Question 922825
if {{{ax^2-bx=ay^2+by}}},then {{{a/b}}}=: move terms with {{{a}}} to one side of equation and terms with {{{b}}} to the other side of equation 

{{{ax^2-ay^2=bx+by}}}.........factor out {{{a}}} and {{{b}}}


{{{a(x^2-y^2)=b(x+y)}}} ...both sides divide by {{{b}}}

{{{a(x^2-y^2)/b=b(x+y)/b}}}

{{{a(x^2-y^2)/b=(x+y)}}}...both sides divide by {{{(x^2-y^2)}}}

{{{a(x^2-y^2)/(b(x^2-y^2))=(x+y)/(x^2-y^2)}}}

{{{a*cross((x^2-y^2))/(b*cross((x^2-y^2)))=(x+y)/(x^2-y^2)}}}

{{{a/b=(x+y)/(x^2-y^2)}}}...since {{{(x^2-y^2)=(x-y)(x+y)}}} we have


{{{a/b=(x+y)/((x-y)(x+y))}}} ...simplify


{{{a/b=cross((x+y))1/((x-y)cross((x+y)))}}}


{{{a/b=1/(x-y)}}}

so, your answer is: (A) {{{1/(x-y)}}}