Question

For what value of $K$ will the following pair of linear equations have infinitely many solutions?
$Kx + 3y - (K - 3) = 0$
$12x + Ky - K = 0$

Answer 1

Hint: Here, in the given question, we are given two equations and using these two equations we need to find the value of $K$. Here we will proceed by using the condition of infinite many solution i.e. $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$. Then we will compare the ratios of the coefficients of the given equation and get the required value of $K$.

Complete step by step answer:
We are given, $Kx + 3y - (K - 3) = 0..........\left( 1 \right)$
$12x + Ky - K = 0..........\left( 2 \right)$
The above equations are of the form:
${a_1}x + {b_1}y + {c_1} = 0$
$\Rightarrow {a_2}x + {b_2}y + {c_2} = 0$
We know that the condition of infinite solution is $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$.
So according to the question,
Here ${a_1} = K$, ${b_1} = 3$ and ${c_1} = - \left( {K - 3} \right)$.
And ${a_2} = 12$, ${b_2} = K$ and ${c_1} = - K$.

Comparing the ratios of the coefficients of given equation, we get,
$ \Rightarrow \dfrac{K}{{12}} = \dfrac{3}{K} = \dfrac{{ - (K - 3)}}{{ - K}}$
On canceling out the negative signs, we get
$ \Rightarrow \dfrac{K}{{12}} = \dfrac{3}{K} = \dfrac{{K - 3}}{K}$
Now we will compare coefficients of $x$ and $y$ i.e., $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}}$.
$ \Rightarrow \dfrac{K}{{12}} = \dfrac{3}{K}$
On cross multiplication, we get
$ \Rightarrow {K^2} = 12 \times 3$
On multiplication of terms, we get
$ \Rightarrow {K^2} = 36$
$ \Rightarrow K = \pm 6$

Now we will compare $\dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$
$ \Rightarrow \dfrac{3}{K} = \dfrac{{K - 3}}{K}$
On cross multiplication, we get
$ \Rightarrow 3K = K(K - 3)$
On multiplication of terms, we get
$ \Rightarrow 3K = {K^2} - 3K$
Shift all terms to one side.
$ \Rightarrow {K^2} - 3K - 3K = 0$
$ \Rightarrow {K^2} - 6K = 0$
On taking $K$ as a common factor we get,
$ \Rightarrow K\left( {K - 6} \right) = 0$
$ \therefore K = 0,6$
Only $6$ satisfies both the equations.

Hence, the given system of equations will have infinitely many solutions if $K = 6$.

Note: We can also use another method to solve these equations, firstly we convert the system of equations into matrix form in terms of $A$, $B$ and $X$. If there’s infinitely many solutions of the system of equations, then the value of adjoint $\left( A \right) \times B = 0$. Further we will use the method of multiplication of two matrices, we will find the value of adjoint $\left( A \right) \times B = 0$. Eventually we will compare the values and get the value of $x$ and $y$.