How do you multiply #(1/(m^2-m)) + (1/m)=5/(m^2-m)#?

Question

6447 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-11T21:49:03

#m = 5#

the common denominator between #m^2 - m# and #m# would be #m^2 - m#, since #m^2 - m# is a multiple of #m#.

#m^2 - m = m(m - 1)#

#1/m = (1(m-1))/(m(m-1))#

#= (m - 1)/(m^2-m)#

#1/(m^2-m) + (m-1)/(m^2-m) = (1 + m - 1)/(m^2-m)#

#= m / (m^2 - m)#

#m / (m^2-m) = 5 / (m^2 - m)#

hence, #m = 5#