Question

Find the value of $k$ for which the system of equations $kx–y=2$and $6x–2y=3$ has a unique solution.

Answer 1

Find the value of $k$ in the equation for the given condition

Give system of equations is,

$$\begin{gathered} kx-y=2 \\ \Rightarrow kx–y-2=0 \end{gathered}$$

$$\begin{gathered} 6x–2y=3 \\ \Rightarrow 6x–2y-3=0 \end{gathered}$$

We know that

If the system of equations $a_{1}x+b_{1}y+c=0$ and $a_{2}x+b_{2}y+c=0$ has a unique solution, then $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$.

Here, $a_1=k$, $b_1=-1$, $a_2=6$ and $b_2=-2$

So for the given system of equations to have unique solution, it must be that

$\begin{array}{rcl}& & \begin{array}{cc}& \frac{k}{6}\ne \frac{-1}{-2}\\ \Rightarrow & \frac{k}{6}\ne \frac{1}{2}\\ \Rightarrow & k\ne \frac{6}{2}\\ \Rightarrow & k\ne 3\end{array}\end{array}$

Hence, for $k\ne 3$, the given system of equations has a unique solution.