Question

Solve by quadratic equations method.
$find \: the \: value \: of \: x \\ \ \frac{1}{(x - 1)(x - 2)} + \frac{1}{(x - 2)(x - 3)} = \frac{2}{3}$
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Answer 1

EXPLANATION.

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

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

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

$\sf \implies \dfrac{2x - 4}{(x - 1)(x - 2)(x - 3)} \ = \ \dfrac{2}{3}$

⇒ 3(2x - 4) = 2[(x - 1)(x - 2)(x - 3)].

⇒ 6x - 12 = 2[(x² - 2x - x + 2)(x - 3)].

⇒ 6x - 12 = 2[(x² - 3x + 2)(x - 3)].

⇒ 6x - 12 = 2[(x³ - 3x² - 3x² + 9x + 2x - 6].

⇒ 6x - 12 = 2[x³ - 6x² + 11x - 6].

⇒ 6x - 12 = 2x³ - 12x² + 22x - 12.

⇒ 2x³ - 12x² + 22x - 12 - 6x + 12 = 0.

⇒ 2x³ - 12x² + 16x = 0.

⇒ 2x(x² - 6x + 8) = 0.

⇒ x² - 6x + 8 = 0.

Factorizes the equation into middle term splits, we get.

⇒ x² - 4x - 2x + 8 = 0.

⇒ x(x - 4) - 2(x - 4) = 0.

⇒ (x - 2)(x - 4) = 0.

⇒ x = 2 and x = 4.

Answer 2

Given : Expression : $\sf \dfrac{1}{(x - 1)(x - 2)} + \dfrac{1}{(x - 2)(x - 3)} = \dfrac{2}{3}$

Exigency To Find : The value of x .

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⠀⠀⠀⠀⠀$\dag\:\:\bf Expression\:: \dfrac{1}{(x - 1)(x - 2)} + \dfrac{1}{(x - 2)(x - 3)} = \dfrac{2}{3}$

⠀⠀⠀⠀⠀⠀$\underline {\bf{\star\:Now \: By \: Solving \: the \: Given \: Expression \::}}$

⠀⠀⠀⠀⠀$\dag\:\:\bf Expression\:: \dfrac{1}{(x - 1)(x - 2)} + \dfrac{1}{(x - 2)(x - 3)} = \dfrac{2}{3}$

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

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

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

⠀⠀⠀⠀⠀$:\implies \:\:\sf \: \dfrac{2x-4}{(x - 1)(x - 2)(x-3)} = \dfrac{2}{3}$

Now taking 2 as common in numerator :

⠀⠀⠀⠀⠀$:\implies \:\:\sf \: \dfrac{2x-4}{(x - 1)(x - 2)(x-3)} = \dfrac{2}{3}$

⠀⠀⠀⠀⠀$:\implies \:\:\sf \: \dfrac{2(x-2)}{(x - 1)(x - 2)(x-3)} = \dfrac{2}{3}$

Now by shifting 2 :

⠀⠀⠀⠀⠀$:\implies \:\:\sf \: \dfrac{x-2}{(x - 1)(x - 2)(x-3)} = \dfrac{\cancel {2}}{3\times \cancel {2} }$

⠀⠀⠀⠀⠀$:\implies \:\:\sf \: \dfrac{x -2}{(x - 1)(x - 2)(x-3)} = \dfrac{1}{3}$

Now by Solving Denominator :

⠀⠀⠀⠀⠀$:\implies \:\:\sf \: \dfrac{x -2}{(x - 1)(x - 2)(x-3)} = \dfrac{1}{3}$

⠀⠀⠀⠀⠀$:\implies \:\:\sf \: \dfrac{x -2}{(x^2 -2x -x + 2)(x-3)} = \dfrac{1}{3}$

⠀⠀⠀⠀⠀$:\implies \:\:\sf \: \dfrac{x -2}{(x^2 -3x + 2)(x-3)} = \dfrac{1}{3}$

⠀⠀⠀⠀⠀$:\implies \:\:\sf \: \dfrac{x -2}{(x^3 -3x^2-3x^2+9x + 2x-6)} = \dfrac{1}{3}$

⠀⠀⠀⠀⠀$:\implies \:\:\sf \: \dfrac{x -2}{(x^3 -6x^2+11x -6)} = \dfrac{1}{3}$

Now by Cross Multiplication :

⠀⠀⠀⠀⠀$:\implies \:\:\sf \: 1\bigg(x^3 -6x^2+11x -6\bigg)= 3(x-2)$

⠀⠀⠀⠀⠀$:\implies \:\:\sf \: x^3 -6x^2+11x -6= 3x-6$

⠀⠀⠀⠀⠀$:\implies \:\:\sf \: x^3 -6x^2+11x -6-3x + 6 = 0$

⠀⠀⠀⠀⠀$:\implies \:\:\sf \: x^3 -6x^2+11x -3x -6 +6 = 0$

⠀⠀⠀⠀⠀$:\implies \:\:\sf \: x^3 -6x^2+8x = 0$

Now taking x as common :

⠀⠀⠀⠀⠀$:\implies \:\:\sf \: x(x^2 -6x+8) = 0$

Therefore,

• Quadratic Equation : x² - 6x + 8

⠀⠀⠀⠀⠀⠀$\underline {\frak{\star\:Now \: By \: Solving \: the \:Equation \: \::}}$

⠀⠀⠀⠀⠀$:\implies \:\:\sf \: x^2 -6x+8 = 0$

By Splitting middle term :

⠀⠀⠀⠀⠀$:\implies \:\:\sf \: x^2 -4x-2x+8 = 0$

Finding out Common Factor :

⠀⠀⠀⠀⠀$:\implies \:\:\sf \: x(x -4)-2(x-4) = 0$

Rewrite in Factored Term :

⠀⠀⠀⠀⠀$:\implies \:\:\sf \: (x -2)(x-4) = 0$

Therefore,

• x - 2 = 0
• x = 2

And ,

• x -4 = 0
• x = 4

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {  Hence,\:The\:Value \:of\:x \:is\:\bf{2\:\:\&\: 4\:\:.}}}}$

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