Question

Find the value of \[k\] for which each of the following systems of equations have infinitely many solutions:
\[
  x + \left( {k + 1} \right)y = 4 \\
  \left( {k + 1} \right)x + 9y = 5k + 2 \\
\]

Answer 1

Hint: In this question, we will proceed by writing the given system of equations and then comparing them with the standard system of equations. Further use the condition for having infinitely many solutions in the given system of equations to get the value of \[k\].


Complete step by step answer:
The given system of equations is
\[
  x + \left( {k + 1} \right)y = 4 \\
  \left( {k + 1} \right)x + 9y = 5k + 2 \\
\]
Which can be written as
\[
  x + \left( {k + 1} \right)y - 4 = 0 \\
  \left( {k + 1} \right)x + 9y - \left( {5k + 2} \right) = 0 \\
\]
This system of equations is of the form
\[
  {a_1}x + {b_1}y + {c_1} = 0 \\
  {a_2}x + {b_2}y + {c_2} = 0 \\
\]
Where, \[{a_1} = 1,{b_1} = k + 1,{c_1} = - 4\] and \[{a_2} = k + 1,{b_2} = 9,{c_2} = - \left( {5k + 2} \right)\]
We know that for infinitely many solutions, we must have
\[
   \Rightarrow \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}} \\
   \Rightarrow \dfrac{1}{{k + 1}} = \dfrac{{k + 1}}{9} = \dfrac{{ - 4}}{{ - 5k - 2}} \\
\]
Now taking first and second parts of the above equation, we get
\[
   \Rightarrow \dfrac{1}{{k + 1}} = \dfrac{{k + 1}}{9} \\
   \Rightarrow 9 = \left( {k + 1} \right)\left( {k + 1} \right) \\
\]
We can write 9 as \[{3^2}\]. So, we have
\[ \Rightarrow {\left( {k + 1} \right)^2} = {3^2}\]
Rooting on both sides, we have
\[
   \Rightarrow \left( {k + 1} \right) = 3 \\
   \Rightarrow k = 3 - 1 \\
  \therefore k = 2 \\
\]
Thus, the required value of \[k\] is 2.

Note:
The condition for the system equations \[{a_1}x + {b_1}y + {c_1} = 0\] and \[{a_2}x + {b_2}y + {c_2} = 0\] to have infinitely many solutions is \[\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}\] and the condition to have unique solution is \[\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}\].