Question

for what value of k, 2 X + 3 Y = 4 and ( K + 2 ) X + 6 Y is equal to 3 k + 2 will have infinitely many solutions

Answer 1

hope this helps you....

Answer 2

Heya !!

2X + 3Y = 4

2X + 3Y - 4 = 0 ----------(1)

And,

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

( K + 2)X + 6Y - (3K + 2 ) = 0

These equations are of the form A1X + B1Y + C1 = 0 and A2X + B2Y + C1 = 0

Where,

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

And,

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

The given equations have infinitely many solutions.

Then , A1/A2 = B1/B2 = C1/C2

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

=> 12 = 3K + 6

=> 3K = 6

=> K = 6/3

=> K = 2