How do you solve #(3x)/(x+1)+6/(2x)=7/x#?

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Answer 1 · 2016-10-01T16:29:59

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

#(3x)/(x+1)+6/(2x)=7/x# can be rewritten as

#(3x)/(x+1)=7/x-6/(2x)#

or #(3x)/(x+1)=7/x-3/x=4/x#

Hence, #(3x)/((x+1))xx x(x+1)=4/x xx x(x+1)#

or #(3x)/(cancel(x+1))xx xcancel((x+1))=4/cancelx xx cancelx(x+1)#

or #3x^2=4x+4#

or #3x^2-4x-4=0#

or #3x^2-6x+2x-4=0#

or #3x(x-2)+2(x-2)=0#

or #(3x+2)(x-2)=0#

Hence either #3x+2=0# i.e. #x=-2/3#

or #x-2=0# i.e. #x=2#