Could any one tell me how to solve this equation? Thanks :)

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Answer 1 · 2016-11-10T20:45:41

#x = 42/55#

#3 x/2+ 7 2/3x+2=24/2-x" "larr# change to improper fractions

#(6+x)/2 + (23x)/3 +2 = 12-x" "larr xx 6#

#(cancel6^3xx(6+x))/cancel2 + (cancel6^2xx(23x))/cancel3 +6xx2 = 6xx12-x xx6#

#18+3x + 46x +12 =72 -6x#

#49x+6x = 72-18-12#

#55x = 42#

#x = 42/55#