## Algebra: A Combined Approach (4th Edition)

$\dfrac{1}{3}(7y-1)+\dfrac{1}{6}(4y+7)=3y+\dfrac{5}{6}$
$\dfrac{1}{3}(7y-1)+\dfrac{1}{6}(4y+7)$ Multiply the fractions by every term inside its respective parentheses: $\dfrac{1}{3}(7y-1)+\dfrac{1}{6}(4y+7)=\dfrac{7}{3}y-\dfrac{1}{3}+\dfrac{4}{6}y+\dfrac{7}{6}=...$ Simplify the resulting fractions if possible: $...=\dfrac{7}{3}y-\dfrac{1}{3}+\dfrac{2}{3}y+\dfrac{7}{6}=...$ Continue simplifying: $...=\dfrac{9}{3}y+\dfrac{21-6}{18}=\dfrac{9}{3}y+\dfrac{15}{18}=\dfrac{9}{3}y+\dfrac{5}{6}=3y+\dfrac{5}{6}$