How do you solve $\frac{x + 2}{3} = \frac{6}{x - 1}$ $(x +2) / 3 = 6/ (x-1)$ ?

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How do you solve $\frac{x + 2}{3} = \frac{6}{x - 1}$ $(x +2) / 3 = 6/ (x-1)$ ?

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Answer 1 · 2015-05-27T03:45:55

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

cross multiplying:
$(x + 2) (x - 1) = 6.3$ $(x+2)(x-1) = 6.3$

$x^{2} - x + 2 x - 2 = 18$ $x^2 -x +2x - 2 = 18$
$x^{2} + x - 20 = 0$ $x^2 + x - 20=0$

Factorising by splitting the middle term:
$x^{2} + 5 x - 4 x - 20 = 0$ $x^2 + 5x - 4x - 20=0$
$x (x + 5) - 4 (x + 5) = 0$ $x(x+5) - 4 (x+5)=0$
$(x - 4) (x + 5) = 0$ $(x- 4) (x+5)=0$

the solutions for the expression are: $x = 4, x = - 5$ $color(blue)(x=4 , x=-5$

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How do you solve $\frac{x + 2}{3} = \frac{6}{x - 1}$ $(x +2) / 3 = 6/ (x-1)$ ?

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How do you solve $\frac{x + 2}{3} = \frac{6}{x - 1}$ $(x +2) / 3 = 6/ (x-1)$ ?