I notice that this system of linear equations has two variables and two equations, so I can try to use elimination or substitution to solve for one of them. I also notice that the second equation, $m+n=15$, is very simple and easy to isolate either variable. So I will use substitution and solve for $m$ in terms of $n$. I subtract $n$ from both sides of the second equation and get $m=15-n$. Now I can plug this expression for $m$ into the first equation and get $3(15-n)+4n=47$. I simplify this equation by distributing the $3$ and combining like terms. I get $45-3n+4n=47$. I simplify further by subtracting $45$ from both sides and combining the $n$ terms. I get $n=2$. Now that I have $n$, I can plug it back into the expression for $m$ and get $m=15-2$. I simplify this and get $m=13$. 
If $3m+4n=47$, and $m+n=15$, what is $m$?