How do you solve #3-6n-4<17#?

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Answer 1 · 2018-07-21T21:32:23

#n > -3#

#3-6n-4 < 17#

Simplify the left hand side:
#-1 - 6n < 17#

Add #color(blue)(6n)# to both sides of the inequality:
#-1 - 6n quadcolor(blue)(+quad6n) < 17 quadcolor(blue)(+quad6n)#

#-1 < 17 + 6n

Subtract #color(blue)17# from both sides:
#-1 quadcolor(blue)(-quad17) < 17 + 6n quadcolor(blue)(-quad17)#

#-18 < 6n#

Divide both sides by #color(blue)6#:
#(-18)/color(blue)6 < (6n)/color(blue)6#

#-3 < n#

#n > -3#

This can be said as "#n# is greater than #-3#."
enter image source here
(mathwarehouse.com)

The open circle on the #-3# means that #-3# is not a solution (but anything greater than it).

Hope this helps!

Answer 2 · 2018-07-21T23:38:50

#n> -3#

We can simplify left side to #-1-6n#. We now have

#-1-6n<17#

Next, we can add #1# to both sides to get

#-6n<18#

Lastly, divide both sides by #6# to isolate #n#. Recall that since we divided by a negative, the sign flips. We're left with

#n> -3#

Hope this helps!