How do you solve $\frac{2}{x + 1} + \frac{5}{x - 2} = - 2$ $2/(x+1) + 5/(x-2)=-2$ ?

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Answer 1 · 2016-07-24T16:50:34

$x = \frac{1}{2}$ $x=1/2$ or $x = - 3$ $x=-3$

$\frac{2}{x + 1} + \frac{5}{x - 2} = - 2$ $2/(x+1)+5/(x-2)=-2$

$\Leftrightarrow \frac{2 (x - 2) + 5 (x + 1)}{(x + 1) (x - 2)} = - 2$ $hArr(2(x-2)+5(x+1))/((x+1)(x-2))=-2$

$\Leftrightarrow 2 (x - 2) + 5 (x + 1) = - 2 (x + 1) (x - 2)$ $hArr2(x-2)+5(x+1)=-2(x+1)(x-2)$

$\Leftrightarrow 2 x - 4 + 5 x + 5 = - 2 (x^{2} - 2 x + x - 2)$ $hArr2x-4+5x+5=-2(x^2-2x+x-2)$

$\Leftrightarrow 7 x + 1 = - 2 x^{2} + 2 x + 4$ $hArr7x+1=-2x^2+2x+4$

$\Leftrightarrow 2 x^{2} + 5 x - 3 = 0$ $hArr2x^2+5x-3=0$

$\Leftrightarrow 2 x^{2} + 6 x - x - 3 = 0$ $hArr2x^2+6x-x-3=0$

$\Leftrightarrow 2 x (x + 3) - 1 (x + 3) = 0$ $hArr2x(x+3)-1(x+3)=0$

$\Leftrightarrow (2 x - 1) (x + 3) = 0$ $hArr(2x-1)(x+3)=0$

Hence either $2 x - 1 = 0$ $2x-1=0$ i.e. $x = \frac{1}{2}$ $x=1/2$

or $x + 3 = 0$ $x+3=0$ i.e. $x = - 3$ $x=-3$

How do you solve 2/(x+1) + 5/(x-2)=-22x+1+5x−2=−22/(x+1) + 5/(x-2)=-2?