MathematicsMedium244×since 2002Q4615If the system of equations x+y+z=62x+5y+αz=βx+2y+3z=14\begin{aligned} &x+y+z=6 \\ &2 x+5 y+\alpha z=\beta \\ &x+2 y+3 z=14 \end{aligned}x+y+z=62x+5y+αz=βx+2y+3z=14 has infinitely many solutions, then α+β\alpha+\betaα+β is equal toA8B36C44D48Check answerSkip