ \frac{6 x^{-2} y^{2}}{x y^{-3}} \cdot \frac{\left(4 x^{2} y\right)^{-2}}{x y^{2}} $$

Short answer

The simplified expression is \(\frac{3y}{8x^{7}}\)

Step-by-step solution

Simplify the expression by applying the rules of exponents

Take the given expression \(\frac{6 x^{-2} y^{2}}{x y^{-3}} \cdot \frac{\left(4 x^{2} y\right)^{-2}}{x y^{2}} \). Now, apply the rule \(a^{-n} = \frac{1}{a^n} \) for the negative exponents to convert them into positive form. So it would become \(\frac{6}{x^{2}}\frac{y^{2}}{y^{-3}} \cdot \frac{\left(\frac{1}{16x^{4} y^{2}}\right)}{x y^{2}} = \frac{6}{x^{2}} \cdot y^{5} \cdot \frac{1}{16x^{5} y^{4}}\)

Merge the like terms

Now, we will merge like terms to simplify the expression further. The like terms here are the x and y terms. So, merging the x and y terms, we get\(\frac{6}{x^{2}} \cdot y^{5} \cdot \frac{1}{16x^{5} y^{4}} = \frac{6}{16x^{7} y^{-1}}\)

Simplify the expression further

Finally, the expression simplifies to:\(\frac{6}{16x^{7} y^{-1}} = \frac{3}{8x^{7} y^{-1}}\). Apply the rule \(a^{-n} = \frac{1}{a^n} \) to finally get \(\frac{3}{8x^{7} y^{-1}} = \frac{3}{8x^{7}} \cdot y = \frac{3y}{8x^{7}}\)