How do you solve #(x-3)/(x-4)+4=(3x)/x#?

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Answer 1 · 2016-08-05T10:44:18

#x=7/2#

#(x-3)/(x-4)+4=3x/x#
or
#(x-3)/(x-4)+4=3#
or
#(x-3)/(x-4)=-4+3#
or
#(x-3)/(x-4)=-1#
or
#x-3=-x+4#
or
#x+x=4+3#
or
#2x=7#
or
#x=7/2#

Answer 2 · 2016-08-05T11:00:30

#x=7/2#

Reworked to make it simpler

Note that #(3x)/x = 3# giving

#(x-3)/(x-4)+4=3#

#=>(x-3)/(x-4)+1=0#

But 1 can be written as: # (x-4)/(x-4)#

#(x-3)/(x-4)+(x-4)/(x-4)=0#

Multiply through out by #(x-4)#

#(x-3)+(x-4)=0#

#2x-7=0#

#x=7/2#