Question

Find the value of $k$ for which the following system of equations has a unique solution $kx+2y=5,3x+y=1$

• A. Has a unique solution if k equal to $6$.
• B. Has a unique solution if k equal to $2$.
• C. Has a unique solution for all real values of k other than $6$.
• D. Has a unique solution for all real values of k other than $2$.

Answer 1

The correct option is C Has a unique solution for all real values of k other than $6$.

The given system of equations is:

$kx+2y–5=0$

$3x+y–1=0$

The above equations are of the form

a₁ x + b₁ y + c₁ = $0$

a₂ x + b₂ y + c₂ = $0$

Here, a₁ = $k$, b₁ = $2$, c₁ = $-5$

a₂ = $3$, b₂ = $1$, c₂ = $-1$

So according to the question,

For unique solution, the condition is

$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$

$\frac{k}{3}\ne \frac{2}{1}$

⇒ $k\ne 6$

Hence, the given system of equations will have unique solution for all real values of $k$ other than $6$.