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For the following system of equation determine the value of $k$ for which the given system of equation has infinitely many solutions.$(k−3)x+3y=k$ and $kx+ky=12$.
Given: Pair of linear equations: $(k−3)x+3y$; $kx+ky=12$
To do: To find the value of $k$ when the given pair of equations has infinitely many solution.
Solution:
The Given pair of linear equation is :
$(k - 3 ) x + 3y = k$
$kx + ky = 12$
We can write these Equations as :
$( k - 3 ) x + 3y - k = 0……….( 1)$
$kx + ky - 12 = 0 ………….( 2)$
On comparing with General form of a pair of linear equations in two variables $x$ & $y$ is:
$a_1x + b_1y + c_1 = 0$
and $a_2x + b_2y + c_2= 0$
$a_1=k-3,\ b_1=-3,\ c_1=-k$
$a_2= k,\ b_2=k,\ c_2=-12$
$\frac{a_1}{a_2}= \frac{k-3}{k},\ \frac{b_1}{b_2}=\frac{3}{k},\ \frac{c_1}{c_2}=\frac{-k}{-12}=\frac{k}{12}$
As known a pair of linear equations has a infinite solution, if
$\frac{a_1}{a_2} =\frac{b_1}{b_2} = \frac{c_1}{c_2}$
$\Rightarrow \frac{k-3}{k} =\frac{3}{k}= \frac{k}{12}$
Taking the first two terms
$\Rightarrow \frac{a_1}{a_2} = \frac{b_1}{b_2}$
$\Rightarrow \frac{k-3} {k} =\frac{3}{k}$
$\Rightarrow k - 3 = 3$
$\Rightarrow k = 3 + 3$
$\Rightarrow k = 6$
Taking the second and third terms
$\Rightarrow \frac{3}{k}= \frac{k}{12}$
$\Rightarrow k^2 = 36$
$\Rightarrow k =\sqrt{36}$
$\Rightarrow k = 6$
Thus, for $k=6$, the given pair of equations has infinitely many solutions.
Updated on: 10-Oct-2022
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