what value of a makes the equation true? multiply both sides of the equation by 6

Answer:
Line 3 of my work has the values for the "?" boxes, but I didn't know if you needed the whole thing solved or not, so I did the whole problem out.
Step-by-step explanation:
[tex]\frac{1}{3} (4a+1)=\frac{1}{2} a\\6*\frac{1}{3} (4a+1)=6*\frac{1}{2} a\\2(4a+1)=3a\\8a+1=3a\\8a-3a=-1\\5a=-1\\a=-\frac{1}{5}[/tex]
Hope this helps :)