How do you divide #{(16-2m)/(m^2+2m-24)}/{(m-8)/(3m+18)}#?

Question

1232 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-29T06:13:46

#(6(8-m))/(m-4)#

To solve #((16−2m)/(m^2+2m−24))/((m−8)/(3m+18))#, first factorize each of the algebraic expression.

#16−2m=2(8-m)=-2(m-8)# and #3m+18=3(m+6)#

#m^2+2m−24=m^2+6m-4m−24#

= #m(m+6)-4(m+6)#=#(m-4)(m+6)#

Now #((16−2m)/(m^2+2m−24))/((m−8)/(3m+18))#

(as #(a/b)/(c/d)-(ad)/(bc)#)

= #((16−2m)/(m^2+2m−24))xx((3m+18)/(m−8))# or

= #(-2(m-8))/((m-4)(m+6))xx(3(m+6))/(m−8)# or

= #(-6(m-8))/(m-4)#

= #(6(8-m))/(m-4)#