Question

The solution of the pair of linear equations: \[ \frac{2x}{3} - \frac{y}{2} = -\frac{1}{6} \quad \text{and} \quad \frac{x}{2} + \frac{2y}{3} = 3 \] is:

• A. x = 2, y = – 3
• B. x = – 2, y = 3
• C. x = 2, y = 3
• D. x = – 2, y = – 3

Answer 1

The Correct Option is C

Step 1: Identify the system of equations:
The given system is:\[\frac{2x}{3} - \frac{y}{2} = -\frac{1}{6} \quad \text{(1)}\]and\[\frac{x}{2} + \frac{2y}{3} = 3 \quad \text{(2)}\]Objective: Solve for \( x \) and \( y \).

Step 2: Clear fractions from the equations:
Multiply equation (1) by the LCM of its denominators (3 and 2), which is 6:\[6 \times \left( \frac{2x}{3} - \frac{y}{2} \right) = 6 \times \left( -\frac{1}{6} \right)\]This yields:\[4x - 3y = -1 \quad \text{(3)}\]Multiply equation (2) by the LCM of its denominators (2 and 3), which is 6:\[6 \times \left( \frac{x}{2} + \frac{2y}{3} \right) = 6 \times 3\]This yields:\[3x + 4y = 18 \quad \text{(4)}\]

Step 3: Solve the simplified system:
The simplified system is:\[4x - 3y = -1 \quad \text{(3)}\]\[3x + 4y = 18 \quad \text{(4)}\]Use the elimination method. To eliminate \( y \), multiply equation (3) by 4 and equation (4) by 3:\[4(4x - 3y) = 4(-1) \quad \Rightarrow \quad 16x - 12y = -4 \quad \text{(5)}\]\[3(3x + 4y) = 3(18) \quad \Rightarrow \quad 9x + 12y = 54 \quad \text{(6)}\]Add equation (5) and equation (6):\[(16x - 12y) + (9x + 12y) = -4 + 54\]\[25x = 50\]Solve for \( x \):\[x = \frac{50}{25} = 2\]

Step 4: Substitute \( x \) to find \( y \):
Substitute \( x = 2 \) into equation (3):\[4(2) - 3y = -1\]\[8 - 3y = -1\]Solve for \( y \):\[-3y = -1 - 8\]\[-3y = -9\]\[y = \frac{-9}{-3} = 3\]

Step 5: Final Answer:
The solution is \( x = 2 \) and \( y = 3 \). The final answer is \( \boxed{x = 2, y = 3} \).