Question

1/(x-1)(x-3) + 1/(x-3)(x-5) + 1/(x-5)(x-7) = 1/9 solve it​

Answer 1

ANSWER:

Given:

• 1/(x-1)(x-3) + 1/(x-3)(x-5) + 1/(x-5)(x-7) = 1/9

To Find:

• Value of x

Solution:

$\text{We are given that,}\\\\:\longrightarrow\dfrac{1}{(x-1)(x-3)}+\dfrac{1}{(x-3)(x-5)}+\dfrac{1}{(x-5)(x-7)}=\dfrac{1}{9}\\\\\text{Taking LCM,}\\\\:\implies\dfrac{(x-5)(x-7)+(x-1)(x-7)+(x-1)(x-3)}{(x-1)(x-3)(x-5)(x-7)}=\dfrac{1}{9}\\\\:\implies\dfrac{(x^2-12x+35)+(x^2-8x+7)+(x^2-4x+3)}{(x-1)(x-3)(x-5)(x-7)}=\dfrac{1}{9}\\\\:\implies\dfrac{x^2-12x+35+x^2-8x+7+x^2-4x+3}{(x-1)(x-3)(x-5)(x-7)}=\dfrac{1}{9}\\\\\text{Solving the like terms,}\\\\:\implies\dfrac{3x^2-24x+45}{(x-1)(x-3)(x-5)(x-7)}=\dfrac{1}{9}\\\\\text{Taking 3 common,}\\\\:\implies\dfrac{3(x^2-8x+15)}{(x-1)(x-3)(x-5)(x-7)}=\dfrac{1}{9}$

$\text{On splitting the middle term,}\\\\:\implies\dfrac{3(x^2-5x-3x+15)}{(x-1)(x-3)(x-5)(x-7)}=\dfrac{1}{9}\\\\:\implies\dfrac{3(x-3)(x-5)}{(x-1)(x-3)(x-5)(x-7)}=\dfrac{1}{9}\\\\\text{On cancelling (x-3)(x-5),}\\\\:\implies\dfrac{3}{(x-1)(x-7)}=\dfrac{1}{9}\\\\\text{On cross-multiplying,}\\\\:\implies3\times9=x^2-8x+7\\\\\text{Transposing LHS to RHS,}\\\\:\implies0=x^2-8x+7-27\\\\:\implies x^2-8x-20=0\\\\\text{On splitting the middle term,}\\\\:\implies x^2-10x+2x-20=0\\\\:\implies x(x-10)+2(x-10)=0\\\\:\implies(x-10)(x+2)=0\\\\\bf{:\implies x=10\:\:\:and\:\:\:x=-2}$

Answer 2

Step-by-step explanation:

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