Question 133358
{{{((x^2-9*y^2)/(7*x^2-21*y))/(x^2+3*x*y)}}} Start with the given expression




{{{((x^2-9*y^2)/(7*x^2-21*y))*(1/(x^2+3*x*y))}}} Multiply the first fraction by the reciprocal of the second fraction




{{{(x^2-9*y^2)/((7*x^2-21*y)*(x^2+3*x*y))}}} Combine the fractions



{{{((x+3y)(x-3y))/((7*x^2-21*y)*(x^2+3*x*y))}}} Factor {{{x^2-9*y^2}}} to get {{{(x+3y)(x-3y)}}}




{{{((x+3y)(x-3y))/(7(x^2-3y)*(x^2+3*x*y))}}} Factor out the GCF 7 from {{{7*x^2-21*y}}}



{{{((x+3y)(x-3y))/(7x(x^2-3y)*(x+3y))}}} Factor out the GCF x from {{{x^2+3*x*y}}}



{{{(cross((x+3y))(x-3y))/(7x(x^2-3y)*cross((x+3y)))}}} Cancel like terms




{{{(x-3y)/(7x(x^2-3y))}}} Simplify



-----------------------------------------------------------

Answer:



So {{{((x^2-9*y^2)/(7*x^2-21*y))/(x^2+3*x*y)}}} simplifies to {{{(x-3y)/(7x(x^2-3y))}}}