# Thread: How to solve |2x/3 + 5| ≥ 6 ??

1. ## 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.

2. 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.

3. 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

4. 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

gosh...why do i feel like its wrong?? anyhow please let me know if its correct or wrong thanks!

5. 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

6. 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.

7. 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.

8. 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

9. 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)?

10. 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)?
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!

11. 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

12. I actually meant to say,
|x|+|x+1|=1

13. As a reference, ThePerfectHacker posted this problem here, where you can also find a solution.

14. ## THE pd

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.
$
|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$

15. |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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