adavydov7 wrote:
Bunuel, thanks for the explanation and the links. Between the two I think I pieced together the complete correct answer (I think the one you provide is correct but not entirely complete). I will outline the complete answer below (as I understand) and highlight what was missing from your explanation in bold so others can see how to arrive at the answer rigorously. If I go wrong anywhere or my understanding is incorrect please be sure to correct me.
abs(x-4)=4-x
Step 1: Set conditions (consequently the step I was skipping when I first did this problem)
a) x-4>=0
b) x-4<0 (you state that this should be x-4<=0 and state that this is where the answer comes from, in actuality this condition is strictly < not <=, in order to get the = part we need to solve (a) as I do below in Step 2 and ensure that it satisfies the condition set out above in Step 1a as I do in Step 3a)
Step 2: Solve the original equations
a) x-4=4-x solves to x=4
b) -(x-4)=4-x solves to 0=0
Step 3: Check the solutions to the equations in Step 2 against their respective conditions in Step 1
a) plug x=4 into x-4>=0 which yields 0>=0 which is TRUE hence 4 is a solution to the problem, i.e. x=4
b) because the equation in Step 2b completely cancels out we are left the with the condition from Step 1b as the solution, i.e. x-4<0 or x<4
Combining the possible solutions from Step 3 for both (a) (x=4) and (b) (x<4) (since we have verified that both ARE solutions to the original problem by checking them against the conditions set in Step 1) we come up with the answer to the problem, or x<=4.
I hope you see what I mean when I say that your method wasn't entirely complete. I just want to be very rigorous in solving these is all. However, if my approach is incorrect please do let me know as I will have to go back and re-read the rules on solving these things to make sure I have it down.
Thanks again!
It seems that you need to brush up fundamentals on absolute value. So, I do advice to follow the links in my previous posts for that.
As for the solution: it's correct. \(|x-4|=4-x\) to hold true LHS
must not be positive which means: \(x-4\leq{0}\) --> \(x\leq{4}\). Exactly as I wrote.
Usually when you expand absolute value you should check when the expression in it is <=0 and >0 (you can put = sign for either of case). But x-4>0 can not hold true as in this case RHS=4-x<0 and we would have that LHS=|x-4|<0 which is cannot possibly be correct as absolute value is always non-negative.
Hope it's clear.
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20 Jan 2012, 14:01
