Question

For which value(s) of $k$ will the pair of equations have no solution?
$kx + 3y = k + 3;12x + ky = k;k \ne 0$

Answer 1

Here we will first modify the given equations by taking the constant terms to the left side of the equation. Then we will compare the equations to the standard form of the equation to get the value of a, b and c of the respective equations. Then we will put these values in the condition for the equations with no solution to get the required value of $k$.

Complete step by step solution:
The given equations are $kx + 3y = k + 3;12x + ky = k$.
First, we will write the constant terms of the given equation on the left side of the equation. Therefore, the equation becomes
$\begin{array}{l}kx + 3y - k + 3 = 0\\\12x + ky - k = 0\end{array}$
Now we will compare the equation $ax + by + c = 0$ to get the values of a, b and c of the given equations. Therefore, we get
$\begin{array}{l}{a_1} = k\\\\{b_1} = 3\\\\{c_1} = - k + 3\\\\{a_2} = 12\\\\{b_2} = k\\\\{c_2} = - k\end{array}$
Now we will write the conditions for the pair of equations to have no solution.
Therefore, the equations must satisfy the condition $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$ to have no solution
Now we will put the respective values in the above equation. Therefore, we get
$\dfrac{k}{{12}} = \dfrac{3}{k} \ne \dfrac{{ - k + 3}}{{ - k}}$
So, for this condition to be true $\dfrac{k}{{12}} = \dfrac{3}{k}$ and $\dfrac{3}{k} \ne \dfrac{{ - k + 3}}{{ - k}}$ must be true.
Now we will solve these equations to get the value of $k$. Therefore, we get
$\Rightarrow {k^2} = 36$ and $3 \ne k - 3$
Now we will simplify the above equation, we get
$\Rightarrow k = \pm 6$ and $k \ne 6$
$\Rightarrow k = - 6$

Hence, the pair of equations have no solution when the value of $k$ is equal to $- 6$ i.e. $k = - 6$.

Note:
We should know the condition of the infinitely many solutions of the homogeneous system of equations also the condition for the no solution of the given pair of equations.. But if the determinant of the coefficient matrix is not equal to 0 then the system of equations will have a unique solution.
Real value is a number which has some real or discrete or possible value. But imaginary value is the number with a real number multiplied with an imaginary part $i$.
A variable of an equation can have both the values i.e. real value and the imaginary values like in our example we get one real value and two imaginary values.