If the system of equations
x+y+z=6x+y+z=6x+y+z=62x+5y+αz=β2x+5y+\alpha z=\beta2x+5y+αz=βx+2y+3z=14x+2y+3z=14x+2y+3z=14
has infinitely many solutions, then α + β is equal to
Official answer: C
Dependence. Row 2 must equal a(row 1)+b(row 3)a(\text{row }1)+b(\text{row }3)a(row 1)+b(row 3): a+b=2a+b=2a+b=2, a+2b=5a+2b=5a+2b=5 give a=−1a=-1a=−1, b=3b=3b=3.
So α=−1+9=8\alpha=-1+9=8α=−1+9=8, β=−6+42=36\beta=-6+42=36β=−6+42=36, and α+β=44\alpha+\beta=44α+β=44.
(C)