Question

        plse sove 2nd bit and 6th bit . plse plse solve it . it was very urgent ​

Answer 1

SOLUTION:--

ii)

$\sf \: let \: \bigg( \dfrac{4x - 3}{3x + 1} \bigg) \: be \: y$

Then the equation becomes

$\sf \implies \: y- \dfrac{10}{y} = 3$

$\sf \implies \: \dfrac{ {y}^{2} - 10}{y} = 3$

$\sf \implies \: { {y}^{2} - 10}{y} = 3y$

$\sf \implies \: {y}^{2} - 3y - 10y= 0$

By using mid-term splitting method

$\sf \implies \: {y}^{2} - 5y+ 2y- 10y= 0$

$\sf \implies \: {x}(x - 5) + 2(x - 5) = 0$

$\sf \implies \: (y + 2)(y - 5) = 0$

$\sf \implies \: y = - 2 \: or \: y = 5$

Now

$\sf \implies \: \bigg( \dfrac{4x - 3}{3x + 1} \bigg) \: = y$

$\sf \implies \: \bigg( \dfrac{4x - 3}{3x + 1} \bigg) \: = - 2$

$\sf \implies \: {4x - 3} = - 2(3x + 1)$

$\sf \implies \: {4x - 3} = -6x - 2$

$\sf \implies \: 4x + 6x = - 2 + 3$

$\implies \sf \: 10x = 1$

$\boxed{ \implies \sf \: x = \dfrac{1}{10}}$

Also

$\sf \implies \: \bigg( \dfrac{4x - 3}{3x + 1} \bigg) \: = 5$

$\sf \implies \: {4x - 3} \: = 5(3x + 1)$

$\implies \sf \: 4x + 3 = 15x + 5$

$\implies \sf - 11x = 2$

$\boxed{ \implies \sf \: x = \dfrac{ - 11}{2} }$

vi)

We have

$\implies \sf \dfrac{1}{x + 1} + \dfrac{2}{x + 2} = \dfrac{4}{x + 4}$

$\implies \sf \dfrac{(x + 2) + 2(x + 1)}{(x + 1)(x + 2)} = \dfrac{4}{x + 4}$

$\implies \sf \dfrac{3x + 4}{x^{2} + 3x + 2} = \dfrac{4}{x + 4}$

$\implies \sf \: (3x + 4)(x + 4) = 4( {x}^{2} + 3x + 2)$

$\implies \sf \: 3 {x}^{2} + 12x + 4x +16 = 4 {x}^{2} + 12x + 8$

$\implies \sf \: {3x}^{2} + 16x + 16 = {4x}^{2} + 12x + 8$

$\implies \sf \: { - x}^{2} + 4x + 8 = 0$

$\implies \sf {x}^{2} - 4x - 8 = 0$

Here a=1 ,b=-4 and c=-8