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1/(x-1)+2/(x-2)=3/(x-3) solve this equation pls

Answer»

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Question :

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



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