# How to solve |2x/3 + 5| ≥ 6 ??

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• Sep 17th 2006, 05:01 AM
Cherry
How to solve |2x/3 + 5| ≥ 6 ??
|2x/3 + 5| ≥ 6

I cant figure out how to solve this one...I tried (im not sure if it's correct) andi couldnt finish it off.

• Sep 17th 2006, 05:08 AM
Quick
Quote:

Originally Posted by Cherry
|2x/3 + 5| ≥ 6

I cant figure out how to solve this one...I tried (im not sure if it's corrent) but i couldnt finish it off.

when you have an absolute value inequality you have to realize that |-6|=6

so that when you have an inequality like |x-2|>3 you can solve by rewriting it to be: x-2>3 and/or x-2<-3 thus x would be x>5 and/or x<1

now you try solving your inequality and show us the work.
• Sep 17th 2006, 05:24 AM
ThePerfectHacker
Quote:

Originally Posted by Quick
when you have an absolute value inequality you have to realize that |-6|=6

so that when you have an inequality like |x-2|>3 you can solve by rewriting it to be: x-2>3 and/or x-2<-3 thus x would be x>5 and/or x<1

now you try solving your inequality and show us the work.

What Quick said works here, but there are times such that does not work, here is a problem I posed on the forums because of the unsual solution:
|x|+|x+1|=0
• Sep 17th 2006, 05:33 AM
Cherry
okay thanks guys..

here i tried to solve it...i hope its correct...

|2x/3 + 5| ≥ 6

2x + 15 divided by 3 ≥ 6

2x + 5 ≥ 6

2x ≥ 6 - 5

x ≥ 1/2

or
|2x/3 + 5| ≥ -6

= x ≥ -11/2

:confused: gosh...why do i feel like its wrong?? anyhow please let me know if its correct or wrong thanks!
• Sep 17th 2006, 05:38 AM
ThePerfectHacker
Quote:

Originally Posted by Cherry
okay thanks guys..

here i tried to solve it...i hope its correct...

|2x/3 + 5| ≥ 6

2x + 15 divided by 3 ≥ 6

But you divide 15 by 3 but not 2, you "canceled out of a sum". In fact, you combined fractions from the first step and then uncombined fractions in the second step, why? In the second line you should have multiplied both sides by 3 thus,
|2x+15| ≥ 18
• Sep 17th 2006, 06:00 AM
Cherry
Oh oh how stupid of me..I cancelled out the sum!!

so..

|2x+15| ≥ 18

2x ≥ 18 - 15

2x ≥3

x ≥ 3/2

it's correct, right?

Thanks ThePerfectHacker. I didnt pay attention to the fact that 3 is divided by 2x + 15 not only 15.
• Sep 17th 2006, 06:18 AM
ThePerfectHacker
Quote:

Originally Posted by Cherry
Oh oh how stupid of me..I cancelled out the sum!!

so..

|2x+15| ≥ 18

2x ≥ 18 - 15

2x ≥3

x ≥ 3/2

it's correct, right?

Thanks ThePerfectHacker. I didnt pay attention to the fact that 3 is divided by 2x + 15 not only 15.

Not exactly you forgot the thing Quick said to add a plus minus in from of the 18 (I said it is usually wrong method but in high schoold problems it works) I see no plus-minus sign where you are working.
• Sep 17th 2006, 06:38 AM
Cherry
Yeah I know, I figured out how to do the other one :) and yep you're right I better add the signs.

Thanks again, and thanks to Quick. You guys are so much better than my teacher who perfers to ignore than help out pfft. You guys rock :D
• Sep 17th 2006, 07:55 AM
Quick
Quote:

Originally Posted by ThePerfectHacker
What Quick said works here, but there are times such that does not work, here is a problem I posed on the forums because of the unsual solution:
|x|+|x+1|=0

How do you solve for x in that problem (in fact I can't see how two positive numbers form 0)? :confused:
• Sep 17th 2006, 11:41 AM
ThePerfectHacker
Quote:

Originally Posted by Quick
How do you solve for x in that problem (in fact I can't see how two positive numbers form 0)? :confused:

It is a question that tests your understanding of math. I dare you to ask it to your stupid physics teacher no way she can get it.
Did you try graphing them?
y=abs(x) and y=abx(x+1) and see where they interesect.
---
The solution to that equation (and this is why I like it so much besides for it being my problem) is that,
-1<=x<=0
There are infinitely many solutions!
• Sep 17th 2006, 12:59 PM
Quick
Quote:

Originally Posted by ThePerfectHacker
The solution to that equation (and this is why I like it so much besides for it being my problem) is that,
-1<=x<=0
There are infinitely many solutions!

|x|+|x+1|=0

substitute: |-1|+|-1+1|=0

thus: 1=0

as said before, I don't think there is any possible value for x
• Sep 17th 2006, 02:09 PM
ThePerfectHacker
I actually meant to say,
|x|+|x+1|=1
• Sep 18th 2006, 02:29 AM
TD!
As a reference, ThePerfectHacker posted this problem here, where you can also find a solution.
• Jul 12th 2007, 11:38 AM
THE pd
THE pd
Quote:

Originally Posted by Cherry
|2x/3 + 5| ≥ 6

I cant figure out how to solve this one...I tried (im not sure if it's correct) andi couldnt finish it off.

Ok,here I am giving u the solution of this.:rolleyes:
$
|2x/3 + 5| ≥ 6
So,(2x/3+5)≥6 or,-(2x/3+5)≥6
or,2x/3≥6-5 or,(2x/3+5)<=-6
or,2x/3 ≥1 or,2x/3<=-11
or,x≥3/2 or,x<=-33/2
So,the result is x>=3/2 or,x<=-33/2$
• Nov 10th 2007, 07:11 AM
qspeechc
|2x/3+5| ≥ 6

I think it is important to remember:

|x| = x if x ≥ o
and
|x| = -x if x 0

So:
If 2x/3 + 5 ≥ 0, then:

2x/3+ 5 ≥ 6

And solve:
x ≥ 3/2

Now, if 2x/3 + 5 ≤ 0, then:

-(2x/3 + 5) ≥ 6

2x/3 + 5 -6

2x/3 -11

x ≤ -33/2
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