16 < 3x + 1 < 4 Part 1: Solve the inequality above. Part 2: Describe the graph of the solution.
Think of it as an inequality with 3 "sides." Whatever you do to one side, you must do to all 3. You want x, so first, subtract 1 from all 3 sides. Then divide all 3 sides by 3. Try it and write what you get.
15 < 3x < 3 5 < 1
16 < 3x + 1 < 4 16 - 1 < 3x < 4 - 1 15 < 3x < 3 15/3 < x < 3/3 5 < x < 1
Thank you, could you help me with Part 2?
5 < x < 1 is not possible
:S
Yeah, it's pretty obvious now since 16 ≮ 4
You don't have to worry about graphing this one.
Is it "obvious" to you @ahoward79 ?
No
Then you learned nothing.
Well, don't insult her @skullpatrol, teach her
I understand the first part just not the second.
You already GAVE her the answer.
@skullpatrol You have not helped me on ONE question, if you arent even going to help me PLEASE, LEAVE ME ALONE. Thank you :)
Can a number be BOTH greater than 5 and less than 1 ?
I was showing her the steps @skullpatrol
No.
She had already tried to do it, but just forgot the x in between
Furthermore, I was pointing out that 5 < x < 1 is not possible. You really need to get over yourself bro.
So what is the solution set for this question? @ahoward79
@skullpatrol, I'm about tired of your antics. Don't respond to posts I post in anymore, seriously
16 < 3x + 1 < 4 You did the next step correctly: Subtract 1 from all three sides: 15 < 3x < 3 The next step is to divide all 3 sides by 3 5 < x < 1 Here's the problem. A compound inequality like this means 5 < x and x < 1 x < 5 means x > 5 That means x > 5 and x < 1 It's not possible for a number to be less than 1 and greater than 5. That's why this can't be graphed. there is no number that can be a solution to this inequality.
Thank you soooo much @Hero you dont know how many times Ive tried to get him to leave me alone.
Why don't you tell me how many times?
don't worry, I already abuse-reported him
Are you going to answer this question by stating the solution set and Part 2: Describe the graph of the solution.
@skullpatrol, leave her alone
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