Question

If the system of equation 3x - 2y - 7 = 0 and 6x + ky +11 = 0 has unique solution then​

Answer 1

Step-by-step explanation:

hope this helps you better

Answer 2

Complete 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:

Option (4) k is not equal to -4

Explanation:

Given that, 3x - 2y -7 = 0 and  6x + ky + 11 = 0.

• If and only if, the number of unknowns and the number of equations is equal, all equations are consistent, and there is no linear dependence between any two or more equations, all equations are independent, then there exists a unique solution to a set of linear simultaneous equations.

Condition for equation has a unique solution :

Let $a_1 x + b_1y + c_1 = 0$ and $a_2x +b_2y + c_2 = 0$ are two equations that

⇒if   $\frac{a_1}{a_2}$ ≠ $\frac{b_1}{b_2}$  it has a unique solution.

Step 1:

We have, 3x - 2y - 7 = 0 and 6x + ky + 11 = 0

so, $a_1 = 3, b_1 = -2 and c_1 = -7$ and $a_ 2 = 6, b_2 = k and c_2 = 11$

⇒ $\frac{3}{6} \neq \frac{-2}{k}$

⇒ $\frac{1}{2}$ ≠ $\frac{-2}{k}$

⇒ $k$ ≠ -4

Final answer:

Hence, k is not equal to -4 is the correct answer.

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