Answer
$\dfrac{1}{3}(7y-1)+\dfrac{1}{6}(4y+7)=3y+\dfrac{5}{6}$
Work Step by Step
$\dfrac{1}{3}(7y-1)+\dfrac{1}{6}(4y+7)$
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:
$...=\dfrac{9}{3}y+\dfrac{21-6}{18}=\dfrac{9}{3}y+\dfrac{15}{18}=\dfrac{9}{3}y+\dfrac{5}{6}=3y+\dfrac{5}{6}$
