Question

If the system of equation 3x-2y-7 =0 and 6x+ky+11=0 has unique solution,then
1)k=4
2)k not equal to 4
3)k= -4
4)k is not equal to -4​

Answer 1

AnswEr:-

Your Answer is Option 2:- k is not equal to -4.

(Note:- The answer of this question is also option,1).

ExplanaTion:-

Given pair of linear equations:-

$\tt\longrightarrow3x - 2y + 7 = 0...(1) \\ \\ \tt\longrightarrow6x + ky + 11 = 0...(2)$

• Also it is given that both the equations has unique solution.

And we have to find out the correct option.

Concept Used:-

If a pair of linear equation has unique solution then,$\tt\dfrac{a_1}{a_2}$ sould not be equal to $\tt\dfrac{b_1}{b_2}$.

Where,

• $\tt a_1\:\:And\:\:a_2$ are the coefficient of x of eq(1) and eq(2) respectively.
• $\tt b_1\:\:And\:\:b_2$ are the coefficient of y of eq(1) and eq(2) respectively.

So Here,

$\tt\longrightarrow a_1 = 3. \\ \\ \tt\longrightarrow a_2 = 6. \\ \\ \tt\longrightarrow b_1 = -2 .\\ \\ \tt\longrightarrow b_2 = k.$.

Thus,

It is given that the pair of linear equation have a unique solution so,

$\tt\mapsto\dfrac{\cancel3}{\cancel6} \:Should \:\:Not = \dfrac{-2}{k}. \\ \\ \tt\mapsto\dfrac{1}{2} \:Should \:\:Not = \dfrac{-2}{k}. \\ \\ \tt\mapsto 1(k) \:Should \:\:Not = -2(2) .\\ \\ \tt\mapsto k \:Should \:\:Not = -4.$

$\tt\therefore$ All values of k are possible except -4.

So here both option 1 and option 2 are correct.

Answer 2

Given : system of equations 3x-2y-7=0 and 6x+ky+11=0 has unique solution  

To Find : Correct option :

1)k=4

2)k not equal to 4

3)k= -4

4)k is not equal to -4​

Solution:

Pair of linear equations

a₁x  +  b₁y + c₁  =  0

a₂x  +  b₂y + c₂  =  0

Consistent

if a₁/a₂ ≠ b₁/b₂   (unique solution  and lines intersects each others)

  a₁/a₂ = b₁/b₂ = c₁/c₂   (infinite solutions and line coincide each other )

Inconsistent

if  a₁/a₂ = b₁/b₂ ≠  c₁/c₂  ( No solution , lines are parallel to each other)

3x-2y-7=0

6x+ky+11=0

3/6  ≠  - 2/k

=> 1/2  ≠  - 2/k

=> k ≠  -4

k not equal to -4 is correct option

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