Question

$\frac{1}{x} + \frac{1}{x - 8} = \frac{1}{12}$
How to Solve this​

Answer 1

Consider the given equation.

$\dfrac {1}{x}+\dfrac {1}{x-8}=\dfrac {1}{12}$

We know how to add two fractions with different denominators but having numerator 1.

$\dfrac {1}{a}+\dfrac {1}{b}=\dfrac{a+b}{ab}$

Thus, the LHS of the equation becomes,

$\dfrac {x+x-8}{x(x-8)}=\dfrac {1}{12}\\\\\\\dfrac {2x-8}{x(x-8)}=\dfrac {1}{12}\\\\\\\dfrac {2(x-4)}{x(x-8)}=\dfrac {1}{12}\\\\\\\dfrac {x-4}{x(x-8)}=\dfrac {1}{24}$

Now we can perform cross multiplication.

$24(x-4)=x(x-8)\\\\24x-96=x^2-8x\\\\x^2-32x+96=0\\\\x^2-(16+4\sqrt {10})x-(16-4\sqrt {10})x+96=0\\\\x(x-16-4\sqrt {10})-(16-4\sqrt {10})(x-16-4\sqrt {10})=0\\\\(x-(16+4\sqrt {10}))(x-(16-4\sqrt {10}))=0$

This implies,

$\Large\boxed {\mathbf {x=16\pm4\sqrt {10}}}$

#answerwithquality

#BAL