Question

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

Answer 1

Solve for x with a method other than graphing?

Answer 2

That's what I thought, but there is a solution...

Answer 3

what is it ?

Answer 4

\[-1\le x \le 1\]

Which is practically 0, but NO SOLUTION isn't correct either

Answer 5

\[\Large - 1 \le x \le 1\]

Answer 6

How?

Answer 7

How does that work?

Answer 8

It doesn't work ;)

Answer 9

If it doesn't work, how do you do it?

Answer 10

By blunderous error. It's not correct.

FFM, can't we only imply that:
\[\Large \begin{array}{l}
\left| { - x} \right| \le 1\\
\left| x \right| \le 1\\
x \le 1
\end{array}\]

Answer 11

Here is my attempt. using what FSM stated,

\( |1-x| \le 2 \implies -1 \le x \le 3 \)

and

\( |1+x| \le 2 \implies -3\le x \le 1 \)

now if we take the union we get \(-1\le x \le 1 \)

Is this correct?

Answer 12

Hmm. Seems to be so.

Answer 13

intersection...
then notice that on \(-1\le x \le 1\)

|x+1|=x+1
and

|1-x|=1-x

thus

\[|x+1|+|1-x|=2 \Leftrightarrow x+1+1-x=2 \Leftrightarrow 2=2\]

thus \(-1\le x \le 1\) is the solution

Answer 14

but if 2=2, how does that suggest -1<x<1?

Answer 15

yes Intersection,Thanks Zarkon.

Answer 16

2=2 is a tautology ...thus the equation works for all x between, and including, -1 and 1

Answer 17

btw x is real here ?

Answer 18

Yes, x is real

Answer 19

Haven't really learned that method.. but always a good way to find out new things. So.. tautology in math means?

Answer 20

you could always use \[|x|=\sqrt{x^2}\]

and start squaring both sides

Answer 21

Let's just assume x is real ;)

How about...
\[\Large \begin{array}{l}
\left| {x + 1} \right| + \left| {1 - x} \right| = 2\\
\sqrt {{{\left( {x + 1} \right)}^2}} \le 2\\
- 3 \le x \le 1\\
\sqrt {{{\left( {1 - x} \right)}^2}} \le 2\\
- 1 \le x \le 1
\end{array}\]
Could you say that since there are two restrictions but since one is more restrictive than the other, it must be obeyed?

Answer 22

http://en.wikipedia.org/wiki/Tautology_%28logic%29

Answer 23

I tried that... but it didn't work.... Yes, it gave me two answers... so I don't think that method works so well for this... I don't know why the first one doesn't apply..

Answer 24

I have a similar equation... Can you see if you can solve that one instead? And see if the method works?

Answer 25

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

Answer 26

\( x \ge 1 \)?

Answer 27

yes :)

Answer 28

How did you get that? Using the squaring method, or?

Answer 29

I don't understand the squaring method :( I did it like before.

Answer 30

I would do cases

Answer 31

Ok... can you give me the equation again? I like to compare different ones to help me understand...

Answer 32

waht is "cases", Zarkon ?

Answer 33

|x+1|=x+1 if \(x\ge-1\)
|x+1|=-(x+1) if \(x\le -1\)

do that for both abs...solve each equation

Answer 34

Yes, I know that method, and I know that those are the values.. but I'm not sure how to plug those into an equation in order to find the solutions (values)?

Answer 35

|1-x|=-(x+1) if x≥1
|x+1|=x+1 if x≤1

so for example if \(x\ge 1\]
|x+1|-|1-x|=2

turns into

(x+1)-(-(1-x))=2
x+1+1-x=2
2=2

so \(x\ge 1\) is part of the solution


check the others...

Answer 36

Um, that's a bit confusing...

Answer 37

if \[x\ge 1\]
|x+1|-|1-x|=2

turns into

(x+1)-(-(1-x))=2
x+1+1-x=2
2=2

so \[x\ge 1\] is part of the solution

Answer 38

Ok, thanks :) that's clearer

Answer 39

now you need to check

\[-1<x<1\] and \[x\le -1\]

Answer 40

so the first one would be -(x-1) -(1+x) = 2
for -1<x<1?

Answer 41

and that is -2=2? What value is that in < or >? I'm not used to tautology,,,

Answer 42

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

-1<x<1

gives


x+1-(1-x)=2
2x=2
which has no solution on -1<x<1

Answer 43

when x>=1, the condition is satisfied for x=1
the condition is satisfied for every value of x between -1 and 1
when x<=-1, this is satisfied by x=-1
so, the answer is:
\[x \in [-1,1]\]

Answer 44

Ok... why is it not -(x-1) -(1+x) = 2?

Answer 45

on -1<x<1

|x+1|=x+1
|1-x|=1-x

Answer 46

Hmm.. I thought since x<1, it would be -(x+1) since it's less than 1?

Answer 47

on \(x\ge 1\)
|x+1|=x+1
|1-x|=-(1-x)

on -1<x<1
|x+1|=x+1
|1-x|=1-x

on \(x\le -1\)
|x+1|=-(x+1)
|1-x|=1-x

Answer 48

I was just a bit confused, because my book says that, when
x<n in this case, x<1
that if the equation is |3x-2|
it would be -(3x-2)?

Answer 49

3x-2=0
3x=2
x=2/3
so 3x-2<0 when x<2/3

thus

|3x-2|=3x-2 if \(x\ge 2/3\)
and
|3x-2|=-(3x-2) if x<2/3

Answer 50

I understand that..., but it also says:
The modulus of x, denoted by |x|, is defined by

|x|=x if x>=0

|x|=-x if x<0....

So, does that only apply to 0?

Answer 51

what do you mean "So, does that only apply to 0?"

Answer 52

it says x<0 or x>= 0

Does this rule only apply to it being less than or greater than 0?

Answer 53

if what is in the abs is positive then you get back what you started with

if what is in the abs is negative then you get back the opposite of what you started with

Answer 54

Um... Ok, so what about -|x+1| + |1-x|=2?
This is only for learning purposes... How would you solve this, step by step?
I'm studying w/o a teacher.. and only a text book, so could you maybe, explain it like I'm one of those people reading "Modulus Functions for Dummies?"
:)

Answer 55

find the places where x+1 is positive and negative

and where 1-x is positive and negative

...when do cases

Answer 56

I need to go soon so I wont be much help from here on out

Answer 57

Ok, well, that's for trying!! I'm still not sure of this... so if anyone else is there...?

Answer 58

just one last note:

it is possible to get erroneous solutions when using this technique so after finding a possible solution you should test any number within your interval to see if it satisfies the original equation.

Answer 59

Thanks for the tip!

Answer 60

Here's a picture of how I would try to solve these equations.

Answer 61

thanks!