Question

12. Find the roots of the equation
$\frac{1}{x + 4} - \frac{1}{x - 7} = \frac{11}{30}$
, x # - 4 , 7.

Class:- 10

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Answer 1

$= \ \textgreater \ \frac{1}{x + 4} - \frac{1}{x - 7} = \frac{11}{30}$

LCM = (x + 4)(x - 7)

= > 30(x - 7) - 30(x + 4)  = 11(x + 4)(x - 7)

= > 30x - 210 - 30x - 120 = 11(x^2 - 7x + 4x - 28)

= > -330 = 11(x^2 - 3x - 28)

= > -330 = 11x^2 - 33x - 308

= > 11x^2 - 33x + 22 = 0

= > x^2 - 3x + 2 = 0

= > x^2 - 2x - x + 2 = 0

= > x(x - 2) - (x - 2) = 0

= > (x - 1)(x - 2) = 0

= > x = 1, x = 2.



Therefore the values of x = 1, 2.



Hope this helps!

Answer 2

Here we simplify the problem into a quadratic equation first. Then we solve the equation to get the roots.


$\displaystyle \frac{1}{x+4} - \frac{1}{x-7} = \frac{11}{30} \\ \\ \\ \implies \frac{x-7-(x+4)}{(x+4)(x-7)}=\frac{11}{30} \\ \\ \\ \implies \frac{x-7-x-4}{(x+4)(x-7)} = \frac{11}{30} \\ \\ \\ \implies \frac{-\cancel{11}}{(x+4)(x-7)} = \frac{\cancel{11}}{30} \\ \\ \\ \implies \frac{-1}{(x+4)(x-7)} = \frac{1}{30} \\ \\ \\ \implies -30 = (x+4)(x-7) \\ \\ \\ \implies -30 = x^2-7x+4x-28 \\ \\ \\ \implies x^2-3x+2=0$

$\displaystyle \implies x^2-x-2x+2=0 \\ \\ \\ \implies x(x-1)-2(x-1)=0 \\ \\ \\ \implies (x-1)(x-2)=0 \\ \\ \\ \implies x-1=0 \quad OR \quad x-2=0 \\ \\ \\ \implies \boxed{\bold{x=1}} \quad OR \quad \boxed{\bold{x=2}}$



Thus, The roots of the equation are x = 1 and x = 2