## Intermediate Algebra for College Students (7th Edition)

Published by Pearson

# Chapter 1 - Section 1.6 - Properties of Integral Exponents - Exercise Set - Page 80: 120

#### Answer

$$\frac{(x^{12})(4y^{8}+1)}{y^{12}}$$

#### Work Step by Step

$$(\frac{2^{-1}x^{-2}y}{x^{4}y^{-1}})^{-2}+ (\frac{xy^{-3}}{x^{-3}y})^{3}$$ Simplify the first term: $(\frac{2^{-1}x^{-2}y}{x^{4}y^{-1}})^{-2}$ Recall the power rule: $(a^{m})^{n}=a^{mn}$ Thus, $$(\frac{2^{-1}x^{-2}y}{x^{4}y^{-1}})^{-2} = \frac{2^{-1(-2)}x^{-2(-2)}y^{-2}}{x^{4(-2)}y^{-1(-2)}} = \frac{2^{2}x^{4}y^{-2}}{x^{-8}y^{2}}$$ Simplify the second term: $(\frac{xy^{-3}}{x^{-3}y})^{3}$ Recall the power rule: $(a^{m})^{n}=a^{mn}$ Thus, $$(\frac{xy^{-3}}{x^{-3}y})^{3} = \frac{x^{3}y^{-3(3)}}{x^{-3(3)}y^{3}} = \frac{x^{3}y^{-9}}{x^{-9}y^{3}}$$ Rewrite the equation: $$\frac{2^{2}x^{4}y^{-2}}{x^{-8}y^{2}} + \frac{x^{3}y^{-9}}{x^{-9}y^{3}}$$ $$=\frac{4x^{4}y^{-2}}{x^{-8}y^{2}} +\frac{x^{3}y^{-9}}{x^{-9}y^{3}}$$ Recall the quotient rule: $\frac{a^{m}}{a^{n}}=a^{m+n}$ and $\frac{a^{n}}{a^{m}}=\frac{1}{a^{m+n}}$ if $m>n$ Thus, $$=\frac{4x^{4}y^{-2}}{x^{-8}y^{2}} +\frac{x^{3}y^{-9}}{x^{-9}y^{3}}$$ $$=\frac{4x^{4+8}}{y^{2+2}} +\frac{x^{3+9}}{y^{3+9}}$$ $$=\frac{4x^{12}}{y^{4}} +\frac{x^{12}}{y^{12}}$$ Get the LCD through prime factorization: $y^{4}: y^{2}\cdot y^{2}$ $y^{12}: y^{4}\cdot y^{3} = y^{2}\cdot y^{2}\cdot y^{3}$ --> LCD Adjust the fractions based on the LCD. $$=\frac{4x^{12}}{y^{4}} +\frac{x^{12}}{y^{12}}$$ $$=\frac{4x^{12}y^{8}}{y^{12}} +\frac{x^{12}}{y^{12}}$$ Combine the terms. $$=\frac{(4x^{12}y^{8})+x^{12}}{y^{12}}$$ Factor $x^{12}$. $$=\frac{(x^{12})(4y^{8}+1)}{y^{12}}$$

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