College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Are you Ready for College Algebra? - A. Diagnostic Test:: Real Numbers and Exponents - Page XVII: 1

Answer

a) $\frac{5}{6}$ b) $\frac{19}{12}$ c) $\frac{16}{3}$ d) $8$

Work Step by Step

a) Simplify the following: $$\frac{1}{3} + \frac{1}{2}$$ Put $\frac{1}{3} + \frac{1}{2}$ over the common denominator 6. $\frac{1}{3} + \frac{1}{2} = \frac{2}{6} + \frac{3}{6}$: $$\frac{2}{6} + \frac{3}{6}$$ $\frac{2}{6} + \frac{3}{6} = \frac{2 + 3}{6}$: $$\frac{2 + 3}{6}$$ $2 + 3 = 5$: Answer: $$\frac{5}{6}$$ b) Simplify the following: $$2 - \frac{2}{3} + \frac{1}{4}$$ Put $2 - \frac{2}{3} + \frac{1}{4}$ over the common denominator $12$. $2 - \frac{2}{3} + \frac{1}{4} = \frac{12 \times 2}{12} + \frac{4(-2)}{12} + \frac{3}{12}$: $$\frac{12 \times 2}{12} + \frac{4(-2)}{12} + \frac{3}{12}$$ $12 \times 2 = 24$: $$\frac{24}{12} + \frac{4(-2)}{12} + \frac{3}{12}$$ $4(-2) = -8$: $$\frac{24}{12} + \frac{-8}{12} + \frac{3}{12}$$ $\frac{24}{12} - \frac{8}{12} + \frac{3}{12} = \frac{24 - 8 + 3}{12}$: $$\frac{24 - 8 + 3}{12}$$ $24 + 3 = 27$: $$\frac{27 - 8}{12}$$ $27 - 8 = 19$: Answer: $$\frac{19}{12}$$ c) Simplify the following: $$4\left(2 - \frac{2}{3}\right)$$ Put $2 - \frac{2}{3}$ over the common denominator $3$. $2 - \frac{2}{3} = \frac{3 \times 2}{3} - \frac{2}{3}$: $$4\left(\frac{3 \times 2}{3} - \frac{2}{3}\right)$$ $3 \times 2 = 6$: $$4\left(\frac{6}{3} - \frac{2}{3}\right)$$ $\frac{6}{3} - \frac{2}{3} = \frac{6 - 2}{3}$: $$4\left(\frac{6-2}{3}\right)$$ $6 - 2 = 4$: $$4 \times \frac{4}{3}$$ $4 \times \frac{4}{3} = \frac{4 \times 4}{3}$: $$ \frac{4 \times 4}{3} $$ $4 \times 4 = 16$: Answer: $$\frac{16}{3}$$ d) Simplify the following: $$\frac{12}{\frac{4}{3} + \frac{1}{6}}$$ Put $\frac{4}{3} + \frac{1}{6}$ over the common denominator 6. $\frac{4}{3} + \frac{1}{6} = \frac{2 \times 4}{6} + \frac{1}{6}$: $$\left(\frac{2 \times 4}{6} + \frac{1}{6}\right)$$ $2 \times 4 = 8$: $$\frac{12}{\frac{8}{6} + \frac{1}{6}}$$ $\frac{8}{6} + \frac{1}{6} = \frac{8 + 1}{6}$: $$\frac{12}{\left(\frac{8 + 1}{6}\right)}$$ $8 + 1 = 9$: $$\frac{12}{\frac{9}{6}}$$ Multiply the numerator of $\frac{12}{\frac{9}{6}}$ by the reciprocal of the denominator $\frac{12}{\frac{9}{6}} = \frac{12 \times 6}{9}$: $$\frac{12 \times 6}{9}$$ The gcd of $12$ and $9$ is $3$, so $\frac{12 \times 6}{9} = \frac{(3 \times 4)6}{3 \times 3} = \frac{3}{3} \times \frac{4 \times 6}{3} = \frac{4 \times 6}{3}$: $$\frac{4 \times 6}{3}$$ $\frac{6}{3} = \frac{3 \times 2}{3} = 2$: $$4 \times 2$$ $4 \times 2 = 8$: Answer: $$8$$
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