Question

Fkr what value k does the system of equations x +2y =3 and 5x+ky+7=0 have no solution ​

Answer 1

Explanation:

Given system of linear equations are:

`x + 2y = 3\\</p><p>\therefore\: x + 2y - 3 = 0......... (1)\\</p><p>\& \:5x + ky + 7 = 0........ (2)\\`

Equating equations (1) & (2) by

`a_1x+b_1y + c_1=0\: \&\:\\ a_2x+b_2y + c_2=0`

We find:

`a_1=1, \:\:b_1=2, \:\: c_1=-3\: \\\&\:\\ a_2=5, \:\:b_2=k, \:\: c_2=7\\</p><p>`

The condition for system of linear equations having no solution is given as:

`(a_1)/(a_2) =(b_1)/(b_2) \\eq(c_1)/(c_2) \\\\</p><p>\therefore (1)/(5) =(2)/(k) \\eq(-3)/(7) \\\\\therefore (1)/(5) =(2)/(k) \\\\</p><p>\therefore k = 5* 2\\\\</p><p>\huge \purple {\boxed {\therefore k = 10}}`

Thus, for k = 10 given system of linear equations have no solution.