The radical axis of two circles $S_1=0$ and $S_2=0$ is given by $S_1 - S_2 = 0$. Given $S_1: x^2+y^2+2x+3y-7=0$ and $S_2: x^2+y^2-2x-y+1=0$. Subtracting $S_2$ from $S_1$: $(x^2+y^2+2x+3y-7) - (x^2+y^2-2x-y+1) = 0$. This simplifies to $x^2+y^2+2x+3y-7 - x^2-y^2+2x+y-1 = 0$. Combining terms results in $(2x+2x) + (3y+y) + (-7-1) = 0$, which is $4x+4y-8=0$. Dividing by 4, the radical axis is $x+y-2=0$.