Solving several Absolute Values in |x+2|- 2|x+1|+ 2|x-1|- 2|x-2|= x-2

I

icandoit

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Heeeeyyyy guys, how are you? :D:D:D
Hope it's been a beautiful Friday for you guys. I have a nasty sort of question that's bugging me for a while:

Determine all possible solutions of the equation

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


So far, what I've done is to solve them by parts:

1. |x+2| = x-2

x+2=x-2 , -(x+2)=x-2
x-x=-2-2 , -x-x=-2+2
0=-4 , x=0
imposssible , x=0 is the solution

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

-2(x+1)= x-2 , -2 [-(x+1)]= x-2
-2x-2=x-2 , 2(x+1)=x-2
-3x=0 , 2x+2=x-2
x=0 , x=-4


3. 2|x-1|= x-2
2(x-1)= x-2 , 2[-(x-1)]= x-2
2x-2=x-2 , -2x+2=x-2
2x-x=-2+2 , -2x-x=-2-2
x=0 , x= 4/3


4. -2|x-2|= x-2
-2(x-2)= x-2 , -2[-(x-2)]= x-2
-2x+4=x-2 , 2x-4=x-2
x=2 , x=2


Then from the x-values, I find what is common. So x= 0, 2.
I feel like something is not correct. I thought of doing variations of the equation (and then testing the obtained x-values), e.g., one absolute value term is negative, then the others positive.
i.e. -(x+2)- 2(x+1)+ 2(x-1)- 2(x-2)= x-2

other variations: (x+2)- 2[-(x+1)]+ 2(x-1)- 2(x-2)= x-2
(x+2)- 2(x+1)+ 2[-(x-1)]- 2(x-2)= x-2

etc.


I'm honestly stuck here, I don't even know the answers to it, so I never know if my working was correct. Please help! Thanks so so so much! :)
 
icandoit said:
Heeeeyyyy guys, how are you? :D:D:D
Hope it's been a beautiful Friday for you guys. I have a nasty sort of question that's bugging me for a while:

Determine all possible solutions of the equation

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


So far, what I've done is to solve them by parts:

1. |x+2| = x-2

x+2=x-2 , -(x+2)=x-2
x-x=-2-2 , -x-x=-2+2
0=-4 , x=0
imposssible , x=0 is the solution

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

-2(x+1)= x-2 , -2 [-(x+1)]= x-2
-2x-2=x-2 , 2(x+1)=x-2
-3x=0 , 2x+2=x-2
x=0 , x=-4


3. 2|x-1|= x-2
2(x-1)= x-2 , 2[-(x-1)]= x-2
2x-2=x-2 , -2x+2=x-2
2x-x=-2+2 , -2x-x=-2-2
x=0 , x= 4/3


4. -2|x-2|= x-2
-2(x-2)= x-2 , -2[-(x-2)]= x-2
-2x+4=x-2 , 2x-4=x-2
x=2 , x=2


Then from the x-values, I find what is common. So x= 0, 2.
I feel like something is not correct. I thought of doing variations of the equation (and then testing the obtained x-values), e.g., one absolute value term is negative, then the others positive.
i.e. -(x+2)- 2(x+1)+ 2(x-1)- 2(x-2)= x-2

other variations: (x+2)- 2[-(x+1)]+ 2(x-1)- 2(x-2)= x-2
(x+2)- 2(x+1)+ 2[-(x-1)]- 2(x-2)= x-2

etc.


I'm honestly stuck here, I don't even know the answers to it, so I never know if my working was correct. Please help! Thanks so so so much! :)
Click to expand...

In my opinion, the best way to do these problems is through graphing

Plot

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

in your graphing calculator and observe the behavior of the function. That might give you a clue as to the path to solve the problem.
 
icandoit said:
Determine all possible solutions of the equation

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

So far, what I've done is to solve them by parts:

1. |x+2| = x-2...
Click to expand...
In no way is this the correct method for solving this sort of equation! :shock:

You cannot solve "x^2 + 3 = x" by solving "x^2 = x or 3 = x"; so also, you cannot solve this summed-term equation by setting individual terms from one side "equal" to the other side. Instead, try dealing with the boundaries set up by the absolute values (like here).

You have these absolute-value terms:

. . . . .\(\displaystyle \lvert\, x\, +\, 2\, \rvert\)

. . . . .\(\displaystyle \lvert\, x\, +\, 1\, \rvert\)

. . . . .\(\displaystyle \lvert\, x\, -\, 1\, \rvert\)

. . . . .\(\displaystyle \lvert\, x\, -\, 2\, \rvert\)

You know, from working with absolute values, that:

. . . . .\(\displaystyle \lvert\, y\, \rvert\, =\, \begin{cases}\, y & \mbox{ for }\, y\, \geq\, 0 \\ -y &\, \mbox{ for }\, y\, <\, 0\end{cases}\)

This gives us:

. . . . .\(\displaystyle \lvert\, x\, +\, 2\, \rvert\, =\, \begin{cases} x\, +\, 2 & \mbox{ for }\, x\, \geq\, -2 \\ -x\, -\, 2 & \mbox{ for }\, x\, <\, -2 \end{cases}\)

