Question

find the values of k so that the pair of equation x+2y = 5 and 3x+ky+15 = 0 has a unique solutions ​

Answer 1

Answer:

Step-by-step explanation:

a1/a2≠b1/b2

1/3≠2/k

So. K≠6.

i.e. K must not be equal to 6.

Answer 2

Answer: The value of k can't be 6.

Step-by-step explanation:

Since we have given that

x+2y=5-------(1)

3x+ky=15---------(2)

We know that it has a unique solution.

As we know that "Intersecting lines have unique solution".

Conditions of intersecting lines is :

$\frac{a_1}{a_2}\neq \frac{b_1}{b_2}\neq \frac{c_1}{c_2}\\\\\frac{1}{3}\neq \frac{2}{k}\neq \frac{5}{15}\\\\\frac{1}{3}\neq \frac{2}{k}\neq \frac{1}{3}\\\\\implies \frac{1}{3}\neq \frac{2}{k}\\\\k\neq 2\times 3\\\\k\neq 6$

Hence, the value of k cant be 6.