|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.
Thankssss in advance :-)
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|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.
Thankssss in advance :-)
when you have an absolute value inequality you have to realize that |-6|=6Quote:
Originally Posted by Cherry
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:Quote:
Originally Posted by Quick
|x|+|x+1|=0
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!
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,Quote:
Originally Posted by Cherry
|2x+15| ≥ 18
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.Quote:
Originally Posted by 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
How do you solve for x in that problem (in fact I can't see how two positive numbers form 0)? :confused:Quote:
Originally Posted by ThePerfectHacker
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.Quote:
Originally Posted by Quick
Did you try graphing them?
y=abs(x) and y=abx(x+1) and see where they interesect.
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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|=0Quote:
Originally Posted by ThePerfectHacker
substitute: |-1|+|-1+1|=0
thus: 1=0
as said before, I don't think there is any possible value for x
I actually meant to say,
|x|+|x+1|=1
As a reference, ThePerfectHacker posted this problem here, where you can also find a solution.
Ok,here I am giving u the solution of this.:rolleyes:Quote:
Originally Posted by Cherry
|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