Solve x - 1 x - 2 + x - 3 x - 4 = 3 1 3 .

The given equation is x - 1 x - 2 + x - 3 x - 4 = 3 1 3 Taking LCM we get, ( x - 1 ) ( x - 4 ) + ( x - 2 ) ( x - 3 ) ( x - 2 ) ( x - 4 ) = 10 3 2 x 2 - 10 x + 10 x 2 - 6 x + 8 = 10 3 Cross multiplying we get, 6 x 2 - 30 x + 30 = 10 x 2 - 60 x + 80 4 x 2 - 30 x + 50 = 0 is a quadratic equation whose roots are, x = - - 30 ± - 30 2 - 4 × 4 × 50 2 × 4 x = 30 ± 100 8 x = 30 ± 10 8 Therefore, x = 5 , 2 . 5 are the required solutions.

Solve x - 1 x - 2 + x - 3 x - 4 = 3 1 3 .

The given equation is x - 1 x - 2 + x - 3 x - 4 = 3 1 3 Taking LCM we get, ( x - 1 ) ( x - 4 ) + ( x - 2 ) ( x - 3 ) ( x - 2 ) ( x - 4 ) = 10 3 2 x 2 - 10 x + 10 x 2 - 6 x + 8 = 10 3 Cross multiplying we get, 6 x 2 - 30 x + 30 = 10 x 2 - 60 x + 80 4 x 2 - 30 x + 50 = 0 is a quadratic equation whose roots are, x = - - 30 ± - 30 2 - 4 × 4 × 50 2 × 4 x = 30 ± 100 8 x = 30 ± 10 8 Therefore, x = 5 , 2 . 5 are the required solutions.

Solve 1 x + 1 - 2 x + 2 = 3 x + 3 - 4 x + 4 .

The given equation is 1 x + 1 - 2 x + 2 = 3 x + 3 - 4 x + 4 Taking LCM in both L.H.S and R.H.S we get, x + 2 - 2 x - 2 ( x + 1 ) ( x + 2 ) = 3 x + 12 - 4 x - 12 ( x + 3 ) ( x + 4 ) - x ( x + 1 ) ( x + 2 ) = - x ( x + 3 ) ( x + 4 ) Cancel the numerators and get an equation as, ( x + 1 ) ( x + 2 ) = ( x + 3 ) ( x + 4 ) x 2 + 2 x + 1 x + 2 = x 2 + 4 x + 3 x + 12 3 x + 2 = 7 x + 12 4 x + 10 = 0 ⇒ x = - 10 4 ∴ x = - 2 . 5 is the required solution.

Solve the equation x 2 3 - x 1 3 - 2 = 0 .

The given equation is x 2 3 - x 1 3 - 2 = 0 Let us assume x 1 3 = t So we get, t 2 - t - 2 = 0 is a quadratic whose roots are, t = 1 ± 1 + 4 × 2 2 t = 1 ± 3 2 t = 2 or - 1 t = 2 ⇒ x 1 3 = 2 So cubing both sides we get x = 8 Similarly t = - 1 ⇒ x 1 3 = - 1 ⇒ x = - 1 So x = - 1 , 8 are the required solutions.