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

While doing problem sets, I came across this question:

Given the linear system:

$\displaystyle x + 2y = 10$

$\displaystyle 3x + (6+t)y = 30$

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:

$\displaystyle x + 2y = 10$

$\displaystyle 3x + 6y= 30$

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?

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

Quote:

Originally Posted by

**Pupil** ...

That would imply that:

$\displaystyle x + 2y = 10$

$\displaystyle 3x + 6y= 30$

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)

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

Quote:

Originally Posted by

**Pupil** While doing problem sets, I came across this question:

Given the linear system:

$\displaystyle x + 2y = 10$

$\displaystyle 3x + (6+t)y = 30$

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:

$\displaystyle x + 2y = 10$

$\displaystyle 3x + 6y= 30$

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, $\displaystyle x+2y=10\mbox{ and }3x+6y=30$ 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 $\displaystyle x+2y=10$ So there are infinity many solutions.

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

$\displaystyle 2x+4y=3\mbox{ and }x+2y=1$

Using the first equation you can get, $\displaystyle x+2y=\frac{3}{2}$. And if you compare this with the second equation it is clear that no values which satisfies both equations. If there is, $\displaystyle \frac{3}{2}=1$ which is contradictory.

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

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