Question

Find the value of k for which each of the following systems of equation has no solution:
$x + 2y = 0$
$2x + ky = 5$

Answer 1

Hint – Here we will proceed by using the no solution condition for linear equation that is $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$. By applying this condition, we will get our answer.

Complete Step-by-Step solution:
If we have linear equation in the following form,
$
  {a_1}x + {b_1}y + {c_1} = 0 \\
  {a_2}x + {b_2}y + {c_1} = 0 \\
 $
For these type of equations, condition of no solution will be,
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$ (If this condition is satisfied then we can say that equations has no solution)
Given equation is
$x + 2y = 0$ …. (1)
$2x + ky = 5$;
We can also write it as,
$2x + ky - 5 = 0$ …. (2)
Now, by comparing equation (1) and (2) by $
  {a_1}x + {b_1}y + {c_1} = 0 \\
  {a_2}x + {b_2}y + {c_1} = 0 \\
 $we will get,
For, equation (1)
${a_1} = 1,{b_1} = 2,{c_1} = 0$
   For, equation (2),
${a_2} = 2,{b_2} = k,{c_2} = - 5$
Now we will apply the no solution condition here that is $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$ we will get
$\dfrac{1}{2} = \dfrac{2}{k} = \dfrac{0}{5}$ …. (3)
$\dfrac{1}{2} \ne 0$ (It is always true)
From equation 3 we can say that,
If, $\dfrac{1}{2} = \dfrac{2}{k}$ Then, $\dfrac{1}{2} = \dfrac{2}{k}$$ \ne $ $\dfrac{0}{5}$
Now, we will simplify the equation,
$\dfrac{1}{2} = \dfrac{2}{k}$
$1 \times k = 2 \times 2$
$ \Rightarrow k = 4$
So, we can say that, if the value of $k = 4$ then no solution will exist for this system.

Note – Whenever we come up with this type of question, one must know that a system of equations is called an inconsistent system of equations if there is no solution because the lines are parallel. If two or more equations that are impossible to solve based on using one set of values of the variables.