## Algebra 1: Common Core (15th Edition)

$\frac{3t-5}{7t^{2}}, with$ $t\ne0$
Given : $\frac{7t^{2}-28t}{2t^{2}-5t-12} \times \frac{6t^{2}-t-15}{49t{3}}=\frac{7t(t-4)}{(2t+3)(t-4)}\times \frac{(2t+3)(3t-5)}{7t(7t^{2})}$ This becomes : $\frac{3t-5}{7t^{2}}$ (After dividing out the commonf factors $7t$, $(2t+3)$ and $(t-4)$)