Question 1199376
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You mentioned
"b=p-2l/2 is the same as b=p/2-l"
which I'm assuming you are saying
{{{b = (p-2L)/2}}} is the same as {{{b = p/2 - L}}}


This claim is true because of these steps
{{{b = (p-2L)/2}}}


{{{b = p/2-(2L)/2}}} Break up the fraction


{{{b = p/2 - L}}} The '2's cancel for the second fraction.


Similar cancellations occur for the other equations.


Eg:
To go from {{{L = (2s-an)/n}}} to {{{L = (2s)/n - a}}}, the 'n's cancel in the 2nd fraction.


To go from {{{d = (c-16e)/16}}} to {{{d = c/16 - e}}}, the '16's cancel in the 2nd fraction.


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Side note: 
when wanting to divide all of something like p-2L over 2, then you should use parenthesis.
You should say (p-2L)/2 instead of p-2L/2


p-2L/2 means {{{p - (2L)/2}}}


(p-2L)/2 means {{{(p - 2L)/2}}}
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