Have two lines given by $p(t) = (-2, -1) + t(1,1)$ and $q(t) = (5,3) +s(4,1)$, and need find their intersection point.
My approach is :
Equating $x$ and $y$ coordinates for each line gives $-2 +t = 5 +4s$ and $-1 +t = 3 +s$, giving rise to the system:
$$ \begin{align} t-4s =& \ 7 \\ t -s =& \ 4 \\ \end{align} $$
Solving the system gives $t= 3, s=-1$, so the solution is (3,1).
Calculation is ok but the solution should be (1,2).
$$p(3) = (-2, -1) + 3(1,1)=q(-1) = (5,3) -(4,1)=(1,2)$$