How do you solve #4 /( y-1) + 2/3 = 6 /( y-1)#? Algebra Rational Equations and Functions Clearing Denominators in Rational Equations 1 Answer Roopali Apr 15, 2018 #y=cancel1, y=4# Explanation: common denominator #(3/3)(4/(y-1)) + ((y-1)/(y-1))(2/3) = (3/3)(6/(y-1))# #12/(3(y-1)) + (2(y-1))/(3(y-1)) = 18/(3(y-1))# #(12+2y-2)/(3(y-1)) = 18/(3(y-1))# #(2y + 10)/(3(y-1)) = 18/(3(y-1))# #(2y+10)(3y-3) = 18(3y-3)# #6y^2 + 24y -30 = 54y - 54# #6y^2 +24y- 54y -30 +54 = 0# #6y^2 - 30y +24 = 0# #(3y-3)(2y-8)# #y=1, y=4# 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 1705 views around the world You can reuse this answer Creative Commons License