Question

1/2
135%
III. Find the value of k for which the equations have infinitely many
solutions.
1) kx - 2y + 6 = 0; 4x - 3y + 9 = 0
2) 2x – 3y = 7; (k + 2) X – (2k + 1) y = 3 (2k – 1)
3) kx + 3y = 2k + 1; 2 (k + 1)X + 9y = 7k + 1
4) 2x + 3y = k; (k - 1) x + ( + 2)y = 3k​

Answer 1

Answer:

1) solution:

for infinitely many solution

the equation should be in the form a1/a2=b1/b2=c1/c2

therefore ,putting the value

k/4=(-2)/(-3)=6/9

value of k=8/3

2)here,a1=2

a2=(k+2)

b1= -3

b2=(2k+1)

c1=7

c2=3(2k-1) or, 6k-3

doing all the steps mention in above equation

2/(k+2)=3/(2k+1) =7/6k-3

or k=4

3) k/2k+2=3/9=2k+1/7k+1

k=2