Answer
Solution set = $\emptyset$.
(no solution)
Work Step by Step
First, we exclude those values of x that yield a zero in any of the denominators.
$x\displaystyle \not\in\{4,\frac{7}{2} \}\qquad (*)$
Multiply the equation with the LCD=$(x-4)(2x-7)$
$(3x+1)(2x-7)=(6x+5)(x-4)\qquad$ ... simplify (distribute)
$6x^{2}-21x+2x-7=6x^{2}-24x+5x-20$
$ 6x^{2}-19x-7=6x^{2}-19x-20\qquad$ ... add $7+6x^{2}+19x-2x$
$-19x+19x=-20+7$
$0=-13$
this is neveer true (no x exists such that this is true)
Solution set = $\emptyset$.
(no solution)
