mikaa_toxica13 2 years ago 16 < 3x + 1 < 4 Part 1: Solve the inequality above. Part 2: Describe the graph

1. fernando12

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

2. satellite73

another county heard from

3. mikaa_toxica13

@satellite73 what ??

4. satellite73

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

5. mikaa_toxica13

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

6. mikaa_toxica13

@satellite73

7. lyzzy7

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

8. lyzzy7

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.