For what #x# is #(3x)/(x+3)>3/(x-2)#? Algebra Linear Inequalities and Absolute Value Inequalities with Multiplication and Division 1 Answer De Rono Nov 1, 2015 The answer is #x > 2 " or " x < 3/2# Explanation: Note that #(3x)/(x+3)>3/(x-2)# #implies 3x*(x-2)> (x-2)* (x+3 )# #implies 3x^2-6x > x^2+x-6# #implies 2x^2 -7x +6 >0# #implies 2 x^2 - 4x -3x +6 >0# #implies 2x*(x-2) -3 (x-2)>0# #implies( 2x - 3 )( x - 2 )>0# #implies x > 2 " or " x < 3/2# Answer link Related questions How do you solve inequalities using multiplication and division? How do you solve two step inequalities? Why do you change the inequality symbol when you multiply or divide by a negative? How do you solve for x in #-10x > 250#? How do you solve and graph #\frac{x}{5} > - \frac{3}{10}#? What is the solution to #\frac{x}{-7} \ge 9# written in set notation? Why do you not change the inequality sign when solving #9x > - \frac{3}{4}#? How do you graph #\frac{k}{-14} \le 1# on a number line? How do you graph #8d < 24#? How do you graph #-8d < 24#? See all questions in Inequalities with Multiplication and Division Impact of this question 1648 views around the world You can reuse this answer Creative Commons License