Question 154270
# 1


{{{((24x^3)/(6y^2))/((8xz^4)/(3y^4))}}} Start with the given expression



{{{((24x^3)/(6y^2))*((3y^4)/(8xz^4))}}} Multiply the first fraction by the reciprocal of the second fraction.



{{{((24x^3)(3y^4))/((6y^2)(8xz^4))}}} Combine the fractions.



{{{(72x^3y^4)/(48xy^2z^4)}}} Multiply



{{{(24*3*x*x*x*y*y*y*y)/(24*2*x*y*y*z*z*z*z)}}} Factor. In this case, {{{72x^3y^4=24*3*x*x*x*y*y*y*y}}} and {{{48xy^2z^4=24*2*x*y*y*z*z*z*z}}}



{{{(highlight(24)*3*highlight(x)*x*x*highlight(y)*highlight(y)*y*y)/(highlight(24)*2*highlight(x)*highlight(y)*highlight(y)*z*z*z*z)}}} Highlight the common terms.



{{{(cross(24)*3*cross(x)*x*x*cross(y)*cross(y)*y*y)/(cross(24)*2*cross(x)*cross(y)*cross(y)*z*z*z*z)}}} Cancel out the common terms.



{{{(3*x*x*y*y)/(2*z*z*z*z)}}} Simplify.



{{{(3*x^2*y^2)/(2*z^4)}}} Regroup.



So {{{(72x^3y^4)/(48xy^2z^4)}}} simplifies to {{{(3*x^2*y^2)/(2*z^4)}}}.



In other words, {{{(72x^3y^4)/(48xy^2z^4)=(3*x^2*y^2)/(2*z^4)}}} where {{{x<>0}}}, {{{y<>0}}}, and {{{z<>0}}}



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# 2




{{{((z^2-10z+21)/(z^2-13z+36))/((z^2-11z+24)/(z^2-9z+14))}}} Start with the given expression.



{{{((z^2-10z+21)/(z^2-13z+36))((z^2-9z+14)/(z^2-11z+24))}}} Multiply the first fraction {{{(z^2-10z+21)/(z^2-13z+36)}}} by the reciprocal of the second fraction {{{(z^2-11z+24)/(z^2-9z+14)}}}.



{{{(((z-3)*(z-7))/(z^2-13z+36))((z^2-9z+14)/(z^2-11z+24))}}} Factor {{{z^2-10z+21}}} to get {{{(z-3)*(z-7)}}}.



{{{(((z-3)*(z-7))/((z-4)*(z-9)))((z^2-9z+14)/(z^2-11z+24))}}} Factor {{{z^2-13z+36}}} to get {{{(z-4)*(z-9)}}}.



{{{(((z-3)*(z-7))/((z-4)*(z-9)))(((z-2)*(z-7))/(z^2-11z+24))}}} Factor {{{z^2-9z+14}}} to get {{{(z-2)*(z-7)}}}.



{{{(((z-3)*(z-7))/((z-4)*(z-9)))(((z-2)*(z-7))/((z-3)*(z-8)))}}} Factor {{{z^2-11z+24}}} to get {{{(z-3)*(z-8)}}}.



{{{((z-3)*(z-7)(z-2)*(z-7))/((z-4)*(z-9)(z-3)*(z-8))}}} Combine the fractions. 



{{{(highlight((z-3))(z-7)(z-2)(z-7))/((z-4)(z-9)highlight((z-3))(z-8))}}} Highlight the common terms. 



{{{(cross((z-3))(z-7)(z-2)(z-7))/((z-4)(z-9)cross((z-3))(z-8))}}} Cancel out the common terms. 



{{{((z-7)(z-2)(z-7))/((z-4)(z-9)(z-8))}}} Simplify. 



{{{((z-2)(z-7)^2)/((z-4)(z-9)(z-8))}}} Regroup the terms. 



So {{{((z^2-10z+21)/(z^2-13z+36))/((z^2-11z+24)/(z^2-9z+14))}}} simplifies to {{{((z-2)(z-7)^2)/((z-4)(z-9)(z-8))}}}.



In other words, {{{((z^2-10z+21)/(z^2-13z+36))/((z^2-11z+24)/(z^2-9z+14))=((z-2)(z-7)^2)/((z-4)(z-9)(z-8))}}} where {{{z<>3}}}, {{{z<>4}}}, {{{z<>8}}}, or {{{z<>9}}}