How do you solve #x/6 - (3x)/2 = 16/3#? Algebra Rational Equations and Functions Clearing Denominators in Rational Equations 1 Answer Suren Abreu Jan 5, 2017 #x=-4# Explanation: #x/6-(3x)/2=16/3# Multiply all terms by #6#. #(6xxx/6)-(6xx(3x)/2)=6xx16/3# #(cancel6xxx/cancel6)-(3cancel6xx(3x)/cancel2)=2cancel6xx16/cancel3# #x-9x=32# #-8x=32# Divide both sides by #8#. #-x=4# or #x=-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 1616 views around the world You can reuse this answer Creative Commons License