Tuesday, April 2, 2013

Intermediate Algebra, Chapter 5, 5.1, Section 5.1, Problem 51

Write the expression $8^{-1}-3^{-1}$ with only positive exponents. Then, simplify the expression.
Remove the negative exponent by rewriting $8^{-1}$ as $\dfrac{1}{8}$. A negative exponent follows the rule of $a^{-n} = \dfrac{1}{a^n}$
$\dfrac{1}{8}-3^{-1}$
Remove the negative exponent by rewriting $3^{-1}$ as $\dfrac{1}{3}$. A negative exponent follows the rule of $a^{-n} = \dfrac{1}{a^n}$
$\dfrac{1}{8}-\dfrac{1}{3}$

To subtract fractions, the denominators must be equal. The denominators can be made equal by finding the least common denominator (LCD). In this case, the LCD is $24$. Next, multiply each fraction by a factor of $1$ that will create the LCD in each of the fractions.
$\dfrac{1}{8}\cdot\dfrac{3}{3}-\dfrac{1}{3}\cdot\dfrac{8}{8}$
Complete the multiplication to produce a denominator of $24$ in each expression.
$\dfrac{3}{24}-\dfrac{8}{24}$
Combine the numerators of all fractions that have common denominators.
$\dfrac{1}{24}(3-8)$
Subtract $3$ to $8$ to get $-5$
$\dfrac{-5}{24}$

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