How do you solve #[ x/(x+1)] = [1/3] + [(x+2)/(4x)]#?

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Answer 1 · 2016-02-18T10:33:03

#x=3# or #x= -2/5#

first of all equalize all the denominators as if you're doing a standard fraction addition/subtraction

the right-hand side
#x/(x+1)= (4x*1/(3*4x))+(3*(x+2))/(3*4x))#

#x/(x+1)=(4x+3x+6)/(12x)#

#x/(x+1)=(7x+6)/(12x)#

#x/(x+1)-(7x+6)/(12x)=0#

Now the left-hand side

#((x*12x)/(12x*(x+1)))-((x+1)*(7x+6))/((x+1)*12x)=0#

#(12x^2-(7x^2+6x+7x+6))/(12x*(x+1))=0#

#(12x^2-(7x^2+13x+6))/(12x*(x+1))=0#

#(12x^2-7x^2-13x-6)/(12x*(x+1))=0#

#(5x^2-13x-6)/(12x*(x+1))=0#

the nominator of this fractional equation should be 0 for the expression to yield 0.
So solve for x in:

#(5x^2-13x-6)=0#

Use the quadratic formula (or factorize the expression)

#x=-2/5# or #x= 3#

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