1. **Stating the problem:** We have a 3x3 grid with mixed numbers and fractions, with cells labeled (a), (b), (c), and (d) empty. We need to find the values of (a), (b), (c), and (d) based on the given fractions and the vertical fractions 5/9 and 5/4 on the right side. 2. **Understanding the problem:** The grid likely represents a system of equations or relationships between the fractions. The vertical fractions 5/9 and 5/4 might represent sums or ratios related to rows or columns. 3. **Convert mixed numbers to improper fractions:** - 5 1/4 = $\frac{21}{4}$ - 1 3/4 = $\frac{7}{4}$ - 2 1/4 = $\frac{9}{4}$ - 3 1/4 = $\frac{13}{4}$ - 1 1/4 = $\frac{5}{4}$ 4. **Analyze the grid and vertical fractions:** Assuming the vertical fractions 5/9 and 5/4 correspond to sums of columns or rows, we can set up equations. 5. **Set variables for unknowns:** Let (a) = $x$, (b) = $y$, (c) = $z$, (d) = $w$. 6. **Form equations based on the grid:** - First row sum: $x + \frac{21}{4} + \frac{7}{4} = $ sum of first row - Second row sum: $\frac{9}{4} + y + \frac{13}{4} = $ sum of second row - Third row sum: $z + \frac{5}{4} + w = $ sum of third row 7. **Use vertical fractions 5/9 and 5/4:** Assuming these are sums of columns: - First column sum: $x + \frac{9}{4} + z = \frac{5}{9}$ - Second column sum: $\frac{21}{4} + y + \frac{5}{4} = \frac{5}{4}$ 8. **Solve for $y$ from second column sum:** $$\frac{21}{4} + y + \frac{5}{4} = \frac{5}{4}$$ $$y = \frac{5}{4} - \frac{21}{4} - \frac{5}{4} = \frac{5 - 21 - 5}{4} = \frac{-21}{4}$$ 9. **Solve for $x$ and $z$ from first column sum:** $$x + \frac{9}{4} + z = \frac{5}{9}$$ Rearranged: $$x + z = \frac{5}{9} - \frac{9}{4} = \frac{20}{36} - \frac{81}{36} = -\frac{61}{36}$$ 10. **Use third row sum to relate $z$ and $w$:** Assuming third row sum equals some total $T$, but since no total is given, we cannot solve further without more information. **Final answers:** - $y = -\frac{21}{4}$ - $x + z = -\frac{61}{36}$ - $w$ and $z$ cannot be determined uniquely with given data. Without additional information or totals for rows or columns, only partial solutions are possible.