1. **Stating the problem:** We have four triangles each with given fractions inside and at the ends of their bases. We want to find the unknowns, which appear as $x$, $b$, symbols representing unknown values in the third and fourth triangles. 2. **Top-left triangle:** Inside fraction is $\frac{10}{3}$, base fractions are $\frac{2}{x}$ (left end) and $\frac{3}{5}$ (right end). We assume the fraction inside equals the sum or combination of the base fractions. Let's check if the sum matches: $$\frac{2}{x} + \frac{3}{5} = \frac{10}{3}$$ Multiply both sides by $15x$ (LCM of denominators 3, 5, x): $$15x \times \left( \frac{2}{x} + \frac{3}{5} \right) = 15x \times \frac{10}{3}$$ Simplify terms: $$15x \times \frac{2}{x} + 15x \times \frac{3}{5} = 15x \times \frac{10}{3}$$ $$15 \times 2 + 3x \times 3 = 5x \times 10$$ $$30 + 9x = 50x$$ Bring terms to one side: $$30 = 50x - 9x = 41x$$ So, $$x = \frac{30}{41}$$ 3. **Top-right triangle:** Inside fraction is $\frac{7}{4}$, base fractions $\frac{1}{b}$ (left) and $\frac{4}{7}$ (right). Assuming sum: $$\frac{1}{b} + \frac{4}{7} = \frac{7}{4}$$ Multiply both sides by $28b$: $$28b \times \left( \frac{1}{b} + \frac{4}{7} \right) = 28b \times \frac{7}{4}$$ Simplify: $$28 + 16b = 49b$$ $$28 = 33b$$ So, $$b = \frac{28}{33}$$ 4. **Bottom-left triangle:** Inside fraction is $\frac{4}{5}$, right end base is $\frac{2}{3p}$, left end is a square symbol (let's call it $s$). Assuming: $$s + \frac{2}{3p} = \frac{4}{5}$$ So, $$s = \frac{4}{5} - \frac{2}{3p} = \frac{12p - 10}{15p}$$ (After finding common denominator $15p$.) 5. **Bottom-right triangle:** Inside symbol square $s'$, base left and right fractions are $\frac{1}{3}$ and $\frac{4}{3t}$. Assuming sum: $$\frac{1}{3} + \frac{4}{3t} = s'$$ Write as: $$s' = \frac{1}{3} + \frac{4}{3t} = \frac{t+4}{3t}$$ Hence, with given data, $x = \frac{30}{41}$, $b = \frac{28}{33}$, bottom-left symbol $s = \frac{12p-10}{15p}$, bottom-right symbol $s' = \frac{t+4}{3t}$. These are expressions or values for the unknowns based on fraction sums.