Question

2(x-1/x+3) -7 (x+3/2x-1) =5 , x≠-3 and 1/2

Answer 1

Answer:

Step-by-step explanation:

$2(\frac{x - 1 }{x + 3} )- 7(\frac{x + 3}{2x -1}) =5 \\\\\frac{2x - 2 }{x + 3} - \frac{7x + 21}{2x - 1} =5\\\\\frac{(2x - 2)(2x - 1) - (7x + 21)(x + 3)}{(x + 3)(2x - 1 )} = 5 \\\\\frac{4x^{2} -2x -4x + 2 -(7x^{2} + 21x + 21x + 63) }{2x^{2} - x + 6x - 3 } =5\\\frac{4x^{2} - 2x - 4x + 2 -7x^{2} - 21x - 21x - 63) }{2x^{2} + 5x - 3} =5\\\frac{-3x^{2} - 48x -62 }{2x^{2} + 5x - 3} =5\\\\-3x^{2} - 27x -62 = 5(2x^{2} + 5x - 3)\\-3x^{2} - 48x -62 = 10x^{2} + 25x - 15\\ -13x^{2} - 73x -47 = 0$$\\13x^{2} +73x +47=0$

not able to solve it further sorry

Answer 2

Step-by-step explanation:

To solve:

$2 \big( \frac{x - 1}{x + 3} \big) - 7\big( \frac{x + 3}{2x - 1} ) = 5$

Step 1: Take LCM

$\bigg( \frac{2(x - 1)(2x - 1) - 7(x + 3)(x + 3)}{(x + 3)(2x - 1)} \bigg) = 5 \\ \\ \frac{2x - 2)((2x - 1) - 7( {x}^{2} + 6x + 9) }{2 {x}^{2} - x + 6x - 3 } = 5 \\ \\ 4 {x}^{2} - 2x - 4x + 2 - 7 {x}^{2} - 42x - 63 = 5(2 {x}^{2} + 5x - 3) \\ \\ - 3 {x}^{2} - 48 {x} - 61 = 10 {x}^{2} + 25x - 15 \\ \\ - 13 {x}^{2} - 73x - 46 = 0 \\ \\ 13 {x}^{2} + 73x + 46 = 0$

Now,its a quadratic equation,solve to find the value of x.

Apply Quadratic formula

$\boxed{x_{1,2} = \frac{ - b ± \sqrt{ {b}^{2} - 4ac} }{2a}} \\ \\ here \: a = 13 \\ b = 73 \\ c = 46 \\ \\ x_{1,2}= \frac{ - 73 ± \sqrt{ {(73)}^{2} - 4 \times 13 \times 46} }{2 \times 13} \\ \\x_{1,2} = \frac{ - 73 ± \sqrt{ 5329 - 2392} }{26} \\ \\x_1 = \frac{ - 73 + 54.2 }{26} \\ \\ x_1 = \frac{ - 73 + 54.2}{26} \\ \\ x_1 = - 0.72 \\ \\ x_2 = \frac{ - 73 - 54.2}{26} \\ \\ x_2 = -4.89$

Hope it helps you.