Absolute value equation

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Albi

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[MATH] \left | x-4 \right | - \left | x+1 \right | = -5[/MATH]
Guys I solved this equation, and x = 4, but it turns out that the equation works for [MATH]x\geq 4[/MATH] , can someone explain to me how is this possible?
 
Albi said:
[MATH] \left | x-4 \right | - \left | x+1 \right | = -5[/MATH]Guys I solved this equation, and x = 4, but it turns out that the equation works for [MATH]x\geq 4[/MATH] , can someone explain to me how is this possible?
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Write it as \(|x-4|+5=|x+1|\) If \(x\ge 4\) we have \((x-4)+5=(x+1)\text{ OR }1=1\) that is correct.
What if \(x<4~?\)
 
Albi said:
[MATH] \left | x-4 \right | - \left | x+1 \right | = -5[/MATH]
Guys I solved this equation, and x = 4, but it turns out that the equation works for [MATH]x\geq 4[/MATH] , can someone explain to me how is this possible?
Click to expand...
Are you asking "how can you arrive at the conclusion x≥ 1"?
 
Albi said:
[MATH] \left | x-4 \right | - \left | x+1 \right | = -5[/MATH]
Guys I solved this equation, and x = 4, but it turns out that the equation works for [MATH]x\geq 4[/MATH] , can someone explain to me how is this possible?
Click to expand...
Please show us how you solved it and got only x = 4. We can't know how you got it wrong without seeing how you got it.

But pka showed part of the work in a casewise method, namely the part that gives the result you are asking about. Just to take an example, if [MATH]x = 100[/MATH], then [MATH]|x-4|-|x+1| = |96| - |101| = 96 - 101 = -5[/MATH], as required. The same happens for any [MATH]x \ge 4[/MATH].
 
Dr.Peterson said:
Please show us how you solved it and got only x = 4. We can't know how you got it wrong without seeing how you got it.

But pka showed part of the work in a casewise method, namely the part that gives the result you are asking about. Just to take an example, if [MATH]x = 100[/MATH], then [MATH]|x-4|-|x+1| = |96| - |101| = 96 - 101 = -5[/MATH], as required. The same happens for any [MATH]x \ge 4[/MATH].
Click to expand...
I solved it by cases
 

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Albi said:
I solved it by cases
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Then what is your question? Your work shows that \(x\ge 4\).

In the second case, \(-1\le x \lt 4\), you found that \(x=4\), which is not in that case as I interpret it; in the third case, \(x\ge 4\), you found that the equation is always true (since it is equivalent to \(-5=-5\)).

(Incidentally, your list of cases as written seems to omit -1 and 4 themselves; you should make that clearer. And what is that symbol that apparently means infinity?)
 
Dr.Peterson said:
Then what is your question? Your work shows that \(x\ge 4\).

In the second case, \(-1\le x \lt 4\), you found that \(x=4\), which is not in that case as I interpret it; in the third case, \(x\ge 4\), you found that the equation is always true (since it is equivalent to \(-5=-5\)).

(Incidentally, your list of cases as written seems to omit -1 and 4 themselves; you should make that clearer. And what is that symbol that apparently means infinity?)
Click to expand...
I've been used to writing it like that, but I'll try to write it better from now on.
Anyways thank you for the explanation because I was confused with the third case and didn't know whether to include it in my solution or not
 
Correct.
Case 1:
If x≥4 then
|x-4|-|x+1|
=(x-4)-(x+1)
=-5
So the equation |x+4|-|x+1|=-5 is satisfied by all x≥4

Case 2:
If -1<x<4 then
|x-4|-|x+1|
=-(x-4)-(x+1)
=-2x+3
Now -2x+3=-5 is satisfied by x=4 alone.

So the equation |x+4|-|x+1|=-5 is satisfied by no x: -1<x<4

Case 3:
If x≤-1 then
|x-4|-|x+1|
=-(x-4)+(x+1)
=5

So the equation |x+4|-|x+1|=-5 is satisfied by no x: x≤-1

The only solutions of |x+4|-|x+1|=-5 are values of x: x≥4
 
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