Hello!
For this problem, since we are given both the values of [tex]a\\[/tex] and [tex]b[/tex], we'll just want to plug the values in to our equation and we'll get our answer!
[tex]\frac{a^2-5ab}{3a}, a=-4, b=7[/tex]
[tex]\frac{(-4)^2-5(-4)(7)}{3(-4)}[/tex]
[tex]\frac{16+140}{-12}[/tex]
[tex]\frac{156}{-12}[/tex]
[tex]-13[/tex]
Hope this helps!
