Question

solve for x:

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

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Answer 1

$\huge{\mathcal{Hi\: there!}}$

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$\huge{\bold{Solution :}}$

$\mathsf{ \large{ \frac{x - 3 + x - 1}{(x - 1)(x - 2)(x - 3)} = \frac{2}{3}}} \\ \\ \mathsf{ \frac{2x - 4}{x {}^{3} -6x {}^{2} + 11x - 6} = \frac{2}{3} } \\ \\ \mathsf{\frac{2(x - 2)}{x{}^{3} -6x {}^{2} + 11x - 6} = \frac{2}{3} } \\ \\ \mathsf{ \frac{x - 2}{x{}^{3} -6x {}^{2} + 11x - 6} = \frac{1}{3} } \\ \\ \mathsf{ 3x - 6 = x{}^{3} -6x {}^{2} + 11x - 6} \\ \\ \mathsf{{}x^{3} -6x {}^{2} + 8x = 0} \\ \\ \mathsf{x(x {}^{2} - 6x + 8) = 0 } \\ \\ \mathsf{x {}^{2} - 6x + 8 = 0}$

Now, solve this equation, to get the value of x ;

$\mathsf{x {}^{2} - 6x + 8 = 0} \\ \\ \mathsf{x {}^{2} - 4x - 2x + 8 = 0 } \\ \\ \mathsf{x(x - 4) - 2(x - 4) = 0} \\ \\ \mathsf{(x - 2)(x - 4) = 0}$

Hence,

$\mathsf{(x - 2) = 0} \\ \\ \therefore\mathsf{x = 2}$

And,

$\mathsf{(x - 4) = 0} \\ \\ \therefore \mathsf{x = 4}$

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Answer 2

$Given:\frac{1}{(x - 1)(x - 2)} + \frac{1}{(x - 2)(x - 3)} =\frac{2}{3}$

LCM = (x - 1)(x - 2)(x - 3)

⇒ 3(x - 3) + 3(x - 1) = 2(x - 1)(x - 2)(x - 3)

⇒ 3x - 9 + 3x - 3 = 2x^3 - 12x^2 + 22x - 12

⇒ 2x^3 - 12x^2 + 16x = 0

⇒ 2x(x^2 - 6x + 8) = 0

⇒ x^2 - 6x + 8 = 0

⇒ x^2 - 2x - 4x + 8 = 0

⇒ x(x - 2) - 4(x - 2) = 0

⇒ (x - 2)(x - 4) = 0

⇒ x = 2,4.

Therefore, the value of x = 2,4.

Hope this helps!