Question

please help me ❤️❤️❤️​

Answer 1

$\tt \frac{x-1}{x-2} + \frac{x-3}{x-4} = 3\frac{1}{3} \\ \\ \tt \implies \frac{(x-1)(x-4) + (x-2)(x-3)}{(x-2)(x-4)} = \frac{10}{3} \\ \\ \tt \implies \frac{{x}^{2} -4x - x+ 4 + {x}^{2} -2x - 3x + 6 }{{x}^{2} - 2x -4x + 8 } = \frac{10}{3} \\ \\ \tt \implies \frac{2{x}^{2} -10x +10}{{x}^{2}-6x+8} = \frac{10}{3} \\ \\ \tt \implies \frac{2({x}^{2}-5x+5)}{{x}^{2}-6x+8} = \frac{2(5)}{3} \\ \\ \tt \implies \frac{\orange{\cancel{2}}({x}^{2}-5x+5)}{{x}^{2}-6x+8} = \frac{\orange{\cancel{2}}(5)}{3} \\ \\ \tt \implies \frac{{x}^{2}-5x+5}{{x}^{2}-6x+8} = \frac{5}{3} \\ \\ \tt \implies 3({x}^{3}-5x+5) = 5({x}^{2}-6x+8) \\ \\ \tt \implies 3{x}^{2}-15x+15 = 5{x}^{2} - 30x + 40 \\ \\ \tt \implies 2{x}^{2} -15x + 25 = 0 \\ \\ \tt \implies 2{x}^{2} -10x -5x+25 = 0 \\ \\ \tt \implies 2x(x-5)-5(x-5) = 0 \\ \\ \tt \implies (x-5)(2x-5) = 0\\ \\ \tt \implies x = 5 \:\:or\:\: x = \frac{5}{2} \\ \\ \tt \therefore \boxed{\bold{\underline{\red{\tt x = 5 \:\:or\:\: x = \frac{5}{2} }}}}$

Answer 2

ohk pa no probs..njoy pannunga..enaku laan leave illa nu sollitanga..safe ah irunga..tc..time ku sapdunga..byee