Question

the following is a pair of linear equation 2 X + 3 Y = 7 and( K + 1 )X + (2 k - 1) Y = 4 k + 1 write the condition for an infinitely many solutions and find the value of k

Answer 1

Answer:

Step-by-step explanation:

2x+3y-7 = 0

a¹=2 , b¹ =3 , c¹ =-7

(k+1)x + (2k-1)y -(4k+1) = 0

A²=(k+1) , b² = (2k-1) , c² = -(4k+1)

For infinitely many

A¹/a² = b¹/b² = c¹/c²

2/(k+1) = 3/(2k-1) = -7=-(4k+1)

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

On cross multiply

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

4k-2 = 3k +3

4k-3k = 3+2

K=5

Hope it works

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Answer 2

Answer:

Step-by-step explanation:

It has infinity many solution

Therefore

A1/a2=b1/b2=c1/c2

:2/k+1=3/2k-1=-7/-4

=4k-2=3k+3

=k=5