Question 252593


{{{((3x+6)/(4x+12))/((x^2-4)/(x+3))}}} Start with the given expression.



{{{((3x+6)/(4x+12))((x+3)/(x^2-4))}}} Multiply the first fraction {{{(3x+6)/(4x+12)}}} by the reciprocal of the second fraction {{{(x^2-4)/(x+3)}}}.



{{{((3(x+2))/(4x+12))((x+3)/(x^2-4))}}} Factor {{{3x+6}}} to get {{{3(x+2)}}}.



{{{((3(x+2))/(4(x+3)))((x+3)/(x^2-4))}}} Factor {{{4x+12}}} to get {{{4(x+3)}}}.



{{{((3(x+2))/(4(x+3)))((x+3)/((x-2)(x+2)))}}} Factor {{{x^2-4}}} to get {{{(x-2)*(x+2)}}}.



{{{(3(x+2)(x+3))/(4(x+3)(x-2)(x+2))}}} Combine the fractions. 



{{{(3*highlight((x+2))highlight((x+3)))/(4*highlight((x+3))(x-2)highlight((x+2)))}}} Highlight the common terms. 



{{{(3*cross((x+2))cross((x+3)))/(4*cross((x+3))(x-2)cross((x+2)))}}} Cancel out the common terms. 



{{{3/(4(x-2))}}} Simplify. 



{{{3/(4x-8)}}} Distribute.



So {{{((3x+6)/(4x+12))/((x^2-4)/(x+3))}}} simplifies to {{{3/(4x-8)}}}.



In other words, {{{((3x+6)/(4x+12))/((x^2-4)/(x+3))=3/(4x-8)}}}