Question

16 < 3x + 1 < 4

Part 1: Solve the inequality above.
Part 2: Describe the graph

Answer 1

16 < 3x + 1 < 4 ==> This is not possible, if it's not a typo then:
15 < 3x < 3
5 < x < 1 ==> This is an impossible outcome, there is no solution.

But if it is:
4 < 3x + 1 < 16
3 < 3x < 15
1 < x < 5
interval notation: (1 , 5)

Edit: Well what do you know! I never thought I see the day that in a group of nine 55% not only actually believe that 4>16 and 1>5 but they also proved it and wrote it down for those 45% in the group and the rest of the world to see that 16 & 5 are no longer greater than 4 & 1. lol

Answer 2

another county heard from

Answer 3

@satellite73 what ??

Answer 4

unless you have a typo there is no solution to your question, because it would mean that \(16<4\) which is false

Answer 5

well then how about
-7y - 17 > 11

Part 1: Solve the inequality above.
Part 2: Describe the graph of the solution.
i gotta do one or the other

Answer 6

@satellite73

Answer 7

for -7y-17 > 11
add 17 on each side
-7y > 28
then divide each side by -7
y > -4

Answer 8

for the graph it is an open circle on -4 and the numbers to the right of four are greater so you shade in the line to the right.