How do you solve #\frac { 1} { x } + \frac { 1} { x + 18} = \frac { 1} { 12}#?

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Answer 1 · 2017-02-22T05:32:46

#x_1=-12;x_2=18#

It's equivalent to

#(12(x+18)+12x-x(x+18))/(12x(x+18))=0#

Let be #x!=0, x!=-18#

then the equation is

#12(x+18)+12x-x(x+18)=0#

that's

#12x+216+12x-x^2-18x=0#

#-x^2+6x+216=0#

#x^2-6x-216=0#

to solve it let's apply the quadratic formula, then it is:

#x=3+-sqrt(9+216)#

#x=3+-15#

#x_1=-12;x_2=18#