1. ## linear 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)

2. Originally Posted by nmound
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)
I would solve this by finding the inverse of the co-efficent matrix,

Start with $Ax = b \Rightarrow x = A^{-1}b$

otherwise you will have to use muliple substitutions.

3. that isn't how you solve these equations....

4. Originally Posted by nmound
that isn't how you solve these equations....

It isn't the way to solve these equations?

Or it isn't the preferred way you would like to see this system of equations solved?

5. 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

6. 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.

7. 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

8. Originally Posted by nmound
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
I think it's important to look at the second equation

$10x-5y=-20$

Now making y the subject,

$5y=10x+20$

$y=2x+4$

which is the same equation as the first.

This is why you found 0=0.

9. Originally Posted by nmound
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
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."
$x$ can by any real number. Try plugging in any real number for $x$, and the equations will always hold true.
I think in interval notation, this would be $(-\infty, \infty)$ .

10. Originally Posted by lisczz
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."
$x$ can by any real number. Try plugging in any real number for $x$, and the equations will always hold true.
I think in interval notation, this would be $(-\infty, \infty)$ .
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.

11. 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.