Question

I know these type of questions are easy but I guess this one is quite tricky for me.Please Help Me.

Find 'k' for which the system of equations has a 'Unique' solution.

3x-y-5=0; 6x-2y+k=0.

My doubt is that since they have a unique solution a1/a2 not equal to b1/b2. But in this case a1/a2 is equal to b1/b2.

Answer 1

your system is equal 3x-y = 5
and 6x -2y =-k
and find k to have only one solution.
time 2 to the first equation, you have 6x -2y =10 . so if -k =10 you have many solutions for the system, because at that moment y= 3x -5, to every value of x, you have correspondent y. For example, if x =1, y = -2
x=2, y = 1 etc...
so, to avoid it, to get rid of that situation, k must be ???

Answer 2

Thanks. So do you say the set of equations have infinitely many solutions??

Answer 3

if k =-10

Answer 4

Oh yeah! got it ! Thanks!! :D

Cheers.

Answer 5

yw