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 linear equations:
The given system is:
\[\frac{2x}{3} - \frac{y}{2} = -\frac{1}{6} \quad \text{and} \quad \frac{x}{2} + \frac{2y}{3} = 3\]The objective is to determine the values of \( x \) and \( y \).

Step 2: Convert equations to integer coefficients:
To remove fractions, multiply each equation by the least common multiple (LCM) of its denominators.
For the first equation, the LCM of 3 and 2 is 6. Multiplying by 6 yields:
\[6 \times \left( \frac{2x}{3} - \frac{y}{2} \right) = 6 \times \left( -\frac{1}{6} \right)\]\[4x - 3y = -1 \quad \text{(Equation 1)}\]For the second equation, the LCM of 2 and 3 is 6. Multiplying by 6 yields:
\[6 \times \left( \frac{x}{2} + \frac{2y}{3} \right) = 6 \times 3\]\[3x + 4y = 18 \quad \text{(Equation 2)}\]

Step 3: Solve the simplified system:
The simplified system is:
\[4x - 3y = -1 \quad \text{(Equation 1)}\]\[3x + 4y = 18 \quad \text{(Equation 2)}\]Employ the elimination method. Multiply Equation 1 by 4 and Equation 2 by 3 to equalize the \( y \) coefficients:
\[4 \times (4x - 3y) = 4 \times (-1) \quad \Rightarrow \quad 16x - 12y = -4 \quad \text{(Equation 3)}\]\[3 \times (3x + 4y) = 3 \times 18 \quad \Rightarrow \quad 9x + 12y = 54 \quad \text{(Equation 4)}\]Add Equation 3 and Equation 4:\[(16x - 12y) + (9x + 12y) = -4 + 54\]\[25x = 50\]\[x = \frac{50}{25} = 2\]

Step 4: Find the value of \( y \):
Substitute \( x = 2 \) into Equation 1:
\[4x - 3y = -1\]\[4(2) - 3y = -1\]\[8 - 3y = -1\]\[-3y = -1 - 8 = -9\]\[y = \frac{-9}{-3} = 3\]

Conclusion:
The solution is \( x = 2 \) and \( y = 3 \).