Question

If the system of equations
x+y+z=6
2x+5y+αz=β
x+2y+3z=14
has infinitely many solutions, then α + β is equal to

• A. 8
• B. 36
• C. 44
• D. 48

Answer 1

Answer: (C)

44

Answer 2

$\begin{vmatrix}1&1&1\\\2&5&α\\\1&2&3\end{vmatrix}$=1(15−2α)–1(6−α)+1(−1)

=15–2α–6+α−1
=8–α
For infinite solutions,~Δ=0 ⇒α=8
$\triangle_x$=$\begin{vmatrix}6&1&1\\\β&5&8\\\14&2&3\end{vmatrix}$= 6(−1)−1(3β–112)+1(2β−70)

=−6–3β+112+2β–70
=36–β
$\triangle_x$=0
⇒ For β=36
α+β=44
So, the correct option is (C): 44