Question

The system of equations \( x + 3y + 7 = 0 \), \( 3x + 10y - 3z + 18 = 0 \), and \( 3y - 9z + 2 = 0 \) has:

• A. unique solution
• B. infinitely many solutions
• C. no solution
• D. finite number of solutions \bigskip

Hint:

When solving a system of linear equations, a contradiction like \( 0 = 3 \) indicates that the system has no solution.

Answer 1

The Correct Option is C

Step 1: Write the system of equations clearly: \[ x + 3y + 7 = 0 \quad \text{(1)}, \] \[ 3x + 10y - 3z + 18 = 0 \quad \text{(2)}, \] \[ 3y - 9z + 2 = 0 \quad \text{(3)}. \] We will solve this system step by step. 

Step 2: From equation (1), solve for \( x \): \[ x = -3y - 7 \quad \text{(4)}. \] Substitute equation (4) into equations (2) and (3) to simplify the system. \bigskip Step 3: Substituting \( x = -3y - 7 \) into equation (2): \[ 3(-3y - 7) + 10y - 3z + 18 = 0, \] \[ -9y - 21 + 10y - 3z + 18 = 0, \] \[ y - 3z - 3 = 0 \quad \text{(5)}. \] Now substitute equation (5) into equation (3). 

Step 4: Substituting into equation (3): \[ 3y - 9z + 2 = 0, \] \[ y - 3z = -\frac{2}{3} \quad \text{(6)}. \] Now subtract equation (5) from equation (6): \[ 0 = 3 \quad \text{which is a contradiction.} \] 

Step 5: Since we encounter a contradiction, the system of equations has no solution.