Answer:
[tex] \frac{3}{4} (5a - 16) - \frac{1}{3} (6a - 3) \\ \frac{15a}{4} - \frac{48}{4} - \frac{6a}{3} + \frac{3}{3} \\ ( \frac{15a}{4} \frac{ \times3 }{ \times 3} ) - ( \frac{6a}{3} \frac{ \times 4}{ \times 4} ) - 12 + 1 \\ \frac{45a}{12} - \frac{24a}{12} - 11 \\ = \frac{21a}{12} - 11[/tex]
thanks.
hope this helps.
