If $x$ and $y$ are positive integers with $7x+4y=100$, the largest possible value of $x$ is:

$4y=100-7x$ must be positive and a multiple of 4, so $7x$ must leave remainder 0 on division by 4, which needs $x$ to be a multiple of 4. The possible values are $x=4$, 8, 12, and the largest is 12 with $y=(100-84)/4=4.(x=13$ gives $4y=9$, and $x=14$ gives $4y=2$ — neither is a whole number of $y.$)