Question

For which value of k the system of eq. kx +3y = k-3 and 12x + ky =6 have no solution​

Answer 1

Answer :

k = - 6

Knowledge Required :

In a given pair of linear equations

a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0

▶The system of equations will have a unique solution if

$\boxed{\sf{\frac{a_1}{a_2} \neq \frac{b_1}{b_2}}}$

▶ The system of equations will have infinitely many solutions if

$\boxed{\sf{\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}}}$

▶ The system of equations will have no solutions if

$\boxed{\sf{\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq \frac{c_1}{c_2}}}$

Solution :

Given system of equations is

→ kx + 3y = k-3 and

→ 12 x + ky = 6

We have to find the value of k for which this system of equations will have no solution

So,

$\longmapsto\;\sf{\frac{k}{12}=\frac{3}{k}\neq \frac{-(k-3)}{-6}}$

Taking

$\rightarrowtail\sf{\frac{k}{12}=\frac{3}{k}}$

$\rightarrowtail\sf{k^2=36}$

$\rightarrowtail\boxed{\sf{k=6\;or\;-6}}$ ...eqn(1)

Taking

$\rightarrowtail\sf{\frac{k}{12}\neq \frac{k-3}{6}}$

$\rightarrowtail\sf{6k\neq 12k-36}$

$\rightarrowtail\sf{-6k\neq -36}$

$\rightarrowtail\boxed{\sf{k\neq 6}}$ ....eqn(2)

By eqn (1) and (2)

$\bigstar\boxed{\boxed{\sf{k=-6}}}$

Answer 2

$\rule{220}3$

$\Huge{\red{\underline{\textsf{Answer}}}}$

$\large\underline\mathtt\orange{\fbox{k = -6}}$ $\blue\bigstar$

$\large{\purple{\underline{\tt{Solution\::-}}}}$

$\textbf{The\: given \:equations\: :-}$

$\dashrightarrow$ $\tt kx + 3y = 3$

$\dashrightarrow$ $\tt kx + 3y - 3 = 0-----(1)$

$\dashrightarrow$ $\tt 12x + ky = 6$

$\dashrightarrow$ $\tt 12x + ky - 6 = 0-----(2)$

$\textbf{This \:equations \:is \:of \:following \:form\: :-}$

$\tt a_{1}x + b_{1}y + c_{1} = 0, a_{2}x + b_{2}y + c_{2} = 0$

Where, $\tt a_1 = k, b_1 = 3, c_1 = -3\: and\: a_2 = 12, b_2 = k, c_2 = -6$

$\rule{220}3$

In order that the given system has no solution, we must have :-

$\longrightarrow$ $\boxed{\pink{\tt \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}}}$

$\longrightarrow$ $\tt \dfrac{k}{12} = \dfrac{3}{k} \neq \dfrac{-3}{-6}$

$\longrightarrow$ $\tt \dfrac{k}{12}\:and\: \dfrac{3}{k} \neq \dfrac{1}{2}$

$\longrightarrow$ $\tt k^{2} = 36 \:and\: k \neq 6$

$\longrightarrow$ $\tt k = \pm 36 \:and\: k \neq 6$

Hence, the given system of equations has no solution when k = -6.

$\rule{220}3$