The given question is a medium difficulty 3-mark question from the chapter Quadratic Equations. Concept tested: Simplifying an algebraic equation and computing the value of x. This question appeared in Section C of the 2016 CBSE class 10 board paper.

Question 16: Solve for x : \\frac{1}{(x-1)(x-2)}) + \\frac{1}{(x-2)(x-3)}) = \\frac{2}{3}). x ≠ 1, 2, 3

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\\frac{1}{(x-1)(x-2)}) + \\frac{1}{(x-2)(x-3)}) = \\frac{2}{3})

**Take (x - 1)(x - 2)(x - 3) as the common denominator for the LHS of the equation and simplify**

\\frac{(x-3)+(x-1)}{(x-1)(x-2)(x-3)}) = \\frac{2}{3})

**Cross Multiply and Simplify**

3(2x - 4) = 2(x - 1)(x - 2)(x - 3)

3(2(x - 2)) = 2(x - 1)(x - 2)(x - 3)

Divide both sides by 2(x - 2). Possible because x ≠ 2. So, (x - 2) will not be 0.

3 = (x - 1)(x - 3)

**Expand RHS and Reorder the terms**

x^{2} - 4x + 3 = 3

x^{2} - 4x = 0

x(x - 4) = 0

x = 0 or x = 4

Class 10 Maths

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