Question

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

Answer 1

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

(k - 3)x + 3y = k

We have,

a₁ = (k - 3)

b₁ = 3

c₁ = -k

$\fbox{Second\;situation :-}$

kx + ky = 12

★We have,

a₂ = k

b₂ = k

c₂ = -12

★For Infinite solution :-

$\tt{\rightarrow\dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}=\dfrac{c_{1}}{c_{2}}}$

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

$\fbox{Third\;situation :-}$

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

⇒ k(k - 3) = 3k

⇒ k² - 3k = 3k

⇒ k² = 6k

⇒ k = 6

$\fbox{Fourth\;situation :-}$

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

⇒ 36 = 6k

$\tt{\rightarrow\dfrac{36}{6}=k}$

⇒ k = 6

$\fbox{Fifth\;situation :-}$

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

⇒ 12k - 36 = k²

⇒ k² - 12k + 36 = 0

⇒ (k - 6)² = 0

⇒ k = 6

$\Large{\fbox{We\;get\; Infinite\; Solutions}}$