Question

Find the value of k for which each of the following systems of equations have infinitely many solution:
2x+(k-2)y=k6x+(2k-1)y=2k+5

Answer 1

k = 5

Step-by-step explanation:

Given:  

2x + (k-2)y -k = 0

6x + (2k-1)y -(2k+5) = 0

The system of equations has infinitely many solutions.

a1 = 2, b1 = k-2, c1 = -k

a2 = 6, b2 = 2k-1, c2 = -(2k+5)

 So a1/a2 = b1/b2 = c1/c2

2/6 = (k-2) / (2k-1)

2(2k-1) = 6(k-2)

4k -2 = 6k -12

+10 = 2k

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 = 6

b1 = (k - 2), b2 = (2k - 1)

c1 = -k, c2 = -(2k + 5)

________________[Put Values]

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