Find the value of x in (1/(x−1)) + (1/(x−2)) = (3/(x−3)) .


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Question Number 62610 by hovea cw last updated on 23/Jun/19

$Find the value of x in$ $\frac{1}{x - 1} + \frac{1}{x - 2} = \frac{3}{x - 3} .$
Commented by mathmax by abdo last updated on 24/Jun/19
$the set of definition for this (e) is D =R−{1,2,3} (e) ⇔((x−2+x−1)/((x−1)(x−2))) =(3/(x−3)) ⇒((2x−3)/(x^2 −3x+2)) =(3/(x−3)) ⇒ (2x−3)(x−3) =3(x^2 −3x+2) ⇒2x^2 −6x−3x +9 =3x^2 −9x +6 ⇒ 3x^2 −9x+6−2x^2 +6x−9 =0 ⇒x^2 −3x −3 =0 Δ =9−4(−3) =9 +12 =21 ⇒x_1 =((3+(√(21)))/2) and x_2 =((3−(√(21)))/2)$
$t h e s e t o f d e f i n i t i o n f o r t h i s (e) i s D = R - {1, 2, 3}$ $(e) \Leftrightarrow \frac{x - 2 + x - 1}{(x - 1) (x - 2)} = \frac{3}{x - 3} \Rightarrow \frac{2 x - 3}{x^{2} - 3 x + 2} = \frac{3}{x - 3} \Rightarrow$ $(2 x - 3) (x - 3) = 3 (x^{2} - 3 x + 2) \Rightarrow 2 x^{2} - 6 x - 3 x + 9 = 3 x^{2} - 9 x + 6 \Rightarrow$ $3 x^{2} - 9 x + 6 - 2 x^{2} + 6 x - 9 = 0 \Rightarrow x^{2} - 3 x - 3 = 0$ $Δ = 9 - 4 (- 3) = 9 + 12 = 21 \Rightarrow x_{1} = \frac{3 + \sqrt{21}}{2} a n d x_{2} = \frac{3 - \sqrt{21}}{2}$
Commented by $@ty@m last updated on 24/Jun/19

$P l . c h e c k . . .$ $3 x^{2} - 9 x + 6 - 2 x^{2} + 6 x - 9 = 0$ $s h o u l d b e$ $3 x^{2} - 9 x + 6 - 2 x^{2} + 6 x + 3 x - 9 = 0$
	Answered by $@ty@m last updated on 23/Jun/19

	$\frac{1}{x - 1} - \frac{1}{x - 3} = \frac{2}{x - 3} - \frac{1}{x - 2}$ $\frac{x - 3 - x + 1}{(x - 1) (x - 3)} = \frac{2 x - 4 - x + 3}{(x - 3) (x - 2)}$ $\frac{- 2}{x - 1} = \frac{x - 1}{x - 2}$ $- 2 x + 4 = x^{2} - 2 x + 1$ $x^{2} = 3$ $x = \sqrt{3}$
	Answered by MJS last updated on 23/Jun/19

	$x \neq 1 \land x \neq 2 \land x \neq 3$ $(x - 2) (x - 3) + (x - 1) (x - 3) = 3 (x - 1) (x - 2)$ $x^{2} = 3$ $\Rightarrow x = \pm \sqrt{3}$
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