How do you solve #(3x)/(x+1)+6/(2x)=7/x#? Algebra Rational Equations and Functions Clearing Denominators in Rational Equations 1 Answer Shwetank Mauria Oct 1, 2016 #x=-2/3# or #x=2# Explanation: #(3x)/(x+1)+6/(2x)=7/x# can be rewritten as #(3x)/(x+1)=7/x-6/(2x)# or #(3x)/(x+1)=7/x-3/x=4/x# Hence, #(3x)/((x+1))xx x(x+1)=4/x xx x(x+1)# or #(3x)/(cancel(x+1))xx xcancel((x+1))=4/cancelx xx cancelx(x+1)# or #3x^2=4x+4# or #3x^2-4x-4=0# or #3x^2-6x+2x-4=0# or #3x(x-2)+2(x-2)=0# or #(3x+2)(x-2)=0# Hence either #3x+2=0# i.e. #x=-2/3# or #x-2=0# i.e. #x=2# 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 2695 views around the world You can reuse this answer Creative Commons License