Question

find the value of k for which the pair of equations 2x+ (k+1 ) y= 3k + 1 and 4x + 6y=21 has a unique solution​

Answer 1

Answer:

k = 2

Step-by-step explanation:

The given system of equations are

$2x + (k + 1)y \: - (3k - 1) = 0$

$4x + 6y - 21 = 0$

Here, a1 = 2, b1 = ( k + 1 ), c1 = - ( 3k - 1 )

a2 = 4, b2 = 6, c2 = -21

Now, a1/a2 = 2/4

b1/b2=

$\frac{(k + 1)}{6}$

c1/c2=

$\frac{ - (3k - 1)}{ - 21} = \frac{3k - 1}{21}$

The given system of equations have unique solutions ie.

a1/a2 not equal to b1/b2

2/4 not equal to

$\frac{(k + 1)}{6}$

Now, 6 × 2/4 = 12/4 = 3 not equal to k + 1

k not equal to ( 3 - 1 = 2 )