Prealgebra (7th Edition)

Published by Pearson
ISBN 10: 0321955048
ISBN 13: 978-0-32195-504-3

Chapter 7 - Cumulative Review - Page 529: 22



Work Step by Step

$\frac{x}{2}$ +$\frac{1}{5}$=3 -$\frac{x}{5}$ subtract $\frac{1}{5}$ from both sides $\frac{x}{2}$ +$\frac{1}{5}$ -$\frac{1}{5}$=3 -$\frac{x}{5}$ -$\frac{1}{5}$ write 3 like $\frac{15}{5}$ $\frac{x}{2}$ =$\frac{15}{5}$ -$\frac{1}{5}$ -$\frac{x}{5}$ $\frac{x}{2}$ =$\frac{14}{5}$ -$\frac{x}{5}$ add $\frac{x}{5}$ to both sides $\frac{x}{2}$ +$\frac{x}{5}$ =$\frac{14}{5}$ -$\frac{x}{5}$ +$\frac{x}{5}$ $\frac{x}{2}$ +$\frac{x}{5}$ =$\frac{14}{5}$ LCD for 2 and 5 is 10 $\frac{x}{2}$$\times$$\frac{5}{5}$ + $\frac{x}{5}$$\times$$\frac{2}{2}$=$\frac{14}{5}$ $\frac{5x}{10}$+$\frac{2x}{10}$=$\frac{14}{5}$ $\frac{7x}{10}$=$\frac{14}{5}$ x$\times$$\frac{7}{10}$ =$\frac{14}{5}$ divide by $\frac{7}{10}$ both sides x$\times$$\frac{7}{10}$$\div$$\frac{7}{10}$ =$\frac{14}{5}$$\div$$\frac{7}{10}$ x=$\frac{14}{5}$$\times$$\frac{10}{7}$ x=$\frac{14\times10}{5\times7}$ x=$\frac{140}{35}$ x=4 The solution is 4 check $\frac{x}{2}$ +$\frac{1}{5}$=3 -$\frac{x}{5}$ $\frac{4}{2}$ +$\frac{1}{5}$=3 -$\frac{4}{5}$ 2+$\frac{1}{5}$=3-$\frac{4}{5}$ $\frac{10}{5}$+$\frac{1}{5}$=$\frac{15}{5}$-$\frac{4}{5}$ $\frac{11}{5}$=$\frac{11}{5}$
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