If the lines $2x + 3y + 1 = 0$, $3x - y - 4 = 0$, and $x+2y+k=0$ are concurrent, find the value of $k$.

Solution

1. Solve the system of the first two equations: $2x + 3y = -1$ and $3x - y = 4$.
2. Multiplying the second equation by 3 gives $9x - 3y = 12$; adding this to the first equation yields $11x = 11$, so $x=1$.
3. Substituting $x=1$ into $3x - y = 4$ gives $3(1) - y = 4$, which results in $y=-1$.
4. Substitute the intersection point $(1,-1)$ into the third equation $x+2y+k=0$ to get $1 + 2(-1) + k = 0$.
5. Simplify the equation to $1 - 2 + k = 0$, which results in $k-1=0$, so $k=1$.
6. Hence the answer is (A).