Question

For what value of k, the pair of linear equations 3x+y=4 and 6x+ky=8 have infinitely many

solutions?​

Answer 1

Answer:

The equation are :

1. 3x + 2y = 4

--> 3x + 2y -4 = 0

2. 6x + ky = 8

--> 6x + ky - 8 = 0

Here a1 = 3 , b1 = 2 , c1 = -4

and a2 = 6 , b2 = k , c2 = -8

The given solution has infinitely many solutions if :

\frac{a1}{a2} = \frac{b1}{b2} = \frac{c1}{c2}

a2

a1

=

b2

b1

=

c2

c1

- > \frac{3}{6} = \frac{2}{k} = \frac{ - 4}{ - 8}−>

6

3

=

k

2

=

−8

−4

- > \frac{1}{2} = \frac{2}{k} = \frac{1}{2}−>

2

1

=

k

2

=

2

1

On comparing ,

- > \frac{1}{2} = \frac{2}{k}−>

2

1

=

k

2

--> k = 2 × 2

--> k = 4

On comparing ,

\frac{2}{k} = \frac{1}{2}

k

2

=

2

1

--> k = 2 × 2

--> k = 4

Step-by-step explanation:

I hope it helps u dear friend ^_^♡♡

Answer 2

Answer:

K= 4

Hope it helps you

Thanku