How do you solve the rational equation #(y+2)/y = 1/(y-5)#? Algebra Rational Equations and Functions Clearing Denominators in Rational Equations 1 Answer Lucy May 20, 2018 #y=2+-sqrt14# Explanation: #(y+2)/y = 1/(y-5)# #(y+2)(y-5) = y# #y^2-3y-10=y# #y^2-4y-10=0# Using quadratic formula, #y=(-b+-sqrt(b^2-4ac))/(2a)# #y=(4+-sqrt((-4)^2-4(1)(-10)))/(2times1)# #y=(4+-sqrt(16+40))/2# #y=(4+-sqrt56)/2# #y=(4+-2sqrt14)/2# #y=2+-sqrt14# 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 1675 views around the world You can reuse this answer Creative Commons License