Question

Find the value of k for which the given system of equations has no solution.$$kx + 3y = k - 3; 12x + ky = k$$

A
k=5
B
k=2
C
k=3
D
None of these

Solution

The correct option is C None of theseA system of equations, $$a_1x+b_1y+c_1=0$$ and $$a_2x+b_2y+c_2=0$$ has no solution for the following condition:$$\displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$$The equations in the given question are:$$kx+3y-k+3=0$$ and $$a_{1}=k,b_{1}=3,c_{1}=3$$$$12x+ky-k=0$$ and $$a_{2}=12,b_{2}=k,c_{2}=-k$$Substituting the above values in the condition mentioned above for no solution:$$\dfrac{a_{1}}{b_{1}}=\dfrac{a_{2}}{b_{2}}$$$$\dfrac{k}{12}=\dfrac{3}{k}$$$$k^{2}=36$$$$\Rightarrow k=6$$Mathematics

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