Question

If the values \( x = \alpha, y = \beta, z = \gamma \) satisfy all the 3 equations \(x+2y+3z=4\), \(3x+y+z=3\) and \(x+3y+3z=2\), then \(3\alpha + \gamma = \)

• A. Option (1): \(\beta = -2\).
• B. Option (2): \(2\beta = 2(-2) = -4\).
• C. Option (3): \(1-2\beta = 1 - 2(-2) = 1 - (-4) = 1+4 = 5\).
• D. \( 2\beta+1 \)

Hint:

Subtract the first equation from the third to find \(\beta\). (\(\alpha + 3\beta + 3\gamma = 2\)) - (\(\alpha + 2\beta + 3\gamma = 4\)) \(\implies \beta = -2\). From the second equation, \(3\alpha + \gamma = 3 - \beta\). Substitute \(\beta = -2\): \(3\alpha + \gamma = 3 - (-2) = 5\). Check which option yields 5 when \(\beta=-2\). Option (3): \(1-2(-2) = 1+4=5\).

Answer 1

The Correct Option is C

The system of equations is: \begin{align} \label{eq:q5_1} \alpha + 2\beta + 3\gamma &= 4
3\alpha + \beta + \gamma &= 3 \label{eq:q5_2}
\alpha + 3\beta + 3\gamma &= 2 \label{eq:q5_3} \end{align} We want to find the value of \(3\alpha + \gamma\).
From equation \eqref{eq:q5_2}, we can express \(3\alpha + \gamma\) as: \[ 3\alpha + \gamma = 3 - \beta ().
\] To find \(\beta\), subtract equation \eqref{eq:q5_1} from equation \eqref{eq:q5_3}: \[ (\alpha + 3\beta + 3\gamma) - (\alpha + 2\beta + 3\gamma) = 2 - 4 \] \[ \beta = -2.
\] Substitute \(\beta = -2\) into equation (): \[ 3\alpha + \gamma = 3 - (-2) = 3 + 2 = 5.
\] Now, we evaluate the given options using \(\beta = -2\): % Option (A) Option (1): \(\beta = -2\).
% Option (B) Option (2): \(2\beta = 2(-2) = -4\).
% Option (C) Option (3): \(1-2\beta = 1 - 2(-2) = 1 - (-4) = 1+4 = 5\).
% Option (D) Option (4): \(2\beta+1 = 2(-2)+1 = -4+1 = -3\).
Since \(3\alpha + \gamma = 5\), and option (3) also evaluates to 5, option (3) is the correct answer.
\[ \boxed{1-2\beta} \]