Question

If the system of equations 6x - 2y = 3 and Kx - y = 2 has unique solution, then find the value of K

Answer 1

Hi ,

Given system of equations are

6x - 2y = 3

6x - 2y - 3 = 0 ----( 1 )

kx - y = 2

kx - y - 2 = 0 ----( 2 )

Compare above equations with

a1 x + b1 y + c1 = 0 and

a2 x + b2 y + c2 = 0 , we get

a1 = 6 , b1 = -2 , c1 = -3 ;

a2 = k , b2 = -1 , c2 = -2 ;

Now ,

a1/a2 ≠ b1/b2

[ Given they have Unique solution ]

6/k ≠ ( -2 )/( -1 )

6/k ≠ 2

k/6 ≠ 1/2

k ≠ 6/2

k ≠ 3

Therefore ,

For all real values of k , except k≠ 3,

Above equations has unique solution.

I hope this helps you.

: )

Answer 2

Heya !!!


Give equations are;


6X - 2Y = 3



6X - 2Y -3 = 0 -------(1)



And,



KX - Y -2 = 0 ----------(2)




These equations are of the form of A1X+B1Y+C = 0 and A2X + B2Y + C2 = 0



Where,



A1 = 6 , B1 = -2 and C = -3




A2 = K , B2 = -1 and C2 = -2





GIVEN that these equations has unique solution.

For unique solution we must have,


A1/A2 # B1/B2 [ Where # stand for not equal]




6/K # -2/-1




-6 # -2K



K = -6/-2



K # 3



Hence,


K has all real values , other than 3.




HOPE IT WILL HELP YOU..... :-)