Question

If the system of equations:

4x+py=21 and px-2y=15 has unique solutions , then which of the following could be the value of p?

103

105

192

197


Both 103 and 105

Both 105 and 192

The first three

All of the above

Answer 1

Given:

The system of equations: 4x+py=21 and px-2y=15

To find:

If the system of equations: 4x+py=21 and px-2y=15 has unique solutions , then which of the following could be the value of p?

Solution:

Condition for the system of equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0  to have unique solution is,

a1/a2 ≠ b1/b2

From given, we have,

4x + py = 21 and px - 2y = 15

a1 = 4 and a2 = p

b1 = p and b2 = -2

c1 = 21 and c2 = 15

4/p ≠ p/-2

⇒ p × p ≠ 4 × -2

⇒ p² ≠ -8

⇒ p ≠ √(-8)

Therefore, for all the values other than √(-8), the system of equations will have unique solution.

Answer 2

Answer : All of the above

To have a unique solution, the system of equations must satisfy two conditions:

1 .The system must have exactly two equations.

2. The determinant of the coefficient matrix must be nonzero.

In this case, the system has two equations, so we need to check the second condition.

The coefficient matrix of the system is:

| 4 p |

| p -2 |

The determinant of this matrix is:

det = (4)(-2) - (p)(p) = -8 - p²

For the system to have a unique solution, the determinant must be nonzero. Therefore, we need to solve the inequality:

-8 - p² ≠ 0

This simplifies to:

p² + 8 ≠ 0

Since p² + 8 is always positive for any value of p, we can conclude that the system has a unique solution for all values of p. Therefore, any of the given values of p could be the correct answer.

Hence, the answer is option (4) "All of (A), (B), (C) and (D)".