Mrs. Stevens has a full classroom of students with the same number of students in each row. Patricia is sitting in the 2nd row from the front of the room and the 4th row from the back. She is also the 4th student from the left of the room and the 3rd student from the right. How many students are there in Mrs. Stevens' class? Don’t call the students $x$
Let $r$ represent the number of rows in the classroom and $s$ represent the number of students in each row. Since Patricia is the 2nd student from the front and the 4th student from the back, there are $r-1$ rows in front of her and $r-4$ rows behind her.
Since Patricia is the 4th student from the left and the 3rd student from the right, there are $s-3$ students to her right and $s-4$ students to her left.
Thus, there are $(r-1)+(r-4)+1=2r-4$ rows in total and $2(s-3)+1$ students in total.
We know that $2r-4=s+1$ because there is the same number of students in each row as there are rows in the classroom.
Therefore, $2r-4=2(s-3)+1$.
Distributing the $2$ on the left-hand side of the equation gives $2r-4=2s-6+1$.
Combining like terms on the right-hand side gives $2r-4=2s-5$.
Adding $4$ to both sides of the equation gives $2r=2s-1$.
Since $2s-1$ is odd, $2r$ must be odd. So $r$ must be odd.
Since $2r-4=s+1$, there are $(2r-4)-(s+1)=s-3$ students behind Patricia.
Combining like terms on the left-hand side of the equation gives $(2r-s)-5=s-3$.
Adding $s$ and $3$ to both sides of the equation gives $2r-5=2s$.
Substituting $2s=2r-5$ into $2r=2s-1$ gives $2r=2(2r-5)-1$.
Distributing the $2$ on the right-hand side of the equation gives $2r=4r-10-1$.
Combining like terms on the right-hand side gives $2r=4r-11$.
Adding $11$ to both sides of the equation gives $13=2r$.
Therefore, $r=\boxed{13}$ and $s=2r-5=\boxed{21}$. Answer: \boxed{13}.