Question

Find the value of k for which the following system of equations has a unique solution:
4x+ky+8=02x+2y+2=0

Answer 1

k ≠ 4.

Step-by-step explanation:

Given:  

4x +ky +8 = 0

2x +2y +2 = 0

The system of equations has a unique solution.

a1 = 4, b1 = k, c1 = 8

a2 = 2, b2 = 2, c2 = 2

So a1/a2 is not equal to b1/b2

4/2 ≠ k/2

k ≠ 4.

So for all values of k except 4, the system of equations will have a unique solution.

Answer 2

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

As, it is given equations has unique solutions.

We know the case of unique solutions.

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

Where,

a1 = 4, a2 = 2

b1 = k, b2 = 2

c1 = 8, c2 = 2

________________[Put Values]

$\sf{→\frac{4}{2} ≠ \frac{k}{2} ≠ \frac{8}{2}} \\ \\ \sf{→ \frac{4}{2} ≠ \frac{k}{2}} \\ \\ \sf{→k ≠ \frac{4 \times \cancel{2}}{\cancel{2}}} \\ \\ \sf{→k ≠ 4}$

$\Large{\star{\boxed{\rm{k≠4}}}}$