Question

Solve the equation and verify it !
$\sf \large \: \frac{3}{x - 1} - \frac{2}{x - 2} = \frac{1}{x - 3}$
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Answer 1

Step-by-step explanation:

$\sf \: Solution:-$

LCM of the denominators of the terms in the LHS is (x - 1)(x - 2)

$\sf \: So, \frac{3(x - 2) - 2(x - 1)}{(x - 1)(x - 2)} \: = \frac{ 1}{(x - 3)} \\ \\ \sf \: \Rightarrow \frac{3x-6-2x+2}{ x^ 2 -2x-x+2} = \frac{1}{ x-3}$

(Removing the brackets in the numerator and the denominator of the LHS)

$\sf \Rightarrow \frac{ x-4 }{x^ 2 -3x+2 }= \frac{1 }{x-3} \\ \\ \sf \: \red{( by \: cross \: multiplication)} \: \\ \\ \sf \: \Rightarrow(x-4)(x-3)=x^ 2 -3x+2 \\ \\ \sf \Rightarrow x^ 2 -3x-4x+12=x^ 2 -3x+2 \\ \\ \sf \: \red{(On \: removing \: the \: brackets) } \\ \sf \: \Rightarrow \cancel{x^ 2} - \cancel{x^ 2} -7x+3x=+2-12 \\ \\ \sf \red{(Transposing \: x ^ 2 \: - 3x \: to \: the \: LHS \: and \: 12 \: to \: the \: RHS) } \\ \\ \sf \: \Rightarrow \: - 4x = - 10 \\ \\ \sf\Rightarrow \: x = \frac{ - 10}{ - 4} \\ \\ \sf \red{ (Transposing \: -4 \: to \: the \: RHS) }\\ \\ \sf \therefore \: x = 2.5$

Varification

$\sf \Rightarrow \frac{3}{ 2.5-1 }- \frac{2 }{2.5-2} = \frac{ 1}{ 2.5-3} \\ \sf \: \Rightarrow \: \frac{3}{1.5} - \frac{2}{0.5} = \frac{1}{ - 0.5} \\ \sf \: \Rightarrow \: 2 - 4 = - 2 \\ \sf \: \Rightarrow \: - 2 = - 2 \\ \\ \sf \red{Thus, \: LHS \: = RHS \: . Hence, \: verified.}$