1. **State the problem:** We are given three equations: $$\frac{440}{a} = \frac{l - 440}{b}, \quad \frac{396}{a} = \frac{l - 396}{c}, \quad \frac{450}{b} = \frac{l - 450}{c}$$ We need to find the value of $l$. 2. **Rewrite each equation to express $l$ in terms of $a,b,c$: ** From the first equation: $$\frac{440}{a} = \frac{l - 440}{b} \implies b \cdot 440 = a(l - 440)$$ $$440b = al - 440a$$ $$al = 440b + 440a$$ $$l = \frac{440b + 440a}{a} = 440 + \frac{440b}{a}$$ From the second equation: $$\frac{396}{a} = \frac{l - 396}{c} \implies c \cdot 396 = a(l - 396)$$ $$396c = al - 396a$$ $$al = 396c + 396a$$ $$l = 396 + \frac{396c}{a}$$ From the third equation: $$\frac{450}{b} = \frac{l - 450}{c} \implies c \cdot 450 = b(l - 450)$$ $$450c = bl - 450b$$ $$bl = 450c + 450b$$ $$l = 450 + \frac{450c}{b}$$ 3. **Set the expressions for $l$ equal to each other:** From first and second: $$440 + \frac{440b}{a} = 396 + \frac{396c}{a}$$ Multiply both sides by $a$: $$a \cdot 440 + 440b = a \cdot 396 + 396c$$ $$440a + 440b = 396a + 396c$$ Bring terms to one side: $$440a - 396a + 440b - 396c = 0$$ $$44a + 440b - 396c = 0$$ Divide entire equation by 44: $$a + 10b - 9c = 0$$ From first and third: $$440 + \frac{440b}{a} = 450 + \frac{450c}{b}$$ Multiply both sides by $ab$: $$440ab + 440b^2 = 450ab + 450ac$$ Bring all terms to one side: $$440ab + 440b^2 - 450ab - 450ac = 0$$ $$-10ab + 440b^2 - 450ac = 0$$ Divide entire equation by 10: $$-ab + 44b^2 - 45ac = 0$$ 4. **Solve the system:** From step 3 first equation: $$a = 9c - 10b$$ Substitute into second: $$-(9c - 10b)b + 44b^2 - 45(9c - 10b)c = 0$$ $$-9cb + 10b^2 + 44b^2 - 405c^2 + 450bc = 0$$ $$-9cb + 54b^2 - 405c^2 + 450bc = 0$$ Group terms: $$54b^2 + (-9cb + 450bc) - 405c^2 = 0$$ $$54b^2 + 441bc - 405c^2 = 0$$ Divide entire equation by 9: $$6b^2 + 49bc - 45c^2 = 0$$ 5. **Solve quadratic in $b$:** Treat as quadratic in $b$: $$6b^2 + 49bc - 45c^2 = 0$$ Divide by $c^2$ (assuming $c \neq 0$): $$6\left(\frac{b}{c}\right)^2 + 49\left(\frac{b}{c}\right) - 45 = 0$$ Let $x = \frac{b}{c}$: $$6x^2 + 49x - 45 = 0$$ Use quadratic formula: $$x = \frac{-49 \pm \sqrt{49^2 - 4 \cdot 6 \cdot (-45)}}{2 \cdot 6} = \frac{-49 \pm \sqrt{2401 + 1080}}{12} = \frac{-49 \pm \sqrt{3481}}{12}$$ $$\sqrt{3481} = 59$$ So: $$x_1 = \frac{-49 + 59}{12} = \frac{10}{12} = \frac{5}{6}$$ $$x_2 = \frac{-49 - 59}{12} = \frac{-108}{12} = -9$$ 6. **Find corresponding $a$ values:** Recall: $$a = 9c - 10b = 9c - 10xc$$ For $x = \frac{5}{6}$: $$a = 9c - 10 \cdot \frac{5}{6} c = 9c - \frac{50}{6} c = \frac{54}{6} c - \frac{50}{6} c = \frac{4}{6} c = \frac{2}{3} c$$ For $x = -9$: $$a = 9c - 10(-9)c = 9c + 90c = 99c$$ 7. **Find $l$ using first expression:** Recall: $$l = 440 + \frac{440b}{a} = 440 + 440 \cdot \frac{b}{a}$$ For $x = \frac{5}{6}$: $$\frac{b}{a} = \frac{\frac{5}{6} c}{\frac{2}{3} c} = \frac{5}{6} \cdot \frac{3}{2} = \frac{15}{12} = \frac{5}{4}$$ $$l = 440 + 440 \cdot \frac{5}{4} = 440 + 550 = 990$$ For $x = -9$: $$\frac{b}{a} = \frac{-9 c}{99 c} = -\frac{1}{11}$$ $$l = 440 + 440 \cdot \left(-\frac{1}{11}\right) = 440 - 40 = 400$$ 8. **Check which $l$ satisfies all equations:** Check $l=990$ in second expression: $$l = 396 + \frac{396c}{a}$$ For $a = \frac{2}{3} c$: $$\frac{396c}{a} = \frac{396c}{\frac{2}{3} c} = 396 \cdot \frac{3}{2} = 594$$ $$l = 396 + 594 = 990$$ Matches. Check $l=400$: $$l = 396 + \frac{396c}{a} = 396 + \frac{396c}{99c} = 396 + 4 = 400$$ Matches. Check third expression for $l=990$: $$l = 450 + \frac{450c}{b}$$ For $b = \frac{5}{6} c$: $$\frac{450c}{b} = \frac{450c}{\frac{5}{6} c} = 450 \cdot \frac{6}{5} = 540$$ $$l = 450 + 540 = 990$$ Matches. For $l=400$: $$\frac{450c}{b} = \frac{450c}{-9 c} = -50$$ $$l = 450 - 50 = 400$$ Matches. Both $l=990$ and $l=400$ satisfy all equations. **Final answer:** $$\boxed{l = 400 \text{ or } l = 990}$$