## Prealgebra (7th Edition)

$\frac{x}{2}$ +$\frac{1}{5}$=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-$\frac{4}{5}$ $\frac{10}{5}$+$\frac{1}{5}$=$\frac{15}{5}$-$\frac{4}{5}$ $\frac{11}{5}$=$\frac{11}{5}$