Question

x-1/x+2+x-3/x=29/20 solve for x

Answer 1

Answer:

$\frac{x - 1}{x + 2} + \frac{x - 3}{x} = \frac{29}{20}$

$\frac{x(x - 1) + (x + 2)(x - 3)}{x(x + 2)} = \frac{29}{20}$

$\frac{ {x}^{2} - x + {x}^{2} - 3x + 2x - 6 }{ {x}^{2} + 2x} = \frac{29}{20}$

$\frac{ {2x}^{2} - 2x - 6 }{ {x}^{2} + 2x } = \frac{29}{20}$

$20( {2x}^{2} - 2x - 6) = 29( {x}^{2} + 2x)$

${40x}^{2} - 40x - 120 = {29x}^{2} + 58x$

${11x}^{2} - 98x - 120$

${11x}^{2} - 110x + 12x - 120$

$11x(x - 10) + 12(x - 10)$

➠ $(11x + 12)(x - 10)$

here is your answer

hope is helps you..!!!

Answer 2

${\underline{\boxed{\tt{\green{\dfrac{x-1}{x + 2} + \dfrac{x-3}{x} = \dfrac{29}{20} }}}}}$

$\sf\implies \dfrac{x-1}{x + 2} + \dfrac{x-3}{x} = \dfrac{29}{20}$

$\sf\implies\dfrac{ x ( x - 1) + ( x +2) ( x - 3)}{ x ( x +2} = \dfrac{29}{20}$

$\sf\implies\dfrac{x^2 - x + x^2 - 3x + 2x - 6}{x^2 + 2x } = \dfrac{29}{20}$

$\sf\implies \dfrac{2x^2 -2x - 6 }{x^2 + 2x} = \dfrac{29}{20}$

$\sf\implies 20 ( 2x^2 - 2x - 6 ) = 29 ( x^2 + 2x)$

$\sf\implies 40x^2 - 40x - 120 = 29x^2 + 58x$

$\sf\implies 40x^2- 29x^2 - 40x - 58x - 120 = 0$

$\sf\implies 11x^2 - 90x - 120 = 0$

• middle term split

$\sf\implies11x^2 - 110x + 12x - 120 = 0$

$\sf\implies 11x ( x - 10 ) + 12 ( x - 10) = 0$

$\sf\implies ( 11x + 12 ) ( x - 10)$

$\begin{tabular}{|c|c|}\cline{1-2}\sf 11x + 12 = 0 &\sf x - 10 = 0\\\cline{1-2}\sf 11x = -12&\sf x = 10\\\cline{1-2}\sf x = -12\div 11 &\sf x = 10\\\cline{1-2}\end{tabular}$

• since x cannot be negetive

Therefore

${\underline{\boxed{\tt{\green{ x = 10}}}}}$