Question

the solution of 2x+3/2x-1= 3x-1/3x+1

Answer 1

SOLUTION

TO SOLVE

$\displaystyle \sf{ \frac{2x + 3}{2x - 1} = \frac{3x - 1}{3x + 1} }$

EVALUATION

$\displaystyle \sf{ \frac{2x + 3}{2x - 1} = \frac{3x - 1}{3x + 1} }$

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

$\sf{ \implies \:6 {x}^{2} + 2x + 9x + 3 = 6 {x}^{2} - 2x - 3x + 1 \: }$

$\sf{ \implies \:11x + 3 = - 5x + 1 \: }$

$\sf{ \implies \:11x + 5x = -3 + 1 \: }$

$\sf{ \implies \:16x = -2\: }$

$\displaystyle \sf{ \implies \:x = \frac{ - 2}{16} \: }$

$\displaystyle \sf{ \implies \:x = - \frac{1}{8} \: }$

FINAL ANSWER

The required solution is

$\displaystyle \sf{ \:x = - \frac{1}{8} \: }$

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Answer 2

(2x-1)(3x-1)=(3x+1)(2x-3)

6x^2+2x-9x-3=6x^2-3x-2x+1

6x^2-7x-3x=6x^2-5x+1

6x^2-6x^2=7x-5x+3+1

0=2x-4

x=-4/2=-2