Question

For which value of the given system of equations have the infinitely many solution,
$(k - 3)x + 3y = k{\text{ and }}kx + ky = 12$
A. $ - 6$
B. $6$
C. $0$
D. $3$

Answer 1

Hint: Here we need to know the concept that when we have two equations and we need to prove that they are having infinite solutions, we need to equate the ratios of the coefficients of the variable as well as the constant term. If the general system of equations ${a_1}x + {b_1}y + {c_1} = 0{\text{ and }}{{\text{a}}_2}x + {b_2}y + {c_2} = 0$ have the infinitely many solutions then we can say that:
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$

Complete step-by-step answer:
Here we are given the system of the equations in two variables which have infinitely many solutions. So we must know that if the two equations have infinitely many solutions then we need to equate the ratios of the coefficients of the variable as well as the constant term. If the general system of equations ${a_1}x + {b_1}y + {c_1} = 0{\text{ and }}{{\text{a}}_2}x + {b_2}y + {c_2} = 0$ have the infinitely many solutions then we can say that:
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$
Here we are given the equations:
$(k - 3)x + 3y = k$
$kx + ky = 12$
We can write these two equations in the form of:
$(k - 3)x + 3y - k = 0 - - - - - (1)$
$kx + ky - 12 = 0$$ - - - - (2)$
So by comparing these two equations (1) and (2) with the general form of system of the equations ${a_1}x + {b_1}y + {c_1} = 0{\text{ and }}{{\text{a}}_2}x + {b_2}y + {c_2} = 0$ we can say that:
$
  {a_1} = \left( {k - 3} \right) \\
\Rightarrow {b_1} = 3 \\
\Rightarrow {c_1} = - k \\
 $
$
  {a_2} = k \\
\Rightarrow {b_2} = k \\
\Rightarrow {c_2} = - 12 \\
 $
So we can now apply the condition for the existence of infinitely many solutions which is:
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$
Now we need to just substitute the values and we will get:
$\dfrac{{k - 3}}{k} = \dfrac{3}{k} = \dfrac{{ - k}}{{ - 12}}$
So we can equate the terms:
$
  \dfrac{{k - 3}}{k} = \dfrac{3}{k} \\
\Rightarrow k - 3 = 3 \\
\Rightarrow k = 3 + 3 = 6 \\
 $
So we get $k = 6$
We can also equate:
$
  \dfrac{3}{k} = \dfrac{{ - k}}{{ - 12}} \\
\Rightarrow {k^2} = 36 \\
\Rightarrow k = \sqrt {36} = \pm 6 \\
 $
But we need to take the common value. So we can say that $k = 6$
Hence B is the correct option.

Note: Here in these types of problems, students must try not to confuse between the general equation and the equation given. Students must take all the variables and the constant term to the left hand side and then compare with the general equation as it will avoid them from making mistakes in the further calculations.