Question

1. The pair of linear equations3a +4b = k, 9a + 12b = 6 have infinitely many solutions when,a) K = -2b) K = 3c) K = 2d) K = -3​

Answer 1

Answer:

k = - 2 is the answer

3a + 4b - k = 0

9a + 12b - 6 = 0

If these equations have infinetly many solutions then ;

a1/a2 = b1/b2 = c1/c2

3/9 = 4/12 = -k/-6

1/3 = 1/3 = -2/-6

1/3 = 1/3 = 1/3

I hope you got it!!!

Thanks ❣️

Answer 2

Step-by-step explanation:

The equations are

3x+4y−k=0

9x+12y−6=0

Here, a1=3,b1=4,c1=−k

and a2=9,b2=12,c2=−6

For the system to have infinite solution if

$\frac{ a_{1} }{a_{2}} = \frac{ b_{1} }{b_{2}} = \frac{ c_{1} }{c_{2}}$

$\frac{3}{9 } = \frac{1}{12} = \frac{ - k}{ - 6}$

⇒

⇒3k=6

⇒k=2

hope it helps you

by rakhithakur