Math question

Thanks Alan,  I don't disagree with your answer but I don't understand why you introduced 'a'.

I'm going to do it my way. Plus I am experimenting with LaTex.  I wrote this using texmaker and then tried to copy it straight over into the forum.  It didn't work.  I used begin{align}and also the begin{equarray} and I don't think either of these are recognised commands in the forum. So I have copied the output to a .jpg file and then uploaded it using Tiny pic.  This worked okay.  

Lots of experimenting happening here. 

Actually, I am very familiar with equarray but I have not used align very much.  I would have preferred it if the output was closer together.  I wonder if Rom can help me with this.  I think he is our most capable user of LaTex.

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Let ax be -x, so ax must be positive, given that x is negative.  So we have l-ax-3l = 5 or l-(ax+3)l = 5

The term in brackets is positive, and the absolute value operator means that the minus sign vanishes, so we are left with ax+3 = 5, which means that ax = 2.  This in turn means that x = -2.  

Similarly, let ay be -y. Then we have l-ay+7l = 15 or l-(ay-7)l = 15 (and ay must be larger than 7), so ay-7 = 15 , so ay = 22.  This means y = -22.

Therefore x - y = -2 - (-22) = -2 + 22 = 20

I think better with positive values!!  My 'a' isn't meant to be a multiplier, I just used ax as a two letter name for -x.

However, I think your method is much clearer!

Thanks Alan