Question

the linear of equation (3K+1)X+3y-5 and 2x-3y+5=0 have infinite solution.find the value of k

Answer 1

Given : (3k+1)x + 3y - 2 = 0 ; (k2+1)x+(k−2)y−5=0
(
k
2
+
1
)
x
+
(
k
−
2
)
y
−
5
=
0
For a system which has no solution,
a1a2=b1b2≠c1c2
a
1
a
2
=
b
1
b
2
≠
c
1
c
2
∴3k+1k2+1=3k−2≠25
∴
3
k
+
1
k
2
+
1
=
3
k
−
2
≠
2
5
⟹3k+1k2+1(3k+1)(k−2)⟹3k2−6k+k−2−5k∴k=3k−2=3

Answer 2

Hi ,

I there is an errand in the problem ,

It may be like this ,

Compare ( 3k+1)x+3y+5 = 0 and

2x+3y+5 = 0 with a1x+b1y+c1 = 0

a2x+b2y+c2 = 0 ,

a1 = 3k + 1 , b1 = 3 , c1 = 5 ;

a2 = 2 , b2 = 3 , c2 = 5 ;

a1/a2 = b1/b2 = c1/c2

[ given infinite solutions ]

( 3k+1 )/2 = 3/3

( 3k + 1 )/2 = 1

3k + 1 = 2

3k = 1

k = 1/3

I hope this helps you.

: )