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

|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

|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

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.

san2009 wrote: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

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.

kevincanspain wrote: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?

kevincanspain wrote: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

rishimaharaj wrote: 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.

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