Question

Find the value of k for which each of the following systems of equations have infinitely many solution:
kx+3y=2k+12(k+1)x+9y=7k+1

Answer 1

k = 1/2

Step-by-step explanation:

Given:  

kx + 3y = 2k+12

(k+1)x + 9y = 7k+1

The system of equations has infinitely many solutions.

a1 = k, b1 = 3, c1 = -(2k+12)

a2 = k+1, b2 = 9, c2 = -(7k+1)

 So  a1/a2 = b1/b2 = c1/c2

k/ k+1 = 3/9

9k = 3 (k+1)

9k = 3k + 3

6k = 3

Therefore k = 1/2

Answer 2

$\Large{\underline{\underline{\bf{Solution :}}}}$

As, it is given equations has infinite many solutions.

We know the case of infinite many solutions.

$\Large{\star{\boxed{\rm{\frac{a_{1}} {a_{2}} = \frac{b_1}{b_2} = \frac{c_1}{c_2}}}}}$

Where,

a1 = k, a2 = (k + 1)

b1 = 3, b2 = 9

c1 = -(2k + 12), c2 = -(7k + 1)

________________[Put Values]

$\sf{→\frac{k}{(k + 1)} = \frac{3}{9} = \frac{-(2k + 12)}{-(7k + 1)}} \\ \\ \sf{→\frac{k}{(k + 1)} = \frac{\cancel{3}}{\cancel{9}}} \\ \\ \sf{→k + 1 = 3k} \\ \\ \sf{→3k - k = 1} \\ \\ \sf{→2k = 1} \\ \\ \sf{→k = \frac{1}{2}} \\ \\ \Large{\star{\boxed{\rm{k = \frac{1}{2}}}}}$