Question

Determine the value of k for which the system of linear equations has infinitely many solution.
(k-3)x+3y = k
kx + ky = 12​

Answer 1

Answer:-

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• Given:-

$(k-3)x + 3y = k$

$kx + ky = 12$

Rearranging the terms:-

$\longrightarrow(k-3)x + 3y - k = 0$

$\longrightarrow kx + ky - 12 = 0$

We have,

$a_1 = (k - 3)$

$b_1 = 3$

$c_1 = - k$

• Second Situation:-

$\implies kx + ky = 12$

$\implies kx + ky - 12 = 0$

We have,

$a_2 = k$

$b_2 = k$

$c_2 = -12$

• For Infinite solution :-

$\bf\boxed{\dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}=\dfrac{c_{1}}{c_{2}}}$

$\implies{\dfrac{k-3}{k}=\dfrac{3}{k}=\dfrac{k}{12}}$

• Taking third Situation:-

${\dfrac{k-3}{k}=\dfrac{3}{k}}$

$\longrightarrow k(k - 3) = 3k$

$\longrightarrow k^2 - 3k = 3k$

$\longrightarrow k^2 = 6k$

$\bf\implies{k = 6}$

• Taking fourth Situation:-

${\dfrac{3}{k}=\dfrac{6}{12}}$

$\longrightarrow 36 = 6k$

$\longrightarrow{\dfrac{36}{6}=k}$

$\longrightarrow\cancel{\dfrac{36}{6}}=k$

$\bf\implies{k = 6}$

• Taking fifth Situation:-

${\dfrac{k-3}{k}=\dfrac{k}{12}}$

$\longrightarrow 12k - 36 = k^2$

$\longrightarrow k^2 - 12k + 36 = 0$

$\longrightarrow(k - 6)^2 = 0$

$\bf\implies{k = 6}$

$\therefore$ Therefore,

We get Infinite Solutions.

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