Answer:
\(\frac{3a+8}{3(2a+1)(3a-5)}\)
Step-by-step explanation:
We want to subtract \(\frac{1}{6a+3}\) from \(\frac{1}{3a-5}\).
We set up the subtraction problem as follows;
\(\frac{1}{3a-5}-\frac{1}{6a+3}\)
We collect LCM of the denominators to obtain
\(=\frac{6a+3-(3a-5)}{(6a+3)(3a-5)}\)
Expand the bracket in the numerator to obtain;
\(=\frac{6a+3-3a+5}{(6a+3)(3a-5)}\)
This simplifies to;
\(=\frac{3a+8}{3(2a+1)(3a-5)}\)
