Question

Refer to the attachment for question

Please solve

Answer 1

HEY THERE!!

Method Of Solution;

$\:2 ( \frac{2x - 1}{x + 3}) - 3( \frac{x + 3}{2x - 1}) \\ \\ \\ \: on \: putting \: \frac{2x - 1}{x + 3} = y \: the \: given \: equation \\ \\ \\ \implies \: 2y - \frac{3}{y} = 5 \\ \\ \\ \\ \implies \: {2y}^{2} - 3 = 5y \\ \\ \implies \: {2y}^{2} - 5y - 3 = 0 \\ \\ \implies \: {2y}^{2} - 6y + y - 3 = 0 \\ \\ \\ \implies \: 2y(y - 3) + (y - 3) = 0 \\ \\ \\ \implies \: (y - 3)(2y + 1) = 0 \\ \\ \implies \: y = 3 \: \: and \: \: y = \frac{ - 1}{2} \\ \\ \\ \: substitute \: y \: in \: equation \\ \\ \\ \\ \\ y = 3 \implies \: \frac{2x - 1}{x + 3} = 3 \\ \\ \implies \: 2x - 1 = 3x + 9 \\ \\ \implies \: - 1 - 9 = 3x - 2x \\ \\ \implies \: x = - 10 \\ \\ \\ \\ \\ \: y = \frac{ - 1}{2} \\ \\ \\ \implies \: \frac{2x - 1}{x + 3} = \frac{ - 1}{2} \\ \\ \\ \implies \: 2(2x - 1) = - 1(x + 3) \\ \\ \implies \: 4x - 2 = - x - 3 \\ \\ \\ \implies \: 4x + x = - 3 + 2 \\ \\ \\ \implies \: 5x = - 1 \\ \\ \\ \implies \: x = \frac{ - 1}{5}$

Thus, -10 and -1/5 are the roots of the given Equation;

Answer 2

Given Equation : $2 \bigg( \dfrac{2x-1}{x+3} \bigg) - 3 \bigg( \dfrac{x+3}{2x+1}\bigg) = 5$

⇒ $2 \bigg( \dfrac{2x-1}{x+3} \times \dfrac{2x+1}{2x+1} \bigg) - 3 \bigg( \dfrac{x+3}{2x+1} \times \dfrac{x+3}{x+3} \bigg)= 5$

Now denominators of both the given fractions is same.

⇒ $2 \bigg(\dfrac{(2x-1)(2x+1)}{(x+3)(2x+1)} \bigg) -3\bigg( \dfrac{(x+3)(x+3)}{(x+3)(2x+1)} \bigg) = 5$

⇒ $\dfrac{2(2x-1)(2x+1)-3(x+3)(x+3)}{(x+3)(2x+1)} = 5$

In numerator  by using ( a - b )( a + b ) = a^2 - b^2, writing ( 2x - 1 )( 2x + 1 ) as ( 2x )^2 - 1^2 ⇒ 4x^2 - 1 and by using ( a + b )( a + b ) = ( a + b )^2 = a^2 + b^2 + 2ab , writing ( x + 3 )( x + 3 ) as x^2 + 9 + 6x.

⇒ $\dfrac{2(4x^{2} - 1 ) - 3( x^2 + 9 + 6x ) }{ x( 2x + 1 ) + 3( 2x + 1 ) } = 5$

⇒ $\dfrac{ 8x^{2}- 2 - 3x^{2} - 27 - 18x }{ 2x^2 + x + 6x + 3}= 5$

⇒ $\dfrac{5x^{2} - 18x - 29}{2x^2 + 7x + 3} = 5$

⇒ 5x^2 - 18x - 29 = 5( 2x^2 + 7x + 3 )

⇒ 5x^2 -  18x - 29 = 10x^2 + 35x + 15

⇒ 0 = 5x^2 + 53x + 44

On comparing the formed equation with ax^2 + bx + c = 0, we get that a = 5 , b = 53 and c = 44.

         By Shridhara Acharyya's method

⇒ x = $\dfrac{-b \pm \sqrt{ b^2 - 4ac}}{2a}$

⇒ $x = \dfrac{-53 \pm \sqrt{ ( 53 )^{2} - 4( 5 \times 44 )}}{2(5)}$

⇒ $x = \dfrac{-53 \pm \sqrt{ 2809 - 880}}{10}$

⇒ $x = \dfrac{-53 \pm \sqrt{1929}}{10}$

Therefore the values of x are $\dfrac{-53 - \sqrt{1929}}{10} \:\:or\:\:\dfrac{-53 + \sqrt{1929}}{10}$ satisfying the given equation.