We are given the following inequality:
[tex]6a+4b>10[/tex]If we replace b = 2, we get:
[tex]\begin{gathered} 6a+4(2)>10 \\ 6a+8>10 \end{gathered}[/tex]Now we solve for "a" first by subtracting 8 on both sides:
[tex]\begin{gathered} 6a+8-8>10-8 \\ 6a>2 \end{gathered}[/tex]Now we divide both sides by 6
[tex]\frac{6a}{6}>\frac{2}{6}[/tex]Simplifying:
[tex]a>\frac{1}{3}[/tex]Therefore, for b = 2, the possible values of "a" are those that are greater than 1/3
