# CBSE Class 10 Math Question Paper 2016 | Q16

###### CBSE Maths Board Paper Solution | 3 Mark | Quadratic Equations and Polynomials

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)}) = $$frac{2}{3}$. x ≠ 1, 2, 3 ## Target Centum in CBSE 10th Maths #### Free Online CBSE Coaching online.maxtute.com ### Video Explanation ## NCERT Solution to Class 10 Maths #### With Videos ### Explanatory Answer | CBSE 2016 Maths Paper Q16 $\frac{1}{$x-1$(x-2)}) + $$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

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