## College Algebra (10th Edition)

The solution set is $\displaystyle \{-\frac{20}{39}\}$
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 $\displaystyle \{-\frac{20}{39}\}$.