Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Skills Handbook - Adding and Subtracting Fractions - Exercises - Page 791: 5



Work Step by Step

The fractions have to be given a common denominator in $6$$\frac{2}{3}$+$3$$\frac{4}{5}$. The least common multiple of 3 and 5 is 15. This can be found by drawing out a factor tree. Both 3 and 5 are prime to begin with, so they can be simply be multiplied together. So, $\frac{2}{3}$$\times$$\frac{5}{5}$$=$$\frac{10}{15}$ and $\frac{4}{5}$$\times$$\frac{3}{3}$$=$$\frac{12}{15}$ Now, substitute these new fractions into the places of the old fractions in the original question. $6$$\frac{10}{15}$+$3$$\frac{12}{15}$$=$$9$$\frac{22}{15}$ In order to get $\frac{22}{15}$ out of an improper fraction, simplify it into a mixed number. 15 goes into 22 one time and the remainder is 7. So, $\frac{22}{15}$$=$$1$$\frac{7}{15}$ Adding this answer to the whole number 9, $9+1$$\frac{7}{15}$$=10$$\frac{7}{15}$
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