find the values of Alpha for which the system of equation 6 x minus 2 Y is equal to 3 and Y is equal to Alpha x minus 2 has a unique solution
Answers
Answered by
0
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.
: )
Similar questions