Question

What is value of k so that the pair of

linear equations kx-y=2 and 6x-2y=3 has
a unique solution?​

Answer 1

Answer:

pls mark it as the brainliest answer

Step-by-step explanation:

let k = 2 ,x=2,y=2

then kx - y = 2

2(2) - 2 = 2

4-2 = 2

let x = 1 y =3/2

6x - 2y = 3

6(1) - 2(3/2) = 3

6 - 3 = 3

hope it helps u

Answer 2

Step-by-step explanation:

we have two equation

kx-y=2 and 6x-2y =3

for unique solution , we must

$\frac{a1}{a2} \not = \frac{b1}{b2}$

so we have

a1=k ,b1= -1 and c1= -2

a2=6 ,b2 = -2 and c2 = -3

$\frac{k}{6} \not = \frac{ - 1}{ - 2} \\ \implies \frac{k}{6} \not = \frac{1}{2} \\ \implies \: k \not = 3$

hence,the given equation will have a unique solution when

$k \: \not = 3$