Question:
If the system of equations \( 2x + py + 6z = 8 \), \( x + 2y + qz = 5 \) and \( x + y + 3z = 4 \) has infinitely many solutions, then \( p = \)
If the system of equations \( 2x + py + 6z = 8 \), \( x + 2y + qz = 5 \) and \( x + y + 3z = 4 \) has infinitely many solutions, then \( p = \)
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Always look for proportional equations first! If the coefficients of \( x, z \) and the constant term in one equation are a direct multiple of another, the remaining coefficient must follow that same multiple for infinite solutions to be possible.
Updated On: Jun 3, 2026
- \( -1 \)
- \( 2 \)
- \( 3 \)
- \( -3 \)
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The Correct Option is B
Solution and Explanation
Concept:
For a system of linear equations to have infinitely many solutions, the equations must be linearly dependent. This typically means that one equation can be expressed as a linear combination of the others, or the determinant of the coefficient matrix \( (\Delta) \) and the determinants \( \Delta_x, \Delta_y, \Delta_z \) must all be zero.
Step 1: Comparing Equation 1 and Equation 3
Equation 1: \( 2x + py + 6z = 8 \)
Equation 3: \( x + y + 3z = 4 \)
Notice that if we multiply Equation 3 by 2, we get: \[ 2(x + y + 3z) = 2(4) \quad \Rightarrow \quad 2x + 2y + 6z = 8 \]
Step 2: Equating the coefficients to find \( p \).
For the system to have infinitely many solutions, Equation 1 and the modified Equation 3 must represent the same plane (or Equation 1 must be a multiple of Equation 3). Comparing \( 2x + py + 6z = 8 \) with \( 2x + 2y + 6z = 8 \): Matching the coefficients of \( y \): \[ p = 2 \]
Step 3: Verification with Equation 2 (Optional for \( p \)).
Although not required to find \( p \), the value of \( q \) can be found by ensuring Equation 2 is also consistent. Subtracting Eq(3) from Eq(2): \[ (x + 2y + qz) - (x + y + 3z) = 5 - 4 \] \[ y + (q - 3)z = 1 \] Since we found from Eq(1) and Eq(3) that \( y = 1 - 3z + \text{const} \), the system remains consistent. Specifically, \( p=2 \) satisfies the primary dependency.
Step 1: Comparing Equation 1 and Equation 3
Equation 1: \( 2x + py + 6z = 8 \)
Equation 3: \( x + y + 3z = 4 \)
Notice that if we multiply Equation 3 by 2, we get: \[ 2(x + y + 3z) = 2(4) \quad \Rightarrow \quad 2x + 2y + 6z = 8 \]
Step 2: Equating the coefficients to find \( p \).
For the system to have infinitely many solutions, Equation 1 and the modified Equation 3 must represent the same plane (or Equation 1 must be a multiple of Equation 3). Comparing \( 2x + py + 6z = 8 \) with \( 2x + 2y + 6z = 8 \): Matching the coefficients of \( y \): \[ p = 2 \]
Step 3: Verification with Equation 2 (Optional for \( p \)).
Although not required to find \( p \), the value of \( q \) can be found by ensuring Equation 2 is also consistent. Subtracting Eq(3) from Eq(2): \[ (x + 2y + qz) - (x + y + 3z) = 5 - 4 \] \[ y + (q - 3)z = 1 \] Since we found from Eq(1) and Eq(3) that \( y = 1 - 3z + \text{const} \), the system remains consistent. Specifically, \( p=2 \) satisfies the primary dependency.
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