Question

If $(a, b, c)$ is the unique solution of the system of linear equations
$x + y + z = 2, 2x + y - z = 3, 3x + 2y + z = 4$,
then $b^2 + c^2 = ........$

• A. 5
• B. 15
• C. 17
• D. 20

Hint:

To solve systems of linear equations, use substitution or elimination methods to simplify and find the values of variables.

Answer 1

The Correct Option is C

We are given the system of linear equations: \[ x + y + z = 2 (1) 2x + y - z = 3 (2) 3x + 2y + z = 4 (3) \] We need to find the values of $x$, $y$, and $z$. Start by solving this system.
1. From equation (1), solve for $z$: \[ z = 2 - x - y (4) \] 2. Substitute equation (4) into equation (2): \[ 2x + y - (2 - x - y) = 3 \] Simplifying: \[ 2x + y - 2 + x + y = 3 ⇒ 3x + 2y = 5 (5) \] 3. Now, substitute equation (4) into equation (3): \[ 3x + 2y + (2 - x - y) = 4 \] Simplifying: \[ 3x + 2y + 2 - x - y = 4 ⇒ 2x + y = 2 (6) \] 4. Now solve equations (5) and (6): From equation (6), solve for $y$: \[ y = 2 - 2x \] Substitute this into equation (5): \[ 3x + 2(2 - 2x) = 5 ⇒ 3x + 4 - 4x = 5 ⇒ -x = 1 ⇒ x = -1 \] 5. Now substitute $x = -1$ into $y = 2 - 2x$: \[ y = 2 - 2(-1) = 4 \] 6. Finally, substitute $x = -1$ and $y = 4$ into equation (4) to find $z$: \[ z = 2 - (-1) - 4 = -1 \] Thus, the solution is $x = -1$, $y = 4$, and $z = -1$. Therefore, $b = 4$ and $c = -1$. Now calculate $b^2 + c^2$: \[ b^2 + c^2 = 4^2 + (-1)^2 = 16 + 1 = 17 \] Thus, $b^2 + c^2 = 17$.