Question

find the value of k so that the pair of linear equation has unique solution 2x-ky=11 and x-2y=-12

Answer 1

the required answer is here.

Answer 2

Answer:

The value of k is any real number other then -4.

Step-by-step explanation:

Given : The pair of linear equation has unique solution $2x-ky=11$ and $x-2y=-12$

To find : The value of k?

Solution :

The condition of unique solution is

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

The system of linear equations are of the form

$a_1x+b_1y+c_1=0$,  $a_2x+b_2y+c_2=0$

On comparison with the given linear equations,

$a_1=2,b_1=-k,c_1=-11\\a_2=1,b_2=-2,c_2=-12$

$2x-ky=11$   $a_1x+b_1y+c_1=0$   a₁x +b₁y+c₁ = 0                  

For these value condition should be satisfied,

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

$2\neq \frac{k}{2}$

$k\neq 4$

Therefore , The value of k is any real number other then -4.