## Introductory Algebra for College Students (7th Edition)

$1\displaystyle \frac{1}{3}$ is a solution.
For the given number to be a solution of the equation, it must hold true when we substitute the variable with the given number. Multiplication comes before addition, you can divide numerators and denominators by common factors before performing multiplication. Convert the given number to an improper fraction, and substitute into the equation. $1\displaystyle \frac{1}{3}=\frac{4}{3}$ LHS =$\displaystyle \frac{2}{3}\cdot\frac{4}{3}+\frac{5}{6}\cdot\frac{4}{3}$ ... reduce,,, $= \displaystyle \frac{2}{3 }\cdot\frac{4}{3}+\frac{5}{6\div 2}\cdot\frac{4\div 2}{3}$ $= \displaystyle \frac{8}{9}+\frac{5\cdot 2}{3\cdot 3}$ $= \displaystyle \frac{8}{9}+\frac{10}{9}$ $=\displaystyle \frac{18}{9}$ $=\displaystyle \frac{18\div 9}{9\div 9}=2$ LHS = RHS, so $1\displaystyle \frac{1}{3}$ is a solution.