Solve the equation by factorisation: 2 x x - 4 + 2 x - 5 x - 3 = 25 3 , x ≠ 3 , 4 .

The given equation is: 2 x x - 4 + 2 x - 5 x - 3 = 25 3 ⇒ 2 x x - 3 + 2 x - 5 x - 4 x - 4 x - 3 = 25 3 ⇒ 2 x 2 - 6 x + 2 x 2 - 8 x - 5 x + 20 x 2 - 3 x - 4 x + 12 = 25 3 ⇒ 4 x 2 - 19 x + 20 x 2 - 7 x + 12 = 25 3 ⇒ 12 x 2 - 57 x + 60 = 25 x 2 - 175 x + 300 ⇒ 13 x 2 - 118 x + 240 = 0 Factorising the above equation by splitting the middle term, we get 13 x 2 - 78 x - 40 x + 240 = 0 ⇒ 13 x x - 6 - 40 x - 6 = 0 ⇒ x - 6 13 x - 40 = 0 ⇒ x - 6 = 0 or 13 x - 40 = 0 ⇒ x = 6 or x = 40 13 Hence, x = 6 and x = 40 13 are the roots of the given equation.

Solve the equation by factorisation: x + 1 x - 1 - x - 1 x + 1 = 5 6 , x ≠ 1 , - 1 .

The given equation is: x + 1 x - 1 - x - 1 x + 1 = 5 6 ⇒ x + 1 2 - x - 1 2 x - 1 x + 1 = 5 6 Using ( a + b ) 2 = a 2 + b 2 + 2 a b to expand ( x + 1 ) 2 , using ( a - b ) 2 = a 2 + b 2 - 2 a b to expand ( x - 1 ) 2 and using ( a + b ) ( a - b ) = a 2 - b 2 in denominator, we get x 2 + 2 x + 1 - x 2 - 2 x + 1 x 2 - 1 = 5 6 ⇒ x 2 + 2 x + 1 - x 2 + 2 x - 1 x 2 - 1 = 5 6 ⇒ 4 x x 2 - 1 = 5 6 ⇒ 24 x = 5 x 2 - 5 ⇒ 5 x 2 - 24 x - 5 = 0 Factorising the above equation by splitting the middle term, we get ⇒ 5 x 2 - 25 x + x - 5 = 0 ⇒ 5 x x - 5 + 1 x - 5 = 0 ⇒ 5 x + 1 x - 5 = 0 ⇒ 5 x + 1 = 0 or x - 5 = 0 ⇒ x = - 1 5 or x = 5 Hence, x = - 1 5 and x = 5 are the roots of the given equation.

Solve the equation by factorisation: 3 x + 1 - 1 2 = 2 3 x - 1 , x ≠ 1 , 1 3 .

The given equation is: 3 x + 1 - 1 2 = 2 3 x - 1 ⇒ 3 x + 1 - 2 3 x - 1 = 1 2 ⇒ 3 3 x - 1 - 2 x + 1 x + 1 3 x - 1 = 1 2 ⇒ 9 x - 3 - 2 x - 2 3 x 2 - x + 3 x - 1 = 1 2 ⇒ 7 x - 5 3 x 2 + 2 x - 1 = 1 2 ⇒ 3 x 2 + 2 x - 1 = 14 x - 10 ⇒ 3 x 2 - 12 x + 9 = 0 ⇒ x 2 - 4 x + 3 = 0 ⇒ x 2 - 3 x - x + 3 = 0 ⇒ x x - 3 - 1 x - 3 = 0 ⇒ x - 3 x - 1 = 0 ⇒ x - 3 = 0 or x - 1 = 0 ⇒ x = 3 or x = 1 Hence, x = 3 and x = 1 are the roots of the given equation.

Solve the equation by factorisation: 1 x - 3 + 2 x - 2 = 8 x , x ≠ 0 , 2 , 3 .

The given equation is: 1 x - 3 + 2 x - 2 = 8 x Taking LCM in the denominator of LHS, we get ⇒ x - 2 + 2 x - 3 x - 3 x - 2 = 8 x ⇒ x - 2 + 2 x - 6 x 2 - 2 x - 3 x + 6 = 8 x ⇒ 3 x - 8 x 2 - 5 x + 6 = 8 x Transposing x 2 - 5 x + 6 from LHS to RHS and x from RHS to LHS, we get ⇒ 8 x 2 - 40 x + 48 = 3 x 2 - 8 x ⇒ 5 x 2 - 32 x + 48 = 0 Factorising above quadratic equation by splitting the middle term, we get ⇒ 5 x 2 - 20 x - 12 x + 48 = 0 ⇒ 5 x x - 4 - 12 x - 4 = 0 ⇒ 5 x - 12 x - 4 = 0 ⇒ 5 x - 12 = 0 or x - 4 = 0 ⇒ 5 x = 12 or x = 4 ⇒ x = 12 5 or x = 4 Hence, x = 4 and x = 12 5 are the roots of the given equation.

Solve the equation by factorisation: a x - a + b x - b = 2 c x - c , x ≠ a , b , c .

The given equation is: a x - a + b x - b = 2 c x - c Taking LCM in the denominator of LHS, we get ⇒ a x - b + b x - a x - a x - b = 2 c x - c ⇒ a x - a b + b x - a b x 2 - b x - a x + a b = 2 c x - c Transposing the denominator of LHS to RHS and RHS to LHS, we get ⇒ 2 c x 2 - b x - a x + a b = x - c a x + b x - 2 a b ⇒ 2 c x 2 - 2 b c x - 2 a c x + 2 a b c = a x 2 + b x 2 - 2 a b x - a c x - b c x + 2 a b c ⇒ a x 2 + b x 2 - 2 c x 2 + 2 b c x + 2 a c x - 2 a b x - a c x - b c x = 0 ⇒ a + b - 2 c x 2 + b c + a c - 2 a b x = 0 ⇒ x a + b - 2 c x + b c + a c - 2 a b = 0 ⇒ x = 0 or a + b - 2 c x + b c + a c - 2 a b = 0 ⇒ x = 0 or a + b - 2 c x = - b c + a c - 2 a b ⇒ x = 0 or a + b - 2 c x = 2 a b - b c - a c ⇒ x = 0 or x = 2 a b - b c - a c a + b - 2 c Hence, x = 0 and x = 2 a b - b c - a c a + b - 2 c are the roots of the given equation. Note: The answer given in the textbook seems to be wrong.