For what value of k y - x = 6 and 3 k x + 2 y = 7 have a unique solution
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Any system of linear equations of the form
a1x+b1y=c1
a2x+b2y=c1
has a unique solution only if the determinant
∣∣∣a1a2b1b2∣∣∣≠0
i.e.
a1b2≠a2b1
So, in order for the given system of equations
x+2y=3
5x+ky=3
to have a unique solution,
∣∣∣152k∣∣∣≠0
i.e.
k≠10
In other words, the given system of equations has a unique solutions for any value of k other than k=10.
Why is there a problem when k=10?
If k=10, then the second equation becomes
5x+10y=3
i.e.
x+2y=35
But the first equation is
x+2y=3
So we would have no solution to the system of equations if
k=10.
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