1. The problem asks to find equivalent fractions for each given fraction with specific denominators. 2. For each fraction $\frac{a}{b}$, find the factors to get the new denominators, then multiply numerator and denominator by the same number to keep the value equivalent. 3. Solve each: - $\frac{2}{3}$ to denominators 6 and 15: $\frac{2}{3} = \frac{2 \times 2}{3 \times 2} = \frac{4}{6} = \frac{2 \times 5}{3 \times 5} = \frac{10}{15}$ - $\frac{2}{7}$ to denominators 21 and 35: $\frac{2}{7} = \frac{2 \times 3}{7 \times 3} = \frac{6}{21} = \frac{2 \times 5}{7 \times 5} = \frac{10}{35}$ - $\frac{5}{12}$ to denominators 36 and 60: $\frac{5}{12} = \frac{5 \times 3}{12 \times 3} = \frac{15}{36} = \frac{5 \times 5}{12 \times 5} = \frac{25}{60}$ - $\frac{6}{11}$ to denominators 22 and 33: $\frac{6}{11} = \frac{6 \times 2}{11 \times 2} = \frac{12}{22} = \frac{6 \times 3}{11 \times 3} = \frac{18}{33}$ - $\frac{4}{15}$ to denominators 60 and 90: $\frac{4}{15} = \frac{4 \times 4}{15 \times 4} = \frac{16}{60} = \frac{4 \times 6}{15 \times 6} = \frac{24}{90}$ - $\frac{3}{10}$ to denominators 20 and 30: $\frac{3}{10} = \frac{3 \times 2}{10 \times 2} = \frac{6}{20} = \frac{3 \times 3}{10 \times 3} = \frac{9}{30}$ 4. Final answers: - $\frac{2}{3} = \frac{4}{6} = \frac{10}{15}$ - $\frac{2}{7} = \frac{6}{21} = \frac{10}{35}$ - $\frac{5}{12} = \frac{15}{36} = \frac{25}{60}$ - $\frac{6}{11} = \frac{12}{22} = \frac{18}{33}$ - $\frac{4}{15} = \frac{16}{60} = \frac{24}{90}$ - $\frac{3}{10} = \frac{6}{20} = \frac{9}{30}$