|x+1| + |x+2| + | x-1|= 20

Question

Please tell me how to generate equations for below -

1) |x+1| + |x+2| + | x-1|= 20
2) |x+1| - |x+2| - | x-1|= 20

I know check points would be -1 -2 and +1 . but then how to proceed with equations ?

for x = +1

I have spend more than twelve hours and checked almost every resource online but couldn't find reasoning behind +ve and -ve signs for the test cases  .

PLEASE HElp !
Regards,
Sandip66612

alphabeta1234 · Answer

Well you got the intervals correct.

So you have three pieces.

1) |x+1| 
2) |x+2|
3) | x-1|

Think/imagine the number line for these three pieces.

Convert them into "distances"

1) |x-(-1)| 
2) |x-(-2)|
3) |x-(1)|

So for 1) imagine the number line --------(-1)--------------

In your first interval
x<=-2 
Ask your self for (1) are we to the LEFT OR RIGHT of -1? The interval is to the left of -1 (x<=-2) hence |x+1|=|x-(-1)|=-(x+1) 
ie you put the negative sign infornt of the expression.
Same thing for (2). Is the interval left of -2 ? Yes, so we put a negative infront of -(x+2)
What about (3)? Is the interval x<=-2 left or right of 1? Left, so we put a negative infront of -(x-1)

So for the interval x<=2 our equation |x+1| + |x+2| + | x-1|= 20
becomes: -(x+1)-(x-2)-(x-1)=20

Another quick one:
-1 <= x <= +1

(1) The interval is to the right of -1. So |x+1|=+(x+1)
(2) The interval is to the right of -2. Hence, |x+2|=+(x+2)
3) The interval is to the left 1. So |x-1|=-(x-1)
Now evaluate and solve.

You have to do this step for all the intervals. It sucks. But nothing we can do about it. Make sure to either plug the solutions into the equations to validate they are correct OR make sure the value x equals is within the interval. Hope this helps

sandip66612 · Answer

Dear AlphaBeta,

I really appreciate your time taking reply. Thank you very much. I think i get it what you are saying. By going that logic - can you confirm below ? ( Just need to confirm this as this has appeared on on of the articles - "ABSOLUTE VALUE" )

IF below cases are correct - then - gmatclubcom/forum/math-absolute-value-modulus-86462.html should be corrected.

|x+3| - |4-x| = | 8 + x |

1) x <= -8

-(x+3)+(4-x) = -(8+x)

2) -8<= x <= -3

-(x+3)+(4-x) = +(8+x)

3) -3<= x <= 4

+(x+3)+(4-x) = +(8+x)

4) x >= 4

+(x+3)-(4-x) = +(8+x)

alphabeta1234 · Answer

You got it dude, the +/-'s on all the pieces are correct.

You have the steps down:

Find the roots for each of the pieces
Rearrange them by size from left to right
Set up the intervals you need to check
Imagine the roots for each piece and ask your self, "Is the interval I am testing is it left or right to the root for that piece? If its left (-), if it's right put (+) on the piece.

Just make sure that once you solve all the pieces you check the answer for x satisfies that interval. Sometimes you will have an interval like x<=-8 and you solve the equation and you get an x=5. Hence a contradiction and not a solution. You can also check by plugging in the value into the equation. Either works.

Regarding Walker's post. He is correct. He didn't do anything wrong and neither did you. Solve the equation and check. 
for x<-=8 
You have +(x-4)
He he has -(4-x).

They are the same thing. When I see |4-x| I immediately flip it to |x-4|=|x-(-4)|. That is what you did. In the interval x<=-8, which is left of -4, |x-4|=-(x-4).

|x+1| + |x+2| + | x-1|= 20

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