Question

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

Answer 1

k = 4

Step-by-step explanation:

Given:  

2x + 3y = k

(k-1)x +(k+2)y = 3k

The system of equations has infinitely many solutions.

a1 = 2, b1 = 3, c1 = k

a2 = k-1, b2 = k+2, c2 = 3k

 So a1/a2 = b1/b2 = c1/c2

2/k-1 = k/3k

3 = k-1

Therefore k = 4

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 = (k + 2)

c1 = k, c2 = 3k

________________[Put Values]

$\sf{→\frac{2}{(k-1)} = \frac{3}{(k + 2)} = \frac{k}{3k}} \\ \\ \sf{→\frac{2}{(k -1)} = \frac{k}{3k}} \\ \\ \sf{→ 2 \times 3\cancel{k} = \cancel{k} \times (k - 1)} \\ \\ \sf{→3 = k - 1} \\ \\ \sf{→k = 3 + 1} \\ \\ \sf{→k = 4} \\ \\ \Large{\star{\boxed{\sf{k = 4}}}}$