Can anyone tell me how to solve: |x|+|x+2|=3
Hi! :) \(\Large\bf \color{#008353}{\text{Welcome to OpenStudy! :)}}\)
I think I can help you with this, gimme couple minutes. Gotta eat my yogurt -_-
Hi & thank you! You can take your time :)
Ok ok so do you remember this little tidbit about the absolute function?\[\Large |x|\quad=\quad \cases{x,\qquad x>0\\-x,\quad x<0}\]
Well when we have a shift, we have to shift that point where the sign switches also.\[\Large\bf\sf |x+2|\quad=\quad \cases{\qquad x+2,\qquad x+2>0\\-(x+2),\qquad x+2<0}\]Which leads to,\[\Large\bf\sf |x+2|\quad=\quad \cases{\qquad x+2,\qquad x>-2\\-(x+2),\qquad x<-2}\]
So we'll look for solutions in each region. For \(\Large\bf\sf x>0\), \[\Large |x|\quad=\quad \cases{\color{orangered}{x,\qquad x>0}\\-x,\quad x<0}\]\[\Large |x+2|\quad=\quad \cases{\color{orangered}{\qquad x+2,\qquad x>-2}\\-(x+2),\qquad x<-2}\]
Both pieces are positive when x is greater than zero, right? See what I'm doing kinda? D:
So for x>0 we have,\[\Large\bf\sf |x|+|x+2|=3\]Becomes,\[\Large\bf\sf x+x+2=3\]
I understand so far xD Then we have to find the x<0 too?
We'll need to be a little bit more specific than that. We actually have 3 regions to look at. Next we'll look here -2<x<0. x is between 0 and -2.
When -2<x<0, \[\Large |x|\quad=\quad \cases{x,\qquad x>0\\\color{orangered}{-x,\quad x<0}}\] \[\Large |x+2|\quad=\quad \cases{\color{orangered}{\qquad x+2,\qquad x>-2}\\-(x+2),\qquad x<-2}\]
So for -2<x<0 we have,\[\Large\bf\sf |x|+|x+2|=3\]Becomes,\[\Large\bf\sf -x+x+2=3\]
And what can we determine here? :o
Ooh .. So that one doesn't work? I think I understand now. Then for x<-2 : -x-x-2=3 ?
No solution in the second area? Correct! For x<-2? Yes very good!
I understand now. Kinda confused me at first xD; Thank you very much for the detail explanation!! :D
Yah it's kind of a weird problem *|c:/* np
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