Question

Find the value of k for which the system of equations x -2y=3 and 3x+ky=1 has a unique solution

Answer 1

Answer:

For all real values of k , (k≠6) given system of equations have unique solution.

Step-by-step explanation:

Compare given system of equations

x-2y=3=> x-2y-3=0 ---(1)

3x+ky=1=> 3x+ky-1=0 ---(2) with

$a_{1}x+b_{1}y+c_{1}=0$

$a_{2}x+b_{2}y+c_{2}=0,we\:get$

$a_{1}=1,b_{1}=-2,c_{1}=-3\\</p><p>a_{2}=3,b_{2}=k,c_{1}=-1$

$\frac{a_{1}}{a_{2}}≠\frac{b_{1}}{b_{2}}$

/* Given lines have unique solution*/

$\implies \frac{1}{3}≠\frac{-2}{k}$

$\implies 1\times k≠-2 \times 3$

$\implies k≠-6$

Therefore,

For all real values of k , (k≠6) given system of equations have unique solution.

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Answer 2

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