How do you solve #(3x)/9+1/2=x/4+2# and check for extraneous solutions?

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Answer 1 · 2016-09-27T18:30:08

#x=18#

#(3x)/9+1/2=x/4+2" "larr# multiply by the LCM of 36

#(color(red)cancel36^4xx3x)/cancel9+(color(red)cancel36^18xx1)/cancel2=(color(red)cancel36^9xx x)/cancel4+color(red)(36)xx2" "larr# cancel

#12x+18=9x+72" "larr# no fractions!

#12x-9x= 72-18#

#3x = 54#

#x = 18#

This is a linear equation. There are no extraneous solutions.