Question

Please say the solution to the above question in book ​

Answer 1

3/x-1-2/x-2=1/x-3

∴3/x-2/x-3=1/x-3

∴3-2/x-3=1/x-3

=0

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Answer 2

Step-by-step-explanation:-

$\sf\dfrac{3}{x-1}$-$\sf\dfrac{2}{x-2}$=$\sf\dfrac{1}{x-3}$

Take LCM in LHS

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

$\sf\dfrac{3x-6-2x+2}{(x-1)(x-2)}$=$\sf\dfrac{1}{x-3}$

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

Do Cross multiplication

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

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

$\sf{x² - 3x - 4x + 12}$ = $\sf{x² - 2x - x +2}$

$\sf{x²-7x + 12}$ = $\sf{x²-3x+2}$

Transpose the terms to LHS

$\sf{x² -7x + 12 -x² + 3x -2 =0}$

$\sf{-4x + 10 = 0}$

Transpose 10 to RHS

$\sf{-4x = -10}$

x = $\sf\dfrac{-10}{-4}$

x = $\sf\dfrac{10}{4}$

Simplify x Cancel on 2 table

x = $\sf\dfrac{5}{2}$