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Mathematics 14 Online
OpenStudy (anonymous):

Can anyone tell me how to solve: |x|+|x+2|=3

zepdrix (zepdrix):

Hi! :) \(\Large\bf \color{#008353}{\text{Welcome to OpenStudy! :)}}\)

zepdrix (zepdrix):

I think I can help you with this, gimme couple minutes. Gotta eat my yogurt -_-

OpenStudy (anonymous):

Hi & thank you! You can take your time :)

zepdrix (zepdrix):

Ok ok so do you remember this little tidbit about the absolute function?\[\Large |x|\quad=\quad \cases{x,\qquad x>0\\-x,\quad x<0}\]

zepdrix (zepdrix):

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,\[\Large\bf\sf |x+2|\quad=\quad \cases{\qquad x+2,\qquad x>-2\\-(x+2),\qquad x<-2}\]

zepdrix (zepdrix):

So we'll look for solutions in each region. For \(\Large\bf\sf x>0\), \[\Large |x|\quad=\quad \cases{\color{orangered}{x,\qquad x>0}\\-x,\quad x<0}\]\[\Large |x+2|\quad=\quad \cases{\color{orangered}{\qquad x+2,\qquad x>-2}\\-(x+2),\qquad x<-2}\]

zepdrix (zepdrix):

Both pieces are positive when x is greater than zero, right? See what I'm doing kinda? D:

zepdrix (zepdrix):

So for x>0 we have,\[\Large\bf\sf |x|+|x+2|=3\]Becomes,\[\Large\bf\sf x+x+2=3\]

OpenStudy (anonymous):

I understand so far xD Then we have to find the x<0 too?

zepdrix (zepdrix):

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.

zepdrix (zepdrix):

When -2<x<0, \[\Large |x|\quad=\quad \cases{x,\qquad x>0\\\color{orangered}{-x,\quad x<0}}\] \[\Large |x+2|\quad=\quad \cases{\color{orangered}{\qquad x+2,\qquad x>-2}\\-(x+2),\qquad x<-2}\]

zepdrix (zepdrix):

So for -2<x<0 we have,\[\Large\bf\sf |x|+|x+2|=3\]Becomes,\[\Large\bf\sf -x+x+2=3\]

zepdrix (zepdrix):

And what can we determine here? :o

OpenStudy (anonymous):

Ooh .. So that one doesn't work? I think I understand now. Then for x<-2 : -x-x-2=3 ?

zepdrix (zepdrix):

No solution in the second area? Correct! For x<-2? Yes very good!

OpenStudy (anonymous):

I understand now. Kinda confused me at first xD; Thank you very much for the detail explanation!! :D

zepdrix (zepdrix):

Yah it's kind of a weird problem *|c:/* np

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