Question

For what value of K, system of equation 5x-7y-5 =0

and 2x+ky-1=0, have unique solution.​

Answer 1

Given:-

• system of equations 5x-7y-5 =0  and 2x+ky-1=0, have unique solution.​

To find:-

• The value of k.

Concept:-

For a system of equations :

$=> \sf a_{1} x + b_1 y + c_1 = 0 \\\\ & => a_2 x + b_2 y + c_2 = 0$

They will have unique solution if :

$=>\sf \frac{a_1}{a_2} \neq \frac{b_1}{b_2}$

Solution:-

Here,

$=> \sf a_1 = 5 , a_2 = 2 \\\\=> b_1 = -7 , b_2 = k \\\\=> c_1 = -5 & c_2 =-1$

Now,

$\implies\sf \frac{a_1}{b_1} \neq \frac{b_1}{b_2}\\\\\\ \implies \frac{5}{2} = \frac{-7}{k}\\\\\\ \implies 5k \neq -14$

Hence,  the given system of equations will have a unique solution for all real values of k other than -14/5 .

$\therefore \boxed{k \neq -14/5}$

_______________________________

Answer 2

Given:-

system of equations 5x-7y-5 =0  and 2x+ky-1=0, have unique solution.

To find:-

The value of k.

Concept:-

For a system of equations :

$=> \sf a_{1} x + b_1 y + c_1 = 0 \\\\ & => a_2 x + b_2 y + c_2 = 0$

They will have unique solution if :

$=>\sf \frac{a_1}{a_2} \neq \frac{b_1}{b_2}$

Solution:-

Here,

$=> \sf a_1 = 5 , a_2 = 2 \\\\=> b_1 = -7 , b_2 = k \\\\=> c_1 = -5 & c_2 =-1$

Now,

$\implies\sf \frac{a_1}{b_1} \neq \frac{b_1}{b_2}\\\\\\ \implies \frac{5}{2} = \frac{-7}{k}\\\\\\ \implies 5k \neq -14$

Hence,  the given system of equations will have a unique solution for all real values of k other than -14/5 .

$\therefore \boxed{k \neq -14/5}$

_______________________________