Question

Find the value of k for which the quadratic equations given below has a unique soln y-x=6, 3kx+2y=7

Answer 1

If  a1 /  a2 ≠  b1 /  b2 then the pair of linear equations a1x + b1y + c1 = 0

a2x+ b2y + c2 = 0  has a unique solution.

Given pair of  equations K x + 3y = K-3, 12x + Ky = K.

Here a1 = k ,b1 = 3 , a2 = 12 , b2 = k .

k / 12 ≠ 3 / k

k2 ≠ 36.

k ≠ ± 6.

THANKS!!!

Answer 2

Answer:

Value of k can be anything other than-2/3.

Step-by-step explanation:

Given System of Equations are

y - x = 6

3kx + 2y = 7

System of equation has unique solution.

To find: value of k

Rewrite in Standard form of equation,

-x + y -6 = 0

3kx + 2y - 7 = 0

So we have following relationship between ratios of their coefficient for the unique solution,

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

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

3k ≠ -2

k ≠ -2/3

Therefore, Value of k can be anything other than-2/3.