Question

Find the value of k for which each of the following systems of equations have infinitely many solutions.
2x + 3y – 5 = 0
6x + ky – 15 = 0

Answer 1

Hint: 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 solution:
We are given with system of equations is-
2x + 3y – 5 = 0………… (1)
6x + ky – 15 = 0…………… (2)
The above equations are of the form-
$
  {a_1}x + {b_1}y - {c_1} = 0 \\
  {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 ${a_1} = 2,{b_1} = 3,{c_1} = - 5$
And ${a_2} = 6,{b_2} = k,{c_2} = - 15$
Comparing the ratios of the coefficients of given equation,
We get-
$ \Rightarrow \dfrac{2}{6} = \dfrac{3}{k} = \dfrac{{ - 5}}{{ - 15}}$
$\Rightarrow$ 2k = 18
$\Rightarrow$ k = 9

Hence, the given system of equations will have infinitely many solutions if k = 9.

Note: We can also use another method to solve this question, 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.