when is absolute(x-4) equal to 4-x ?

[san2009]

Can someone please share the process/method of solving an absolute equation of this sort?
Q: when is absoluted(x-4) equal to 4-x ?
I realize that one can try positive/negative/fractional values and see what fits the criteria, but I am hoping for a more methodical approach that shows the underlying logic.

[liferocks]

|x|= x when x>0
=0 when x=0
=-x when x<0

applying this in your question we get |x-4|=4-x or -(x-4)
this is the case when (x-4)<0 or x<4

[san2009]

|x|= x when x>0
=0 when x=0
=-x when x<0

applying this in your question we get |x-4|=4-x or -(x-4)

I get the process up till the above line
but then how do you figure out without testing numbers, that x<4?
btw, answer is x<or equal to 4

[liferocks]

x<4 because x-4 <0
this comes directly from the definition that if (x-4)<0 |(x-4)|=-(x-4)

as for the x=4..it will be true as both x-4 and 4-x will be zero .I didn't included that in the ans as I thought you are looking for the only the sign conversion part .Let me know if this clarifies.

[kevincanspain]

Can someone please share the process/method of solving an absolute equation of this sort?
Q: when is absoluted(x-4) equal to 4-x ?
I realize that one can try positive/negative/fractional values and see what fits the criteria, but I am hoping for a more methodical approach that shows the underlying logic.

You can think of |x - y| as the distance between x and y.

Note that the distance between two numbers is always the greater number minus the lesser number. For example, the distance between - 3 and 7 is 7 -(-3) = 10

Thus |x - y| is equal to y - x if and only if y >=x

It is helpful to realize that |x + 6| is the distance between x and - 6 (since x + 6 = x - (-6) ) and that |x| is the distance between x and 0

Also |x - y | = | y - x| for all real numbers x and y
x^2 = y^2 if and only if |x| = |y| i.e. x=y or x= - y

Keep in mind that |x + y| = |x| + |y| if and only if x and y do not have opposite signs.

Additionally, sqrt (x^2) =|x| for all real values of x. Thus sqrt(x^2) = -x if x < 0









Th

[rishimaharaj]

Hello Kevin,

I'm working through the Numbers and Properties book and am trying to wrap my head around some of the absolute value rules and number properties.

Note that the distance between two numbers is always the greater number minus the lesser number. For example, the distance between - 3 and 7 is 7 -(-3) = 10

This leads me to the assumption that all distances between integers on a number line will be represented by a positive number:
Distance between -2 and -8 is -2 - (-8) = -2 + 8 = 6.
Distance between 3 and -8 is 3 - (-8) = 3 + 8 = 11.

I know in Physics, negative distances are possible. Are there any cases where this would be the same with distances on a number line?

It is helpful to realize that |x + 6| is the distance between x and - 6 (since x + 6 = x - (-6) ) and that |x| is the distance between x and 0

Does this mean that |x + 6| is the distance between x and -6 only?
Or does this mean that |x + 6| is the both the distance between x and 6 as well as x and -6?

Thank you for your insights!
--Rishi

[Frankenstein]

Does this mean that |x + 6| is the distance between x and -6 only?
Or does this mean that |x + 6| is the both the distance between x and 6 as well as x and -6?
--Rishi

Hi,
|x+6| = |x-(-6)| or |-6 - x|
This always means distance between x and -6 only and not between x and 6.

[rishimaharaj]

Thank you Frankenstein for the clarification.

Just looking at it was puzzling, so I plugged in the value of 8 for x, which helped me visualize what you said:

If x = 8, then |x + 6| = |8 + 6| = |14| = 14.
If the distance is 14, then it cannot be that x = 8 and 6, because that distance on the number line is 2.

<==-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8==>
______________________________|--2--|


So the value must be -6:
<==-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8==>
_________|----------------14-----------------|

|x -(-6)| = |8 -(-6)| = |8 + 6| = |14| = 14
and
|-6 - x| = |-6 - 8| = |-14| = 14

Thanks again!
--Rishi