. . . . .\(\displaystyle \lvert\, x\, +\, 1\, \rvert\, =\, \begin{cases} x\, +\, 1 & \mbox{ for }\, x\, \geq\, -1 \\ -x\, -\, 1 & \mbox{ for }\, x\, <\, -1 \end{cases}\)

. . . . .\(\displaystyle \lvert\, x\, -\, 1\, \rvert\, =\, \begin{cases} x\, -\, 1 & \mbox{ for }\, x\, \geq\, 1 \\ -x\, +\, 1 & \mbox{ for }\, x\, <\, 1 \end{cases}\)

. . . . .\(\displaystyle \lvert\, x\, -\, 2\, \rvert\, =\, \begin{cases} x\, -\, 2 & \mbox{ for }\, x\, \geq\, 2 \\ -x\, +\, 2 & \mbox{ for }\, x\, <\, 2 \end{cases}\)

The breakpoints of each of the individual absolute values split the number line into the following intervals:

. . . . .\(\displaystyle (-\infty,\, -2),\, [-2,\, -1),\, [-1,\, 1),\, [1,\, 2),\, \mbox{ and }\, [2,\, \infty)\)

Now consider each interval by turn. For instance, on the first interval, each of the absolute values evaluates to the negative (the "-y" for the "|y|"), so the original equation can be restated as:

. . . . .\(\displaystyle -(x\, +\, 2)\, -\, 2\, \left(-(x\, +\, 1)\right)\, +\, 2\, \left(-(x\, -\, 1)\right)\, -\, 2\, \left(-(x\, -\, 2)\right)\, =\, x\, -\, 2\)

In this manner, all of the absolute-value bars have been removed and, on the interval for "everything less than negative two", we now have a true equation. Note that this means that any solution we get for this interval's form of the equation must then also be "less than negative two". If, say, we get a solution of "x = 3", then we actually have "no solution" on this particular interval. Solving, we get:

. . . . .\(\displaystyle -x\, -\, 2\, -\, 2\, (-x\, -\, 1)\, +\, 2\, (-x\, +\, 1)\, -\, 2\, (-x\, +\, 2)\, =\, x\, -\, 2\)

. . . . .\(\displaystyle -x\, -\, 2\, +\, 2x\, +\, 2\, -\, 2x\, +\, 2\, +\, 2x\, -\, 4\, =\, x\, -\, 2\)

. . . . .\(\displaystyle -x\, +\, 2x\, -\, 2x\, +\, 2x\, -\, 2\, +\, 2\, +\, 2\, -\, 4\, =\, x\, -\, 2\)

. . . . .\(\displaystyle x\, -\, 2\, =\, x\, -\, 2\)

. . . . .\(\displaystyle x\, -\, x\, -\, 2\, +\, 2\, =\, 0\)

. . . . .\(\displaystyle 0\, =\, 0\)

In this (highly unusual) case, the solution is "all x less than negative two" (that is, all the x-values within the interval).

Now you do the other intervals. You can check your answers by looking at a graph. ;)
 
Subhotosh Khan said:
In my opinion, the best way to do these problems is through graphing

Plot

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

in your graphing calculator and observe the behavior of the function. That might give you a clue as to the path to solve the problem.
Click to expand...


Thank you for your tip!
 
stapel said:
In no way is this the correct method for solving this sort of equation! :shock:

.....

In this manner, all of the absolute-value bars have been removed and, on the interval for "everything less than negative two", we now have a true equation. Note that this means that any solution we get for this interval's form of the equation must then also be "less than negative two". If, say, we get a solution of "x = 3", then we actually have "no solution" on this particular interval. Solving, we get:

. . . . .\(\displaystyle -x\, -\, 2\, -\, 2\, (-x\, -\, 1)\, +\, 2\, (-x\, +\, 1)\, -\, 2\, (-x\, +\, 2)\, =\, x\, -\, 2\)

. . . . .\(\displaystyle -x\, -\, 2\, +\, 2x\, +\, 2\, -\, 2x\, +\, 2\, +\, 2x\, -\, 4\, =\, x\, -\, 2\)

. . . . .\(\displaystyle -x\, +\, 2x\, -\, 2x\, +\, 2x\, -\, 2\, +\, 2\, +\, 2\, -\, 4\, =\, x\, -\, 2\)

. . . . .\(\displaystyle x\, -\, 2\, =\, x\, -\, 2\)

. . . . .\(\displaystyle x\, -\, x\, -\, 2\, +\, 2\, =\, 0\)

. . . . .\(\displaystyle 0\, =\, 0\)

In this (highly unusual) case, the solution is "all x less than negative two" (that is, all the x-values within the interval).

Now you do the other intervals. You can check your answers by looking at a graph. ;)
Click to expand...


Firstly, thanks so much for putting in your time to help me! :)
So, even if we get 0 =0, the solution of "all x less than negative two" is a valid solution? Shouldn't we need an x in the final line to be a valid solution? I.e. x = 0 or x = 5?
 
icandoit said:
..even if we get 0 =0, the solution of "all x less than negative two" is a valid solution?
Click to expand...
For what value of x is "zero equals zero" false? If none, then the solution is "all x on (whichever interval you're working in)". ;)
 
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