CBSE Class 10 Math Question Paper 2016 | Q16

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

Video Explanation

Explanatory Answer | CBSE 2016 Maths Paper Q16

\\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
x² - 4x + 3 = 3
x² - 4x = 0
x(x - 4) = 0
x = 0 or x = 4