Question

If the system of equation 6x+2y=3 and kx+y=2 has a unique solutions find value of k.

Answer 1

Answer:

3

Step-by-step explanation:

first for a unique solution it must follow condition

a1/b1 is not equal to a2/b2

which is here:

6/k is not equal to 2/1

now cross multiply 6/k and 2/1= 6=2k

k=6/2=3

hope u understood

Answer 2

Answer:

$\large{\underline{\boxed{\sf k \ne 3}}}$

Step-by-step explanation:

$\large{\boxed{\sf Given \: that:}}$

6x + 2y = 3...... (i)

kx + y = 2 ..... (ii)

This system of equation has a unique solutions. So, for having unique solution, we know that -

$\sf \dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2}$ .....(iii)

On comparing the given equation with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0.

We get-

a₁ = 6, a₂ = k

b₁ = 2, b₂ = 1

c₁ = 3, c₂ = 2

Substituting the values of above in the equation (iii) -

$\sf \dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2} \\ \\ \implies \sf \frac{6}{k} \ne \frac{2}{1} \\ \\ \implies \sf 2 \times k \ne 6 \times 1 \\ \\ \implies \sf 2k \ne 6 \\ \\ \implies \sf k \ne 3$

Hence, the answer is k ≠ 3.

It implies that, for all real values of k, except k = 3, the system has unique solution.