Answer
The solution set is $\displaystyle \{-\frac{20}{39}\}$
Work Step by Step
STEP 1: Restrictions. There must be no zeros in the denominator, so
$\left[\begin{array}{ll}
4t-1\neq 0 & 2t-4\neq 0\\
4t\neq 1 & 2t\neq 4\\
t\neq 1/4 & t\neq 2
\end{array}\right]$
STEP 2: Fractions.
$2t-4=2(t-2)$
The LCM of all denominators is $2(4t-1)(t-2)$
After multiplying both sides with the LCM,
$2(t-2)(6t+7)=(3t+8)(4t-1)$
STEP 3: Remove all parentheses and simplify.
$12t^{2}-24t+14t-28=12t^{2}-3t+32t-8$
$12t^{2}-10t-28=12t^{2}+29t-8$
STEP 4: Variables on one side, all other terms on the other...
$12t^{2}-10t-28=12t^{2}+29t-8\quad/-12t^{2}-29t+28$
$-39t=20$
STEP 5: Simplify and solve.
$-39t=20\qquad/\div(-39)$
$t=-\displaystyle \frac{20}{39}$
No restriction prevents $-\displaystyle \frac{20}{39}$ being a solution, so the
solution set is $\displaystyle \{-\frac{20}{39}\}$.
