[2x-ky+3=0],[3x+2y-1=0]

If 2x-ky +3=0 ,3x -y+1 =0 are conjugate lines with respect to 5x^(2) -6y^(2) = 15 then k=

For the following system of equations determine the value of k for which the given system of equations has no solution: 2x-ky+3=0,quad 3x+2y-1=0

Find the value of k for which the following system of equation has no solution: 2x-ky+3=0,quad 3x+2y-1=0

Find k,if the following equations are consistent: ky=x+1,2x-3y+5=0,3x+2y+1=0

If the system of equations x-ky+3z=0, 2x+ky-2z=0 and 3x-4y+2z=0 has non - trivial solutions, then the value of (10y)/(x) is equal to

If the system of equations x-ky+3z=0, 2x+ky-2z=0 and 3x-4y+2z=0 has non - trivial solutions, then the value of (10y)/(x) is equal to

Sam went for an outing with his friends.They went to dominos to enjoy the delicious pizza. He was enjoying the pizza with his friends and share with them by slicing it. During slicing the pizza. he noticed that the pair of linear equations formed. (i.e.,straight lines) Let these pair of linear equations be y - 2x = 1 and 5y-x=14. For what value(s) of k, the system of linear equations 2x-ky + 3 = 0 and 3x+ 2y,-1=0 has no solution

L_(1)=x+3y-5=0,L_(2)=3x-ky-1=0,L_(3)=5x+2y-12=0 are concurrent if k=

Consider the lines give by L_1: x+3y -5=0, L_2:3x - ky-1=0, L_3: 5x +2y -12=0