What is #x# in this equation: #(x+6)/5 = 9/5 - 2(x-3)#?

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Answer 1 · 2016-07-28T03:00:39

#x=3#

To|solve #(x+6)/5=9/5-2(x-3)#, let us multiply each side by #5# and we get

#(x+6)/5×5=9/5×5-2(x-3)×5#

= #x+6=9-10(x-3)#

= #x+6=9-10x+30#

Now moving terms containing #x# to left and constant terms to right, we get

#x+10x=9+30-6#

or #11x=33#

or #x=33/11=3#