Question

(a) for what Value of K , 2X +3Y = 4 and (K+2)X+6Y = 3K +2 will have infinitely many solution?

Answer 1

Hey

Here is your answer,

It is given that that the equations have infinitely many solutions.

2/k+2 = 3/6 = 4/3k+2

When cross multiplying ,
3(k+2)=6x2
3k+6=12
3k=6
k=2

Hope it helps you!

Answer 2

Hii friend,


2X +3Y = 4........(1)

(K-2)X +6Y = 3K+2......(2)

These equations are in the form of AX1+BY1+ C2 and AX2+BY2+ C2.

Where,

A1 = 2 , B1 = 3 and C1 = -4

and,

A2= K+2 , B2 = 6 , And C2 = -(3K+2)

for infinitely many solution we must have,

A1/A2 = B1/B2 = C1/C2

=> 2/K+2 = 3/6 = -4/-(3K+2)

=> 3(K+2) = 12

=> 3K +6 = 12

=> 3K = 6

=> K = 6/3 = 2





HOPE IT WILL HELP YOU....