Question

Find k if 2x+3y-5=0,kx-6y-8=0 has a unique solution

Answer 1

here is your answer by Sujeet
for unique solution, we must have,
a1/a2= b1/b2
2/3=k/-6
3k=-12
k=-12/3
k=-4

that's all

Sujeet

Answer 2

Hello,

It is mentioned that the linear equation system is having a unique solution.

The criteria for having a unique solution is

$\frac{a1}{a1} \: \: \: not \: equal \: \: to \: \: \frac{b1}{b2}$
Where,

A is the coefficients of x and b is the coefficient of y terms.

So,
The system of linear equation is as follows :-

$2x + 3y - 5 = 0 \\ \\ kx - 6y - 8 = 0$

Now
Using the formula and putting the value in it
We get :-

$\frac{2}{k} = \frac{3}{( - 6)} \\ \\ \frac{2}{k} = \frac{1}{( - 2)} \\ \\ k = 2 \times ( - 2) \\ \\ k = ( - 4)$

This means that the value of k is equal to (-4)


Hope this will be helping you.....