Question

Find the value of k for which the following system of equation has the unique solution:-
kx + 2y = 5
3x + y = 1.

Answer 1

In order to solve this problem we will compare the coefficient of the two equations such that if the two equations are ${a_1}x + {b_1}y + {c_1} = 0$, ${a_2}x + {b_2}y + {c_2} = 0$. Then we will compare the coefficients such that $\dfrac{{{a_1}}}{{{a_2}}}$, $\dfrac{{{b_1}}}{{{b_2}}}$ and $\dfrac{{{c_1}}}{{{c_2}}}$. If $\dfrac{{{a_1}}}{{{a_2}}}$ $\ne$ $\dfrac{{{b_1}}}{{{b_2}}}$ then the equations have unique solution, if $\dfrac{{{a_1}}}{{{a_2}}}$ = $\dfrac{{{b_1}}}{{{b_2}}}$ = $\dfrac{{{c_1}}}{{{c_2}}}$ then the equations have infinitely many solutions and if $\dfrac{{{a_1}}}{{{a_2}}}$ = $\dfrac{{{b_1}}}{{{b_2}}}$ $\ne$ $\dfrac{{{c_1}}}{{{c_2}}}$ then the equations have no solutions. Doing this will solve your problem and will give you the value of k while solving the equations.

Complete step-by-step answer :
The given equations are,
kx + 2y = 5
3x + y = 1
The above equations can be written as,
kx + 2y – 5 = 0
3x + y – 1 = 0
We need to find the value of k.
So, we know that if the two equations are ${a_1}x + {b_1}y + {c_1} = 0$, ${a_2}x + {b_2}y + {c_2} = 0$. Then we will compare the coefficients such that $\dfrac{{{a_1}}}{{{a_2}}}$, $\dfrac{{{b_1}}}{{{b_2}}}$ and $\dfrac{{{c_1}}}{{{c_2}}}$. If $\dfrac{{{a_1}}}{{{a_2}}}$ $\ne$ $\dfrac{{{b_1}}}{{{b_2}}}$ then the equations have unique solution, if $\dfrac{{{a_1}}}{{{a_2}}}$ = $\dfrac{{{b_1}}}{{{b_2}}}$ = $\dfrac{{{c_1}}}{{{c_2}}}$ then the equations have infinitely many solutions and if $\dfrac{{{a_1}}}{{{a_2}}}$ = $\dfrac{{{b_1}}}{{{b_2}}}$ $\ne$ $\dfrac{{{c_1}}}{{{c_2}}}$ then the equations have no solutions.
Here we can clearly see that,
${a_1} = k,\,{b_1} = 2,\,{c_1} = - 5 \\\ {a_2} = 3,\,{b_2} = 1,\,{c_2} = - 1 \\\$
So, we do, $\dfrac{{{a_1}}}{{{a_2}}},\,\dfrac{{{b_1}}}{{{b_2}}},\,\dfrac{{{c_1}}}{{{c_2}}}$.
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{k}{3},\,\dfrac{{{b_1}}}{{{b_2}}} = \dfrac{2}{1},\,\dfrac{{{c_1}}}{{{c_2}}} = \dfrac{5}{1}$
We know that if the system of equation has unique solution then $\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}$.
So, we solve on putting their values and we get,
$\dfrac{k}{3} \ne \dfrac{2}{1}$
$k \ne 6$
Hence, for the system of equations to have a unique solution the value of k must not be 6.
So, the answer is k can have any value other than 6.

Note : When you get to solve such problems you always need to consider the coefficients of linear equations and perform the operations as done above to get the correct result. Above equations are the equations of two lines and if the lines are parallel then it has no solution because they are not intersecting, if the lines are crossing each other then they have a unique solution and if the lines are coinciding then they have many solutions. Knowing this will solve your problem and will give you the right answers.