Question

Find the value of k such that the system of equations has no solutions.
(2k)x - 2y = 3
8x + (2k + 8)y = 6

Answer 1

if the pair of equation have no solutions then,

2k/8 = (-2)/(2k+8) ≠ 3/6

taking,

2k/8 = (-2)/(2k+8)

• use cross multiplication method

$4 k^2 + 16 k = -16$

$k^2 + 4 k + 4 = 0$

• Now solve for x

$\sf k^2+4k+4=0 \\ \implies\sf k^2+2k+2k+4=0 \\ \implies \sf k (k+2)+2 (k+2)=0 \\ \implies\sf (k+2)(k+2)=0 \\ \implies \sf k+2=0 \\ \implies\sf k=(-2)$

$\therefore$${\underline{\boxed{\bf {k=(-2)}}}}$

Answer 2

Step-by-step explanation:

if the pair of equation have no solutions then,

2k/8 = (-2)/(2k+8) ≠ 3/6

taking,

2k/8 = (-2)/(2k+8)

use cross multiplication method

4 k² + 16 k = -16

k ²+4k+4=0

Now solve for x

⟹k ² +2k+2k+4=0

⟹k(k+2)+2(k+2)=0

⟹(k+2)(k+2)=0

⟹k+2=0

⟹k=(−2)

k=(−2)