What is the value of k except which the given system of equations has a unique solution?
2x + 3y - 5 = 0 and kx - 6y - 8 = 0
a)
4
b)
-4
c)
2
d)
3
Correct answer is option 'B'. Can you explain this answer?

Question

What is the value of k except which the given system of equations has a unique solution?2x + 3y - 5 = 0 and kx - 6y - 8 = 0a)4b)-4c)2d)3Correct answer is option 'B'. Can you explain this answer?

Answer

Here
2x + 3y - 5 = 0
and kx - 6y - 8 = 0
For a unique solution,

Answer

To find the value of k for which the given system of equations has a unique solution, we can use the concept of determinants.

The given system of equations is:
2x + 3y - 5 = 0
kx - 6y - 8 = 0

We can rewrite the equations in the form Ax + By + C = 0, where A, B, and C are the coefficients of x, y, and the constant term respectively.

The given system can be rewritten as:
2x + 3y = 5
kx - 6y = 8

Now, let's write the coefficients of x, y, and the constant term in matrix form.

The coefficient matrix, A, is:
| 2 3 |
| k -6 |

Next, we need to find the determinant of A, denoted as |A|.

|A| = (2)(-6) - (3)(k)
= -12 - 3k
= -3k - 12

For the given system of equations to have a unique solution, the determinant |A| must be non-zero. In other words, -3k - 12 ≠ 0.

Solving the inequality -3k - 12 ≠ 0, we have:
-3k ≠ 12
k ≠ -4

Therefore, the value of k except -4 will give the given system of equations a unique solution.

Hence, the correct answer is option B) -4.

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