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

$\dfrac{1}{5}(9y+2)+\dfrac{1}{10}(2y-1)=2y+\dfrac{3}{10}$
$\dfrac{1}{5}(9y+2)+\dfrac{1}{10}(2y-1)$ Multiply the fractions by every term inside its respective parentheses: $\dfrac{1}{5}(9y+2)+\dfrac{1}{10}(2y-1)=\dfrac{9}{5}y+\dfrac{2}{5}+\dfrac{2}{10}y-\dfrac{1}{10}=...$ $...=\dfrac{9}{5}y+\dfrac{2}{5}+\dfrac{1}{5}y-\dfrac{1}{10}=...$ Simplify like terms: $...=\dfrac{10}{5}y+\dfrac{20-5}{50}=2y+\dfrac{15}{50}=2y+\dfrac{3}{10}$