For what values of 'k' will the pair of equations 3x + 4y = 12 and kx + 12y = 30 not have a unique solution?
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12
-
9
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3
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7.5
-
2.5
Two linear equations have no unique solution when the lines are parallel or coincident, meaning the ratio of x and y coefficients is equal. For no unique solution: 3/4 = k/12, giving k = 9. Option B is correct. When k = 9, the second equation becomes 9x + 12y = 30, or 3x + 4y = 10, which is parallel to the first equation 3x + 4y = 12.
To determine the values of 'k' for which the pair of equations 3x + 4y = 12 and kx + 12y = 30 does not have a unique solution, we need to consider the determinant of the coefficient matrix.
The coefficient matrix for the system of equations is: [ \begin{bmatrix} 3 & 4 \ k & 12 \ \end{bmatrix} ]
The determinant of this matrix is given by: [ \text{det} = (3)(12) - (k)(4) = 36 - 4k ]
For a system of equations to have a unique solution, the determinant must be non-zero. In other words, the determinant cannot equal zero.
So, to find the values of 'k' for which the system does not have a unique solution, we need to solve the equation 36 - 4k = 0.
Simplifying the equation, we have: 36 - 4k = 0 4k = 36 k = 9
Therefore, the values of 'k' for which the pair of equations does not have a unique solution are k = 9.