Question 1197271
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x2+xy all over x2-y2
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<pre>
   {{{(x^2+xy)/(x^2-y^2)}}}.


Factor the numerator    {{{x^2+xy)}}} = x*(x+y).

Factor the denominator  {{{x^2-y^2)}}} =(x+y)*(x-y).


Therefore

    {{{(x^2+xy)/(x^2-y^2)}}} = {{{(x*(x+y))/((x+y)*(x-y))}}}.


Cancel common factors (x+y) in the numerator and denominator

    {{{(x^2+xy)/(x^2-y^2)}}} = {{{(x*(x+y))/((x+y)*(x-y))}}} = {{{x/(x-y)}}}.


<U>ANSWER</U>.  {{{(x^2+xy)/(x^2-y^2)}}} = {{{x/(x-y)}}}, as algebraic expressions.
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Solved and explained, with full step by step explanations.