How do you solve #3+ 2n > 21#?

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Answer 1 · 2018-03-25T23:45:44

The solution is #n>9#.

Treat the inequality like a normal equation; first, subtract #3# from both sides, then divide by #2#. Here's what the process looks like:

#3+2n>21#

#2n+3>21#

#2n+3color(blue)-color(blue)3>21color(blue)-color(blue)3#

#2ncolor(red)cancelcolor(black)(color(black)+3color(blue)-color(blue)3)>21color(blue)-color(blue)3#

#2n>21color(blue)-color(blue)3#

#2n>18#

#color(blue)(color(black)(2n)/2)>color(blue)(color(black)18/2)#

#color(blue)(color(black)(color(red)cancelcolor(black)2n)/color(red)cancelcolor(blue)2)>color(blue)(color(black)18/2)#

#n>color(blue)(color(black)18/2)#

#n>9#

That's the solution. Hope this helped!