Question

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

solve this equation pls

Answer 1

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Question :

$\red \bigstar \bf solve \: the \: following \: equations \\ \frac{1}{(x - 1)} + \frac{2}{(x - 2)} = \frac{3}{(x - 3)}$

To Find :

The value of x=?

Solution :

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

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

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

$\\ \implies \large \sf \: \frac{3x - 4}{ {x}^{2} - 2x - x + 2} = \frac{3}{x - 3}$

$\\ \implies \large \sf \: \frac{3x - 4}{ {x}^{2} - 3x + 2} = \frac{3}{x - 3}$

By cross multiply we get,

$\\ \implies \large \sf \: 3x(x - 3) - 4(x - 3) = 3( {x}^{2} - 3x + 2)$

$\\ \implies \large \sf \: 3 {x}^{2} - 9x - 4x + 12 = 3 {x}^{2} - 9x + 6$

$\\ \implies \large \sf \: 3 {x}^{2} - 13x + 12 = 3 {x}^{2} - 9x + 6$

$\\ \implies \large \sf \: 3 {x}^{2} - 13x + 12 - 3 {x}^{2} + 9x - 6 = 0$

$\\ \implies \large \sf \: 3 {x}^{2} - 3 {x}^{2} - 13x + 9x + 12 - 6 = 0$

$\\ \implies \large \sf \cancel{3 {x}^{2} - 3 {x}^{2} } - 4x + 6 = 0$

$\implies \large \sf \: - 4x + 6 = 0$

$\\ \implies \large \sf \: \cancel- 4x = \cancel - 6$

$\\ \implies \large \sf \: x = \frac{6}{4} = \frac{3}{2}$

$\boxed{ \large{ \sf{ \red \bigstar\: x = \frac{3}{2} \: \: is \: the \: answer}}}$

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Additional information :

(1) - x - = + ➡ (minus) X (minus) =(plus)

(2) + x + =+ ➡ (plus) X (plus) =(plus)

(3) - x + = - ➡ (minus) X (plus) = (minus)

(4) + x - =- ➡ (plus) X (minus) = (minus)