How do you solve #1/2 - 3/2x = 4/x - 5/12#?

1 Answer
Oct 1, 2015

The equation doesn't have a real solution

Explanation:

Multiply both sides by #12x# (which is the LCM of 2, 12 and x)

#6x - 18x^2 = 48 - 5x#

Make one of the sides equal zero,

#-18x^2 + 11x - 48 = 0#
#18x^2 - 11x + 48 = 0#

Use the quadratic formula

#x = (11 +- sqrt(121 -4*18*48))/36 = (11 +- sqrt(-3335))/36#

So the equation doesn't have a real solution. We could expand that root using complex numbers if you want but for Algebra just stopping here should do.