Thread: System of equations: Infinitely many, unique, or no solution.

While doing problem sets, I came across this question:

Given the linear system:



a) Determine a particular value of t so that the system has infinitely many solutions.
b) Determine a particular value of t so that the system has a unique solution.

According to the book, the value for t so that the system has infinitely many solutions is t = 0, and for a unique solution any real number that isn't 0.

That would imply that:


has infinitely many solutions, but if you multiply the first equation by 3 and then proceed to eliminate, nothing would remain. So, wouldn't this be a linear system with no solution (inconsistent system) rather than infinitely many? And for any nonzero number, wouldn't it likewise have no solution?

Originally Posted by Pupil

...
That would imply that:


has infinitely many solutions, but if you multiply the first equation by 3 and then proceed to eliminate, nothing would remain. So, wouldn't this be a linear system with no solution (inconsistent system) rather than infinitely many? And for any nonzero number, wouldn't it likewise have no solution?

The solutions given is correct. You would have 0 = 0 if you eliminated everything, and this is typical of a consistent (dependent) system. Now, if you had something like

0 = 2 (an always false statement), that would imply "no solution".

For your second issue, let's solve the first for x...
x = 10 - 2y
sub into the second.
3(10 - 2y) + (6 + t)y = 30
30 - 6y + 6y + ty = 30.
Everything cancels except for
ty = 0
If t = 0 (as in the above), there are infinite solutions, since 0*y is true for all y.
If t is NOT zero, then ty = 0 has the UNIQUE solution when y = 0 (since we're in a field)

Originally Posted by Pupil

While doing problem sets, I came across this question:

Given the linear system:



a) Determine a particular value of t so that the system has infinitely many solutions.
b) Determine a particular value of t so that the system has a unique solution.

According to the book, the value for t so that the system has infinitely many solutions is t = 0, and for a unique solution any real number that isn't 0.

That would imply that:


has infinitely many solutions, but if you multiply the first equation by 3 and then proceed to eliminate, nothing would remain. So, wouldn't this be a linear system with no solution (inconsistent system) rather than infinitely many? And for any nonzero number, wouldn't it likewise have no solution?

Hi Pupil,

The two equations, are both equivalent. That is when you multiply the first one with 3 you get the second one. I suppose you have mentioned "elimination" without seeing this. Since there is only one equation; x or y cannot be eliminated as you do in solving simultaneous equations. Now if you take any real number for x you can find the value of y. For example, (1,4.5), (2,4), (3,3.5) are some of the solutions of So there are infinity many solutions.

A inconsistent system does not have a solution. For example consider the system,



Using the first equation you can get, . And if you compare this with the second equation it is clear that no values which satisfies both equations. If there is, which is contradictory.

Thanks both of you for all the help. I learned more in those two posts than a 2 hour lecture with my prof.