this is the equation

x-y+3z=-4

4x +z=0

x+2y+z=8

I've solved to this point

4x+z=0

4x+z=0

I know this becomes 0=0, but where do I go from here. What would be the solution(s)

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- Apr 28th 2009, 08:21 PMnmoundlinear systems
this is the equation

x-y+3z=-4

4x +z=0

x+2y+z=8

I've solved to this point

4x+z=0

4x+z=0

I know this becomes 0=0, but where do I go from here. What would be the solution(s) - Apr 28th 2009, 08:35 PMpickslides
- Apr 28th 2009, 08:38 PMnmound
that isn't how you solve these equations....

- Apr 28th 2009, 09:41 PMpickslides
- Apr 28th 2009, 10:00 PMnmound
correction. its not the way that these are supposed to be solved. i used the addition method to get to the point stated, and it is at that point that i am stuck at

- Apr 29th 2009, 02:56 AMHallsofIvy
pickslides,

**If**it were true that deleting y from the equations resulted in the two equations 4x+ z= 0, 4x+ z= 0, then the matrix would NOT be invertible. That is the problem mmounds is compaining about. If that were the case then 4x+ z= 0 would be satisfied for any x, z satisfying z= -4x. Pick any value for x, and calculate z from that. Put that into either of the other two equations and solve for y: there exist an infinite number of solutions.

**However**while one equation certainly is 4x+ z= 0, the other two are x-y+3z=-4 and x+2y+z=8.

Multiply the first equation by 2 to get 2x- 2y+ 6z= -8. Add that to the second equation and y is eliminated: 2x+ x-2y+ 2y+ 6z+ z= -8+ 8 or 3x+ 7z= 0, NOT 4x+ z= 0. It is easy to see that the only x and z satisfying 4x+ z= 0 and 3x+ 7z= 0 is x= z= 0. Putting that into x-y+3z=-4 we get -y= -4 so y= 4. Putting it into x+2y+z=8, 2y= 8 which again is satisfied by y= 4. All three equations are satisfied by x= 0, y= 4, z= 0.

In fact, finding the inverse matrix is seldom the best way to solve matrix equations. Row reduction of the matrix is typically much simpler. - Apr 29th 2009, 04:40 AMnmound
Thank you for the response, i have a similar question

y=2x+4

10x-5y=-20

i know using the substitution method plays out like so...

10x-5(2x+4)=-20

10x-10x-20=-20

10x-10x=0

0=0

I have to express the answer in interval notation....how would I do that, and since the answer is 0=0 is their a value for x and y at all - Apr 29th 2009, 02:11 PMpickslides
- Apr 29th 2009, 06:07 PMlisczz
0 is always equal to 0 right? When solving for a variable and both sides of the equation are equal (like 0 = 0 or 5 = 5), the answer is "all real numbers."

$\displaystyle x$ can by any real number. Try plugging in any real number for $\displaystyle x$, and the equations will always hold true.

I think in interval notation, this would be $\displaystyle (-\infty, \infty)$ . - Apr 29th 2009, 06:16 PMpickslides
I realise that 0=0 but what does this actually say in context of what the problem is asking? I really don't think it offers anything. As the 2 equations are exactly the same then maybe there is really only 1 equation and the set of (x,y) solutions lie on that line itself.

- Apr 30th 2009, 03:05 AMHallsofIvy
If you were to choose

**any**value of x and then calculate y=2x+4, you would get x and y that satisfy the system of equations. For example, taking x= 1, y= 6 which satisfy both y= 2x+ 4 and 10x-5y=-20. If you take x= 5, y= 14 also satisfy both equations.

In general, if, solving a system of equations, you arrive at "0= 0" there are an infinite number of solutions. If you arrive at "a= 0" where a is not 0, there are no solutions.