Q.10. Assertion (A): The value of k for which the system of linear equ...
**Answer:**
**Assertion (A):** The value of k for which the system of linear equations kx + y = 2 and 6x - 2y = 3 has a unique solution is 3.
**Reason (R):** The graph of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 gives a pair of intersecting lines if a₁/a₂ = b₁/b₂.
To determine whether the given system of linear equations has a unique solution, we can solve the equations using any suitable method, such as substitution or elimination.
Let's solve the given system of equations using the method of elimination:
Equation 1: kx + y = 2
Equation 2: 6x - 2y = 3
To eliminate the y-variable, we can multiply Equation 1 by 2 and Equation 2 by 1:
2(kx + y) = 2(2) => 2kx + 2y = 4
1(6x - 2y) = 1(3) => 6x - 2y = 3
Now, we can add the two equations:
(2kx + 2y) + (6x - 2y) = 4 + 3
8kx + 0 = 7
8kx = 7
To have a unique solution, the coefficient of x should not be zero. Therefore, 8k ≠ 0, which implies k ≠ 0.
Now, we can solve for x:
8kx = 7
x = 7/(8k)
Since x is expressed in terms of k, the solution will be unique for all non-zero values of k.
Therefore, the value of k for which the system of linear equations has a unique solution is any non-zero value of k, not just 3.
**Conclusion:** The reason provided is incorrect. The graph of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 gives a pair of intersecting lines if a₁/a₂ = b₁/b₂ is satisfied. However, the reason does not explain why the value of k for a unique solution is 3. The correct reason is that the coefficient of x should not be zero (i.e., k ≠ 0) for a unique solution.
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