anonymous
  • anonymous
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.
Mathematics
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SOLVED
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katieb
  • katieb
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Loser66
  • Loser66
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 ???
anonymous
  • anonymous
Thanks. So do you say the set of equations have infinitely many solutions??
Loser66
  • Loser66
if k =-10

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anonymous
  • anonymous
Oh yeah! got it ! Thanks!! :D Cheers.
Loser66
  • Loser66
yw

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