SOLUTION: Solve for real values of x. absolutevalue(x-3) = 3-x I know to set the absolute value up to equal a positive and negative possiblity and so that makes x-3 = 3-x and x-3 = -

Algebra ->  Absolute-value -> SOLUTION: Solve for real values of x. absolutevalue(x-3) = 3-x I know to set the absolute value up to equal a positive and negative possiblity and so that makes x-3 = 3-x and x-3 = -      Log On


   

Question 204633: Solve for real values of x.
absolutevalue(x-3) = 3-x
I know to set the absolute value up to equal a positive and negative possiblity and so that makes
x-3 = 3-x and x-3 = -3+x
For the first I got x = 3 but the second I got 0 = 0 and that's where I am confused. Did I do something wrong?
Found 2 solutions by jim_thompson5910, Edwin McCravy:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
You have the right answers. What this means is that the right side 3-x is the EXACT same as the negative half of the absolute value (go ahead and graph to confirm). So this means that if you plug in any number less than or equal 3, you'll get 0=0. So this means that the solution set is the set of all x values less than or equal 3.


Answer by Edwin McCravy(20086) About Me  (Show Source):
You can put this solution on YOUR website!

abs%28x-3%29+=+3-x

Let's use the principle abs%28A%29=+abs%28-A%29, to replace
the left side:

abs%28-%28x-3%29%29+=+3-x

abs%28-x%2B3%29+=+3-x

abs%283-x%29+=+3-x

Now we use the principle that abs%28A%29=A if and only if M%3E=0

Therefore abs%283-x%29+=+3-x

is true  if and only if  

3-x%3E=0

Subtract 3 from both sides:

-x%3E=-3

Multiply through by -1, which reverses the inequality:

x%3C=3

Or in interval notation %22%28%22-infinity%22%2C+3%5D%22%22%2C3%5D%22

The equation has infinitely many solutions.

Edwin