Question

Please solve this immediately.....​

Answer 1

Answer

Thus , the value of x is 1 and 9

Solution

$\sf \dfrac{x}{3} + \dfrac{3}{6 - x} = \dfrac{2(6 + x)}{15} \\ \\ \implies \sf \dfrac{x(6 - x) + 3 \times 3}{3(6 - x)} = \dfrac{2(6 + x)}{15} \\ \\ \sf \implies15 \{x(6 - x) + 9 \} = 2(6 + x) \times 3(6 - x) \\ \\ \sf \implies 15(6x - {x}^{2} + 9) = 6( {6}^{2} - {x}^{2} ) \\ \\ \implies \sf 90x - 15 {x}^{2} + 135 = 216 - 6 {x}^{2} \\ \\ \implies \sf - 15 {x}^{2} + 6x + 90x + 135 - 216 = 0 \\ \\ \implies \sf - 9x {}^{2} + 90x - 81 = 0 \\ \\ \implies \sf - 9( {x}^{2} - 10x + 9) = 0 \\ \\ \implies \sf {x}^{2} - 10x + 9 = 0 \\ \\ \implies \sf {x}^{2} - x - 9x + 9 = 0 \\ \\ \implies \sf x(x - 1) - 9(x - 1) = 0 \\ \\ \sf \implies(x - 1)(x - 9) = 0$

Therefore, we have :

$\sf x - 1 = 0 \: \: and \: \: x - 9 = 0 \\ \\ \sf \implies x = 1 \: \: and \: \: \implies x = 9$

Answer 2

Step-by-step explanation:

answer..................