The system 4x + ky = 8 and 2x + 3y = m has no solution when k = 6 and m = 5. For which value of m would the system have infinitely many…

With $k = 6$, the first equation is $4x + 6y = 8$. Dividing both sides by 2 gives $2x + 3y = 4$, which has exactly the same left side as the second equation, $2x + 3y = m$. Two equations with the same left side describe parallel lines: if the right sides differ, the lines never meet (no solution), and if the right sides are equal, the equations describe the same line (infinitely many solutions). So the system has infinitely many solutions only when $m = 4$, choice A. Choices B ($m = 2$), C ($m = 5$), and D ($m = 8$) all make the right sides differ from 4, so each of them gives parallel, distinct lines and no solution at all.