Question

find the value of k for which the system of equation x+y-4=0 and 2x+ky=3,has no solution​

Answer 1

Answer:

k=2

Step-by-step explanation:

x+y-4=0 and 2x+ky=3 i.e. 2x+ky-3=0

Here,

a1= 1 , b1=1 , c1 = -4

and

a2=2, b2=k, c2 = -3

A system of equation doesn't have a solution only if,

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

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

$\huge \boxed{ \orange {k = 2}}$