Question 972471
{{{(x^2-3x-9)/(x-9)}}} + {{{(x^2-7x+27)/(9-x)}}}
right now the denominators are not the same but we can make them the same
Factor out -1 from the 2nd denominator
{{{(x^2-3x-9)/(x-9)}}} + {{{(x^2-7x+27)/(-1(x-9))}}} 
dividing -1 into the numerator, changes the sign so we have
{{{(x^2-3x-9)/(x-9)}}} - {{{(x^2-7x+27)/(x-9)}}}
now we can write it:
{{{((x^2-3x-9)-(x^2-7x+27))/(x-9)}}}
removing the brackets changes the signs after the minus
{{{(x^2-3x-9-x^2+7x-27)/(x-9)}}}
combine like terms
{{{(x^2-x^2-3x+7x-9-27)/(x-9)}}}
now we have
{{{(4x-36)/(x-9)}}}
factor 4 out in the numerator, (x-9) cancels
{{{(4(x-9))/(x-9)}}} = 4