Question

(\frac{9x^{-4}y^{7}}{18x^{6}y^{-2}}cdot\frac{3x^{8}}{y^{3}})

Explanation:

Step1: Simplify coefficients

Simplify the coefficients \(\frac{9}{18}\) and multiply by \(3\). \(\frac{9}{18}\times3=\frac{9\times3}{18}=\frac{27}{18}=\frac{3}{2}\)

Step2: Simplify \(x\)-terms

Use the rule \(a^m\cdot a^n = a^{m + n}\) and \(a^{-m}=\frac{1}{a^m}\). For \(x\)-terms: \(x^{-4}\cdot x^{8}\cdot x^{-6}=x^{-4 + 8-6}=x^{-2}=\frac{1}{x^{2}}\) (Wait, correction: Wait, the first fraction has \(x^{-4}\) and \(x^{6}\) in denominator? Wait, original expression: \(\frac{9x^{-4}y^{7}}{18x^{6}y^{-2}}\cdot\frac{3x^{8}}{y^{3}}\). So for \(x\): numerator \(x^{-4}\times x^{8}\), denominator \(x^{6}\). So \(x^{-4 + 8-6}=x^{-2}\)? Wait no, \(\frac{x^a}{x^b}=x^{a - b}\), and \(x^a\cdot x^b=x^{a + b}\). So first, handle the first fraction: \(\frac{9x^{-4}y^{7}}{18x^{6}y^{-2}}=\frac{9}{18}x^{-4-6}y^{7-(-2)}=\frac{1}{2}x^{-10}y^{9}\). Then multiply by \(\frac{3x^{8}}{y^{3}}\): \(\frac{1}{2}\times3x^{-10 + 8}y^{9-3}=\frac{3}{2}x^{-2}y^{6}\). Wait, I messed up earlier. Let's redo:

First, multiply the numerators: \(9x^{-4}y^{7}\times3x^{8}=27x^{-4 + 8}y^{7}=27x^{4}y^{7}\)

Multiply the denominators: \(18x^{6}y^{-2}\times y^{3}=18x^{6}y^{-2 + 3}=18x^{6}y^{1}\)

Now, the fraction is \(\frac{27x^{4}y^{7}}{18x^{6}y^{1}}\)

Simplify coefficient: \(\frac{27}{18}=\frac{3}{2}\)

Simplify \(x\): \(\frac{x^{4}}{x^{6}}=x^{4-6}=x^{-2}\)

Simplify \(y\): \(\frac{y^{7}}{y^{1}}=y^{7-1}=y^{6}\)

So overall: \(\frac{3}{2}x^{-2}y^{6}=\frac{3y^{6}}{2x^{2}}\)

Wait, let's do it step by step correctly:

1. Multiply the two fractions: \(\frac{9x^{-4}y^{7}\times3x^{8}}{18x^{6}y^{-2}\times y^{3}}\)

1. Coefficients: \(9\times3 = 27\); \(18\) remains. So \(\frac{27}{18}=\frac{3}{2}\)

1. \(x\)-terms: \(x^{-4}\times x^{8}\) in numerator, \(x^{6}\) in denominator. So \(x^{-4 + 8-6}=x^{-2}\) (Wait, \(x^{-4}\times x^{8}=x^{4}\), then divide by \(x^{6}\): \(x^{4-6}=x^{-2}\))

1. \(y\)-terms: \(y^{7}\) in numerator, \(y^{-2}\times y^{3}=y^{1}\) in denominator. So \(y^{7-1}=y^{6}\)

So combining: \(\frac{3}{2}x^{-2}y^{6}=\frac{3y^{6}}{2x^{2}}\)

Wait, another way: use exponent rules for multiplication and division.

For \(x\): \(\frac{x^{-4}\times x^{8}}{x^{6}}=x^{-4 + 8-6}=x^{-2}\) (since when multiplying, add exponents; when dividing, subtract exponents)

For \(y\): \(\frac{y^{7}}{y^{-2}\times y^{3}}=y^{7-(-2)-3}=y^{7 + 2-3}=y^{6}\)

Coefficients: \(\frac{9\times3}{18}=\frac{27}{18}=\frac{3}{2}\)

So putting together: \(\frac{3}{2}x^{-2}y^{6}=\frac{3y^{6}}{2x^{2}}\)

Step3: Simplify \(y\)-terms

For \(y\)-terms: \(y^{7}\times y^{3}\) in numerator? No, wait, original: numerator \(y^{7}\), denominator \(y^{-2}\) in first fraction, then multiply by \(\frac{1}{y^{3}}\). So \(y^{7}\div y^{-2}\div y^{3}=y^{7-(-2)-3}=y^{6}\)

Answer:

\(\boldsymbol{\frac{3y^{6}}{2x^{2}}}\) (or \(\frac{3}{2}x^{-2}y^{6}\))