Expert: Scotto - 2/5/2009

Question
Hi

In the past week my teacher started teaching Absolute values to the class. me being a slow learner didn't get the grip of it. could you help me please.
I was wondering if you could help me answer this question and show exactly how you did it and also how to graph it.
Abs(x-5) + Abs(x-1)=7
also if it were possible could you please tell me any tricks you know in making it easier to understand these Absolute value problems.

Answer
Note that when variables (the x value) occur in | |, you need to
see where the value in the | | is 0.  Once this has been found,
the value above and below it need to be checked.

The equation is |x-5| + |x-1| = 7.
The critical points are 5 (from |x-5|)and 1 (form |x-1|).

Case 1: x<1
Case 2: 1<x<5
Case 3: 5<x

Case 1: x-5 and x-7 are negative, so they are really 5-x and 7-x
to be positive.  Taking 5-x + 7-x = 7 gives 12-2x = 7.  Subtracting
12 from both sides gives -2x = -5, so that says that x = -5/2.
To see if this works, we have to check the start, x<1.  We know
that -5/2 < 1, so that's a good value.

Case 2: Here, only the x-5 is negative, so what we is
x-5 + 7-x = 7.  Adding all of the terms gives 2 = 7, and that's
never true, so case 2 can never be satisfied, so x can't be between
1 and 5.

Case 3: Here, both terms are positive, so we have
x-5 + x-1 = 7.  
That turns into 2x - 6 = 7, or 2x = 13, so x = 6.5.  
For this case, we needed 5<x, which is true, so x = 6.5 works.

From what was just done, the only two cases are x = -2/5 and
x = 6.5.  Either of these would make that equation true.

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Side Note:

The major thing to note is sometimes the solution doesn't fit
the assumption.  For example, suppose we had some problem with
a case where we knew that x > 3.  After working and solving the
equation, we get x=1.  In this case as well, there would be no
solution since the case stated that x had to be greater than 3.
The answer we got, x=1, doesn't meet this requirement.