Question

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

Answer:

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

Taking the LCM :

$\tt {\frac {x-7-x-4}{x^{2}-7x+4x-28} = \frac {11}{30}}$

=》 $\tt {\frac {-11}{x^{2}-3x-28} = \frac {11}{30}}$

Cross multiplying :

=》 $\tt {11x^{2} - 33x - 308 = -330}$

=》 $\tt {11x^{2} - 33x + 22}$

Divide the equation by 11 :

=》 $\tt {x^{2} - 3x + 2=0}$

Find the zeroes of this quadratic polynomial :

=》 $\tt {x^{2} - 2x - x + 2=0}$

=》 $\tt {x = 1,2}$

Answer 2

Answer : x = 1 or x = 2

Step by step explanation :

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

$\tt {\frac{(x - 7) - (x + 4)}{(x + 4)(x - 7)} = \frac{11}{30}}$

$\tt{\frac{x - 7 - x - 4}{(x + 4)(x - 7)}} = \frac{11}{30}$

$\tt{\frac{ - 11}{ {x}^{2} + 4x - 7x - 28} } = \frac{11}{30}$

$\tt{ - 30 = x {}^{2} - 3x - 28}$

$\tt{x {}^{2} - 3x - 28 + 30 = 0}$

$\tt{x {}^{2} - 3x + 2 = 0}$

$\tt { {x}^{2} - 2x - x + 2 = 0}$

$\tt{x(x - 2) - 1(x - 2) = 0}$

$\tt{(x - 1)(x - 2) = 0}$

$\tt{x = 1 \: \: or \: \: x = 2}$

Thus,

Value of x is either 1 or 2.

x = 1 or x = 2