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The value of k so that the system of equation has no solution is k = -4How to find the value of k so that the system of equation has no solution?Since we have the system of equations kx + 6y = 6y = 2/3x - 4For the system of equations to have no solution, the determinant coefficient matrix must be zero.Re-arranging the equations, we havekx + 6y = 6-2/3x + y = - 4So, the coefficient matrix is [tex]\left[\begin{array}{ccc}k&6\\-2/3&1\end{array}\right][/tex]So, the determinant D = k × 1 - (-2/3) × 6 = k + 2/3 × 6 = k + 2 × 2 = k + 4Since the determinant must be zero for the system of equations to have no solution, we have thatD = 0k + 4 = 0k = -4So, the value of k = -4Learn more about system of equations here:https://brainly.com/question/13729904#SPJ1
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