Question

Write the value of k for which the system of equations x+y-4=0 and 2x+ky-3=0has no solution

Answer 1

Step-by-step explanation:

$\sf \: a_1x + b_1y + c_1 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: .....1$

$\sf \: a_2x + b_2y + c_2 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: .....2$

$\color{red} \sf \: No \: solution$

$\implies \sf \frac{a_1}{a_2} = \frac{b_1}{b_2} \: \cancel= \: \frac{c_1}{c_2}$

$\sf \: x + y - 4 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: .....1$

$\sf \: 2x + ky - 3 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: .....</strong><strong>2</strong><strong>$

$\sf \: a_1 = 1 , b_1 = 1 , c_1 = - 4$

$\sf \: a_2 = 2 , b_2 = k , c_2 = - 3$

$\sf \: \frac{a_1}{a_2} = \frac{b_1}{b_2} \: \cancel = \: \frac{c_1}{c_2}$

$\sf \: \frac{1}{2} = \frac{1}{k } \: \cancel = \: \frac{ - 4}{ - 3}$

$\sf \frac{1}{2} \: \cancel = \: \frac{4}{3}$

$\sf \: \frac{1}{2} = \frac{1}{k}$

$\underline{ \underline{ \boxed{ \sf \color{red}k = 2 \: }}}$

Answer 2

Answer:

Refers to above attachment.........