Solve for #x#: #2x-1=(6x^3-5x^2+3x)/(3x^2-x-14)#?

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Answer 1 · 2017-10-19T23:17:02

#x=7/15#

Given:

#2x-1 = (6x^3-5x^2+3x)/(3x^2-x-14)#

Multiply both sides by #(3x^2-x-14)# to find:

#6x^3-5x^2+3x = (2x-1)(3x^2-x-14)#

#color(white)(6x^3-5x^2+3x) = 6x^3-5x^2-27x+14#

Subtract #6x^3-5x^2-27x# from both sides to get:

#30x=14#

Divide both sides by #30# to get:

#x=14/30=7/15#