How do you solve #1/2(x-3)+(3/2-x)=5 x#?

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Answer 1 · 2016-03-12T14:08:17

#x=0#

Multiply out the brackets giving:

#x/2-3/2+3/2-x=5x#

But #-3/2+3/2=0#

#x/2-x=5x#

Add #x# to both sides

#x/2=5x+x#

#x/2=6x#

Subtract #x/2# from both sides

#0=6x-x/2#

#0=11/2 x#

Thus #x=0#