Chirsthoper
59

# What is the product? $\frac{2a-7}{a}* \frac{3a^2}{2a^2-11a+14}$ A. $\frac{3}{a-2}$ B. $\frac{3a}{a-2}$ C. $\frac{3a}{a+2}$ D. $\frac{3}{a+2}$

The denominator of the second fraction can be factored as (a-2)(2a-7), then it becomes doable. You cancel the (2a-7) factors, and are left with: $\frac{3a^2}{a(a-2)} \implies \frac{3a}{a-2}$ Do note that you have erased the fact that a≠0 and a≠7/2, so you should always mention that.