Question

For which value(s) of $k$ will the pair of equations $kx+3y=k-3$ and $12x+ky=k$ have no solution?

Answer 1

Step 1: Convert the equations to standard form

Given pair of linear equations,

$kx+3y=k-3$ and $12x+ky=k$

Converting the given equations to standard form,

$kx+3y=k-3\Rightarrow kx+3y+(3-k)=0$

$12x+ky=k\Rightarrow 12x+ky-k=0$

Step 2: Compare the equations with the standard equation

Comparing the two equations with the standard form, we define

$a_1=k$, $b_1=3$ and $c_1=3-k$

$a_2=12$, $b_2=k$ and $c_2=-k$

Step 3: Define the condition for no solutions

For no solution, the given set of equations must satisfy the condition,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

Thus,

$\frac{k}{12}=\frac{3}{k}\ne \frac{3-k}{-k}$

Step 4: Solve for $k$

Considering the first two parts,

$\begin{array}{rcl}& & \begin{array}{cc}& \frac{k}{12}=\frac{3}{k}\\ \Rightarrow & k^2-36=0\\ \Rightarrow & k=\pm 6\end{array}\end{array}$

Considering the last two parts,

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

Considering the last two results, we can say that for $k=-6$, the set of equations will have no solutions

Therefore, for $k=-6$, the set of equations has no solutions.