Question

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

Answer 1

k = 5

Step-by-step explanation:

Given:  

2x +3y - 2 = 0

(k+1)x +9y -(k+1) = 0

The system of equations has infinitely many solutions.

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

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

 So a1/a2 = b1/b2 = c1/c2

2/(k+1) = 3/9

18 = 3(k+1)

18 - 3 = 3k

15 = 3k

Therefore k = 5

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 = 2, a2 = (k + 1)

b1 = 3, b2 = 9

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

________________[Put Values]

$\sf{→\frac{2}{(k + 1)} = \frac{3}{9} = \frac{2}{-(k + 1)}} \\ \\ \sf{→\frac{2}{(k + 1)} = \frac{\cancel{3}}{\cancel{9}}} \\ \\ \sf{→\frac{1}{k + 1} = \frac{1}{3 \times 2}} \\ \\ \sf{→k + 1 = 6} \\ \\ \sf{→k = 6 - 1} \\ \\ \sf{→k = 5} \\ \\ \Large{\star{\boxed{\sf{k = 5}}}}$