How do you multiply #(1/(m^2-m)) + (1/m)=5/(m^2-m)#? Algebra Rational Equations and Functions Clearing Denominators in Rational Equations 1 Answer LM Apr 11, 2018 #m = 5# Explanation: 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# Answer link Related questions What is Clearing Denominators in Rational Equations? How do you solve rational expressions by multiplying by the least common multiple? How do you solve #5x-\frac{1}{x}=4#? How do you solve #-3 + \frac{1}{x+1}=\frac{2}{x}# by finding the least common multiple? What is the least common multiple for #\frac{x}{x-2}+\frac{x}{x+3}=\frac{1}{x^2+x-6}# and how do... How do you solve #\frac{x}{x^2-36}+\frac{1}{x-6}=\frac{1}{x+6}#? How do you solve by clearing the denominator of #3/x+2/x^2=4#? How do you solve #2/(x^2+2x+1)-3/(x+1)=4#? How do you solve equations with rational expressions #1/x+2/x=10#? How do you solve for y in #(y+5)/ 2 - y/3 =1#? See all questions in Clearing Denominators in Rational Equations Impact of this question 6262 views around the world You can reuse this answer Creative Commons License