Question

Find the value of $k$ for which each of the following system of equations has no solution:
$
  2x - ky + 3 = 0 \\
  3x + 2y - 1 = 0 \\
 $

Answer 1

Hint - We will start solving this question by writing down the given equations. By using the given condition, i.e., equations have no solution, we will use an appropriate formula, i.e., $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$ which will give the required value.

Complete Step-by-Step solution:
Now, the given equations are,
$
  2x - ky + 3 = 0 \\
  3x + 2y - 1 = 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 no solution, then the formula to be used is,
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$…………… (2)
The equations we have are,
$
  2x - ky + 3 = 0 \\
  3x + 2y - 1 = 0 \\
 $
Comparing with (1), we have
$
  {a_1} = 2,b = - k,{c_1} = 3 \\
  {a_2} = 3,{b_2} = 2,{c_2} = - 1 \\
 $
Now, putting these values in (2), we obtain
$\dfrac{2}{3} = \dfrac{{ - k}}{2} \ne \dfrac{3}{{ - 1}}$
$
   \Rightarrow 2 \times 2 = - 3k \\
   \Rightarrow 4 = - 3k \\
   \Rightarrow k = - \dfrac{4}{3} \\
 $
Hence, the value of $k$ is $ - \dfrac{4}{3}$.

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.