Question

Find the value of $k$ for which each of the following system of equations have infinitely many solution:
$
  2x + 3y = k \\
  \left( {k - 1} \right)x + \left( {k + 2} \right)y = 3k \\
 $

Answer 1

Hint - We will start solving this question by writing down the given equations. By using the given condition, i.e., equations have infinitely many solutions, we will use an appropriate formula i.e. $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$ which will give the required value.

Complete Step-by-Step solution:
Now, the given equations are,
$
  2x + 3y = k \\
  \left( {k - 1} \right)x + \left( {k + 2} \right)y = 3k \\
 $
These equations can be written as,
$
  2x + 3y - k = 0 \\
  \left( {k - 1} \right)x + \left( {k + 2} \right)y - 3k = 0 \\
 $
Here, we have the equations in the form of,
$
  {a_1}x + {b_1}y + {c_1} = 0 \\
  {a_2}x + {b_2}y + {c_2} = 0 \\
 $ ……………. (1)
And the equations have infinitely many solution, then the formula to be used is,
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$ …………… (2)
The equations we have are,
$
  2x + 3y - k = 0 \\
  \left( {k - 1} \right)x + \left( {k + 2} \right)y - 3k = 0 \\
 $
Comparing with (1), we have
$
  {a_1} = 2,{b_1} = 3,{c_1} = - k \\
  {a_2} = \left( {k - 1} \right),{b_2} = \left( {k + 2} \right),{c_2} = - 3k \\
 $
Now, putting these values in (2), we obtain,
$
  \dfrac{2}{{\left( {k - 1} \right)}} = \dfrac{3}{{\left( {k + 2} \right)}} = \dfrac{{ - k}}{{ - 3k}} \\
   \Rightarrow \dfrac{2}{{\left( {k - 1} \right)}} = \dfrac{3}{{\left( {k + 2} \right)}} = \dfrac{k}{{3k}} \\
 $
$ \Rightarrow \dfrac{2}{{\left( {k - 1} \right)}} = \dfrac{3}{{\left( {k + 2} \right)}}$…… (3) and $\dfrac{3}{{\left( {k + 2} \right)}} = \dfrac{k}{{3k}}$ …… (4)
First we will solve equation (3),
$
  \dfrac{2}{{\left( {k - 1} \right)}} = \dfrac{3}{{\left( {k + 2} \right)}} \\
   \Rightarrow 2\left( {k + 2} \right) = 3\left( {k - 1} \right) \\
   \Rightarrow 2k + 4 = 3k - 3 \\
   \Rightarrow 2k - 3k = - 3 - 4 \\
   \Rightarrow - k = - 7 \\
   \Rightarrow k = 7 \\
 $
Now, we will solve equation (4),
$
  \dfrac{3}{{\left( {k + 2} \right)}} = \dfrac{k}{{3k}} \\
   \Rightarrow 9k = k\left( {k + 2} \right) \\
   \Rightarrow 9k = {k^2} + 2k \\
   \Rightarrow {k^2} + 2k - 9k = 0 \\
   \Rightarrow {k^2} - 7k = 0 \\
   \Rightarrow k\left( {k - 7} \right) = 0 \\
   \Rightarrow k - 7 = 0 \\
   \Rightarrow k = 7 \\
 $
Hence, the value of $k$ is 7.

Note – A linear equation is an equation in which the highest degree term in the variable or variables is of the first degree and when a graph is plotted gives a straight line. While using this formula make sure that you do not get confused between the formulas of equations having different types of solutions.