Question

For what value of k the system of equation x-2y=3, 3x+ky=1 has a unique solution

Answer 1

Here your answer goes

Step :-1

let ,

x - 2y = 3 ------->> 1

3x + ky = 1 ------->> 2

Step :- 2

Compare equation (1) with $a_{1}x + b_{1}y = c_{1}$ and equation (2) with $a_{2}x +b_{2}y = c_{2}$

Where ,

$a_{1} = 1  ; \  b_1 = -2   ; \  c_1 = 3\\ a_2 = 3 ; \ b_2 = k  ; \ c_2 = 1$

Since , $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$

Equation has infinitely many Solutions

$\frac{1}{3} \neq \frac{-2}{k}$

⇒ k ≠ -6

Therefore , for all values of k expect -6 the system x-2y = 3 , 3x + ky = 1 has a unique Solution

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Answer 2

The two equations are

x−2y=3

3x+ky=1

Multiplying the first equation by 3, we get

3x−6y=9

Now, subtracting the first equation from the second, we get

3x+ky−(3x−6y)=1−9

Which reduces to

(k+6)⋅y=−8

From the above, we can see that, if k=−6, then the equation reduces to 0=−8, which is obviously impossible. Hence k≠−6.

For any other value of k, we can have a corresponding y=−8k+6 that satisfies the relation. And once y is found, x can be found, and its uniqueness can be verified.

So, to sum it up, the above simultaneous equations have a unique solution if and only if k≠−6.

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