How do you solve #6/(3x-4) > x + 1#?

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Answer 1 · 2017-02-17T18:48:27

#-5/3 < x < 2#

Multiply both sides of the inequality by #3x-4#:

#6 > (x+1)(3x-4)#

Distribute the right-side: #6 > (3x^2-4x+3x-4)#

Simplify by adding like terms and subtract the #6# from both sides:
#6-6 > 3x^2-x-4-6#;
# 0 > 3x^2-x-10#

Factor the right-side: #0 > (3x+5)(x-2)# or #(3x+5)(x-2) < 0#

Solve for values of #x# that make the equation negative:
#3x+5 < 0# and #x-2<0#

#3x < -5# and #x < 2#

#x < -5/3#

Therefore: #-5/3 < x < 2#

From the graph of #(3x+5)(x-2) < 0# you can see when #y < 0#:
graph{3x^2-x-10 [-20.97, 19.03, -12.32, 7.68]}