Question

How many solutions does the system of linear equations \( 2x - 3y + 1 = 0 \) and \( 3x + y + 2 = 0 \) have?

• A. one and only one solution
• B. no solution
• C. infinitely many solutions
• D. none of these

Hint:

If the system of linear equations has distinct coefficients and constants, it will always have exactly one solution.

Answer 1

The Correct Option is A

Step 1: Solve the system of equations: \[ 2x - 3y + 1 = 0 \quad \text{(1)} \] \[ 3x + y + 2 = 0 \quad \text{(2)} \] Step 2: Express \( y \) from equation (2): \[ y = -3x - 2 \] Step 3: Substitute \( y = -3x - 2 \) into equation (1): \[ 2x - 3(-3x - 2) + 1 = 0 \] \[ 2x + 9x + 6 + 1 = 0 \] \[ 11x + 7 = 0 \quad \Rightarrow \quad x = -\frac{7}{11} \] Step 4: Substitute \( x = -\frac{7}{11} \) into equation (2) to find \( y \): \[ y = -3\left(-\frac{7}{11}\right) - 2 = \frac{21}{11} - 2 = \frac{-1}{11} \] Thus, the system has one and only one solution: \( x = -\frac{7}{11}, y = -\frac{1}{11} \).