Question

2. Find the value of k for which the system of equations x + y − 4 = 0 and 2x + k y = 3, has no solution.​

Answer 1

Answer:

x+y−4=0

2x+ky=3⇒2x+ky−3=0

a1x+b1y+c1=0

a2x+b2y+c2=0

For no solution

a2a1=b2b1=c2c1

So, 21=k1=−3−4

21=k1

∴k=2