smc

The line with $x$-intercept $a$ and $y$-intercept $b$ is given by the equation $\frac{x}{a}+\frac{y}{b}=1$. We are told $(4,3)$ is on the line so $$\frac{4}{a}+\frac{3}{b}=1⟹ab-4b-3a=0⟹(a-4)(b-3)=12$$ Since $a$ and $b$ are integers, this equation holds only if $(a-4)$ is a factor of $12$. The factors are $1,2,3,4,6,12$ which means $a$ must be one of $5,6,7,8,10,16$. The only members of this list which are prime are $a=5$ and $a=7$, so the number of solutions is $\boxed{2}$.