Question

Find the value of k for which the following system of equations has a unique solution:
$$\begin{gathered} 4x+ky+8=0 \\ x+y+1=0 \end{gathered}$$

Answer 1

The given system of equations is
$$\begin{gathered} 4x+ky+8=0.....\left(i\right) \\ x+y+1=0.....\left(ii\right) \end{gathered}$$
This system is of the form
$$\begin{gathered} a_{1}x+b_{1}y+c_1=0 \\ {a}_{2}x+b_{2}y+c_2=0 \end{gathered}$$
where $a_1=4,b_1=k,c_1=8\mathrm{and}{a}_2=1,b_2=1,c_2=1$
For the given system of equations to have a unique solution, we must have
$$\begin{gathered} \frac{a_1}{a_2}\ne \frac{b_1}{b_2} \\ \Rightarrow \frac{4}{1}\ne \frac{k}{1} \\ \Rightarrow k\ne 4 \end{gathered}$$
Hence, $k\ne 4$.