Question

plz solve this question
(algebraic equations)
plz answer my questions first.​

Answer 1

★Given:

$\frac{2}{x+1} - \frac{6}{2x-1} + \frac{3}{3x+2} =0$

★To find :

Value of x

★Solution :

LCM of (x+1),(2x-1),(3x+2) = (x - 1)(2x-1)(3x+2)

$\frac{2}{x+1} - \frac{6}{2x-1} + \frac{3}{3x+2} = 0 \\ \\ \\ \frac{[2(2x-1)(3x+2)]-[6(x+1)(3x+2)] + [3(x+1)(2x-1)]}{(x+1)(2x-1)(3x+2)} = 0 \\ \\ \\ \ \frac{[(4x-2)(3x+2)]-[(6x+6)(3x+2)]+[(3x+3)(2x-1)]}{(x+1)(2x-1)(3x+2)} = 0\\ \\ \\ by \: cross\: multiplication\\ \\ \\ [4x(3x+2)-2(3x+2)]- [6x(3x+2)-6(3x+2)] + [3x(2x-1)+3(2x-1)] = 0$

$[12x^2 + 8x - 6x -4] - [18x^2 + 12x - 18x -12] + [6x^2 -3x +6x -3] = 0 \\ \\ \\(12x^2 +2x -4)-(18x^2 - 6x -12)+(6x^2 +3x-3) = 0 \\ \\ \\ 12x^2 +2x-4 -18x^2 +6x +12 +6x^2 +3x -3 = 0 \\ \\ \\ (12x^2 -18x^2 +6x^2) + (2x+6x+3x) + (-4+12-3) = 0 \\ \\ \\ (\cancel{18x^2}-\cancel{18x^2}) + 11 x +5 = 0 \\ \\ \\ 11x = -5 \\ \\ \\ {\pink{\boxed{ x = \frac{-5}{11} }}}$

Answer 2

Answer:

$\sf{The \ value \ of \ x \ is \ -\frac{19}{25}}$

Given:

• $\sf{\frac{2}{x+1}-\frac{6}{2x-1}+\frac{3}{3x+2}=0}$

To find:

• The value of x.

Solution:

$\sf{\implies{\frac{2}{x+1}-\frac{6}{2x-1}+\frac{3}{3x+2}=0}}$

$\sf{\implies{\frac{2}{x+1}+\frac{3}{3x+2}=\frac{6}{2x-1}}}$

$\sf{\implies{\frac{(2)(3x+2)+(3)(x+1)}{(x+1)(3x+2)}=\frac{6}{2x-1}}}$

$\sf{\implies{\frac{6x+4+3x+3}{3x^{2}+2x+3x+2}=\frac{6}{2x-1}}}$

$\sf{\implies{\frac{9x+7}{3x^{2}+5x+2}=\frac{6}{2x-1}}}$

$\sf{\implies{6(3x^{2}+5x+2)=(2x-1)(9x+7)}}$

$\sf{\implies{18x^{2}+30x+12=18x^{2}+14x-9x-7}}$

$\sf{\implies{30x+12=5x-7}}$

$\sf{\implies{30x-5x=-7-12}}$

$\sf{\implies{25x=-19}}$

$\boxed{\sf{\implies{x=-\frac{19}{25}}}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ x \ is \ -\frac{19}{25}}}}$