How do you solve #-4(2/3) = (2n/3) + (1/2) + (n/3)#?

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Answer 1 · 2015-06-12T20:48:11

#n=-19/3#

#-4(2/3)=(2n/3)+(1/2)+(n/3)# =

#-8/3=(2n)/3+1/2+n/3# =

Subtract #1/2# from both sides.

#-1/2-8/3=(cancel(3)n)/cancel 3# =

#-1/2-8/3=n#

The common denominator for #-1/2 and -8/3# is #6#.

#-1/2*(3/3)-8/3*(2/2)=n# =

#-3/6-16/3=n# =

#-19/3=n#

Switch sides.

#n=-19/3#