## Algebra 1: Common Core (15th Edition)

4a. b=$\frac{60}{23}$ 4b. m=$\frac{13}{6}$
4a. I chose this method because it is the simplest way of solving a problem with fractions in it. It is an easy way to combine like terms/ $\frac{2b}{5}+\frac{3b}{4}=3$ Multiply both sides by the LCM: 20 $\frac{2b}{5}\times20+\frac{3b}{4}\times20=3\times20$ Simplify: $8b+15b=60$ $23b=60$ Divide both sides by 23: $\frac{23b}{23}=\frac{60}{23}$ Simplify: $b=\frac{60}{23}$ 4b. I chose this method because it is the simplest way of solving a problem with fractions in it. It is an easy way to combine like terms. $\frac{1}{9}=\frac{5}{6}-\frac{m}{3}$ Switch sides: $\frac{5}{6}-\frac{m}{3}=\frac{1}{9}$ Subtract $\frac{5}{6}$ $\frac{5}{6}-\frac{5}{6}-\frac{m}{3}=\frac{1}{9}-\frac{5}{6}$ Simplify: $-\frac{m}{3}=\frac{1}{9}-\frac{5}{6}$ Multiply both sides by 3: $3(-\frac{m}{3})=3(-\frac{13}{18})$ Simplify: $-m=-\frac{13}{6}$ Divide both sides by -1 $\frac{-m}{-1}=\frac{-\frac{13}{6}}{-1}$ Simplify: $m=\frac{13}{6}$