Could you explain to me how the ≤ or ≥ sign comes into play and how I would know where to place them?
thanks!
You need to consider the boundary value points somewhere in the range, as sometimes the nature of the equation might behave differently after and before, and ON the transition point itself.
While in this particular question, this was not an issue as with or without considering it, you could get the right answer.
However, as a rule of thumb we should always involve the = part in one of the ranges to make sure the solution is consistent and not missing on any boundary value conditions.
thanks!
kpali wrote:
WholeLottaLove wrote:
Can someone tell me if this approach is correct?
|x+3|-|4-x|=|8+x|
So we have:
x=-3
x=4
x=-8
x<-8
-(x+3) - (4+x) = -(8+x)
-x-3 - 4 - x = -8-x
-2x-7=-8-x
1=x (fails, as x is 1 when it must be less than -8)
-8<x<-3
-(x+3) - (4-x) = (8+x)
-x-3 -4+x=8+x
-7=8+x
-15=x (fails, as x is -15 when it must be between -8 and -3)
-3<x<4
(x+3)-(4+x)=8+x
-1=8+x
-9=x (fails, as x is -9 when it must be between -3 and 4)
x>4
(x+3) - -(4-x) = (8+x)
x+3 - (-4+x) = (8+x)
x+3 +4-x=8+x
7=8+x
x=-1 (fails, as x=-1 when it must be greater than 4)
Is this correct?
Thanks!
|x+3|-|4-x|=|8+x|
So we have:
x=-3
x=4
x=-8
x<-8
-(x+3) - (4+x) = -(8+x)
-x-3 - 4 - x = -8-x
-2x-7=-8-x
1=x (fails, as x is 1 when it must be less than -8)
-8<x<-3
-(x+3) - (4-x) = (8+x)
-x-3 -4+x=8+x
-7=8+x
-15=x (fails, as x is -15 when it must be between -8 and -3)
-3<x<4
(x+3)-(4+x)=8+x
-1=8+x
-9=x (fails, as x is -9 when it must be between -3 and 4)
x>4
(x+3) - -(4-x) = (8+x)
x+3 - (-4+x) = (8+x)
x+3 +4-x=8+x
7=8+x
x=-1 (fails, as x=-1 when it must be greater than 4)
Is this correct?
Thanks!
You need to consider the boundary value points somewhere in the range, as sometimes the nature of the equation might behave differently after and before, and ON the transition point itself.
While in this particular question, this was not an issue as with or without considering it, you could get the right answer.
However, as a rule of thumb we should always involve the = part in one of the ranges to make sure the solution is consistent and not missing on any boundary value conditions.