1. **State the problem:** We are given six fraction addition problems arranged in boxes, some with missing fractions to find. 2. **Recall the formula for adding fractions:** $$\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}$$ where $a,b,c,d$ are integers and $b,d \neq 0$. 3. **Solve the first problem (given):** $$\frac{2}{3} + \frac{1}{6} = \frac{2 \times 6}{3 \times 6} + \frac{1 \times 3}{6 \times 3} = \frac{12}{18} + \frac{3}{18} = \frac{15}{18} = \frac{5}{6}$$ This confirms the first box. 4. **Second problem:** $$\frac{1}{3} + x = \frac{7}{12}$$ Find $x$: $$x = \frac{7}{12} - \frac{1}{3} = \frac{7}{12} - \frac{4}{12} = \frac{3}{12} = \frac{1}{4}$$ 5. **Third problem:** $$\frac{4}{9} + \frac{1}{2} = y$$ Find $y$: $$y = \frac{4 \times 2}{9 \times 2} + \frac{1 \times 9}{2 \times 9} = \frac{8}{18} + \frac{9}{18} = \frac{17}{18}$$ 6. **Fourth problem:** $$\frac{9}{20} + \frac{2}{5} = z$$ Convert $\frac{2}{5}$ to denominator 20: $$\frac{2}{5} = \frac{2 \times 4}{5 \times 4} = \frac{8}{20}$$ Add: $$z = \frac{9}{20} + \frac{8}{20} = \frac{17}{20}$$ 7. **Fifth problem:** $$w + \frac{3}{4} = \frac{7}{8}$$ Find $w$: $$w = \frac{7}{8} - \frac{3}{4} = \frac{7}{8} - \frac{6}{8} = \frac{1}{8}$$ 8. **Sixth problem:** $$a + b = \frac{3}{4}$$ Since no further info, assume $a = \frac{1}{2}$ and $b = \frac{1}{4}$ as a simple solution: $$\frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4}$$ **Final answers:** - Second box missing fraction: $\frac{1}{4}$ - Third box sum: $\frac{17}{18}$ - Fourth box sum: $\frac{17}{20}$ - Fifth box missing fraction: $\frac{1}{8}$ - Sixth box missing fractions: $\frac{1}{2}$ and $\frac{1}{4}$