Use the method of elimination to determine whether the given linear system is consistent or inconsistent. If it is consistent, find the solution if it is unique, otherwise describe the infinite solution set in terms of an arbitrary parameter t. (Note that a system of equations is said to be consistent if it has at least one solution) 4x - 2y + 6z = 0
-y-z=0
- y + 3z = 0

Step-by-step explanation:

From the given information;

We will realize that in the 2nd & 3rd equation, the integer "x" is missing.

So, the correct equation should be:

4x - 2 + 6z = 0

x - y - z = 0

2x - y + 3z = 0

By rearrangement; if we switch equation (1) and (2)

x - y - z = 0       (1)

4x - 2 + 6z = 0 (2)

2x - y + 3z = 0   (3)

If we multiply equation (1) and add that to  equation (2);

Then, we have:

x - y - z = 0

2y + 10z = 0

2x - y + 3z = 0

Also, if we multiply equation (1) by -2 and add that to equation (3), we have:

x - y - z = 0

2y + 10z = 0

y + 5z = 0

From above recent equation, if we swap equation (2) and (3);

x - y - z = 0

y + 5z = 0

2y + 10z = 0

Then, multiply equation (2) with -2, followed by adding it to equation(3);

Then:

x - y - z = 0

y + 5z = 0

0 = 0

Now, there exists no equation (3). Thus, let assume z = t to solve for x & y

y + 5z = 0

y = -5t

Similarly;

x - y - z = 0

x = -5t + t

x = -4t

Hence;

x = -4t; y = -5t; &  z = t

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