How do you solve #-x / 2 + 1 / (2x) = -4 / x#?

1 Answer
EZ as pi
Jul 4, 2016

#x = +-3#

Explanation:

When we are working with equations which have fractions, we can get rid of the fractions immediately by multiplying each term by the Lowest Common Multiple of the denominators so that we can cancel each denominator.

In this case the LCM is #2x#

#(-x xx color(red)(2x))/2 +(1xx color(red)(2x))/(2x) = -(4xx color(red)(2x))/x#

#(-x xx color(red)(cancel2x))/cancel2 +(1xx color(red)(cancel(2x)))/(cancel(2x)) = -(4xx color(red)(2cancelx))/cancelx#

#-x^2 +1 =-8#

#-x^2 = -9 rArr x^2 = 9 " or " x^2 -9 = 0#

#x = +-sqrt 9 " "(x+3)(x-3) = 0#

#x = +-3#

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