Question 113236
p^2-3p+pq-3q,,,,,,,,,,,p-3
___________ Divided by_____
11p^2-11q^2,,,,,,,,,,,8p-8q 
:
Assume this is the problem:
{{{((p^2-3p+pq-3q))/((11p^2-11q^2))}}}
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{{{((p-3))/((8p-8q))}}}
:
You know that when you divide fractions, you invert the dividing fraction & multiply
{{{((p^2-3p+pq-3q))/((11p^2-11q^2))}}} * {{{((8p-8q))/((p-3))}}}
:
Factor where we can:
{{{(p(p-3)+q(p-3))/(11(p^2-q^2))}}} * {{{(8(p-q))/((p-3))}}}
:
Factor some more
{{{((p-3)+(p+q))/(11(p-q)(p+q))}}} * {{{(8(p-q))/((p-3))}}}
:
Note how everything cancels except {{{8/11}}}