Question

The value of kk for which the system of equation x+2y-3=0x+2y−3=0 and 5x+ky+7=05x+ky+7=0 has no solution, is​

Answer 1

Answer:

all the values of k the system of linear equation has no solution accept k = -14/3

Explanation:

For a system of equations a1​x+b1​y+c1​=0;a2​x+b2​y+c2​=0  to have no solution, the condition to be satisfied is

$\frac{a1}{a2} = \frac{b1}{b2} \neq \frac{c1}{c2}$

So,

$\frac{1}{5} = \frac{2}{k } \neq \frac{-3}{7}$

Lets take

$\frac{1}{5}= \frac{2}{k}\\\k = 10$

Now,

lets take

$\frac{2}{k} \neq \frac{-3}{7} \\-3k \neq 14\\k \neq -14/3$

For all the values of k the system of linear equation has no solution accept k = -14/3

Hope you understand.