Question

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

Answer 1

Step-by-step explanation:

Given system of linear equations are:

$x + 2y = 3\\\therefore\: x + 2y - 3 = 0......... (1)\\\& \:5x + ky + 7 = 0........ (2)$

Equating equations (1) & (2) by

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

We find:

$a_1=1, \:\:b_1=2, \:\: c_1=-3\: \\\&\:\\ a_2=5, \:\:b_2=k, \:\: c_2=7$

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

$\frac{a_1}{a_2} =\frac{b_1}{b_2} \neq\frac{c_1}{c_2} \\\\\therefore \frac{1}{5} =\frac{2}{k} \neq\frac{-3}{7} \\\\\therefore \frac{1}{5} =\frac{2}{k} \\\\\therefore k = 5\times 2\\\\\huge \purple {\boxed {\therefore k = 10}}$

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