Question 1

Solve: (i) 3/x-1 + 1/x-3 = 4/x-2.

Question 2

Answer:

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

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

$\\ \longrightarrow\sf{ (3x - 9 + x - 1)/(x(x - 3) - 1(x - 3)) = (4)/(x - 2)}$

$\\ \longrightarrow\sf{ \frac{3x + x - 9 - 1}{{x}^(2) - 3x - x + 3} = (4)/(x - 2)}$

$\\ \longrightarrow\sf{ \frac{4x - 10}{{x}^(2) - 4x + 3} = (4)/(x - 2)}$

$\\ \longrightarrow\sf{ (4x - 10) (x - 2) = 4({x}^(2) - 4x + 3)}$

$\\ \longrightarrow\sf{4x ( x -2 ) - 10 (x - 2) = 4{x}^(2) - 16x + 12}$

$\\ \longrightarrow\sf{4{x}^(2) - 8x - 10x + 20 = 4{x}^(2) - 16x + 12}$

4x² get cancelled on both sides .

$\\ \longrightarrow\sf{ - 18x + 20 = - 16x + 12}$

$\\ \longrightarrow\sf{ - 18x + 16x = 12 - 20}$

$\\ \longrightarrow\sf{ -2x = - 8}$

$\\ \longrightarrow\sf{ x = (- 8)/(-2)}$

$\\ \longrightarrow\sf{ x = 4}$

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