How do you solve #4( n + 4) = 48#?

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Answer 1 · 2018-07-12T09:15:05

Divide by 4

#n+4=12#

Subtract 4

#n=8#

Answer 2 · 2018-07-12T10:14:42

#n=8#

We start with
#\color(red)(4)(n+4)=48#

Divide by 4 to cancel multiplication:
#(\cancel(\color(red)(4))(n+4))/\cancel(\color(red)(4))=48/\color(red)(4)#
#n+\color(blue)(4)=12#

Subtract 4 to cancel addition:
#n\cancel(+\color(blue)(4))\cancel(-\color(blue)(4))=12-\color(blue)(4)#
#n=8#

You have successfully isolated your variable, solving for its value :-)