Solve the following linear system of equations.

3x-2y+z=2
x-y+z=2
5x+10y-5Z=10

Take the last equation and divide all by 5 to simplify it, so:

3x-2y+z=2 ...(1)
x-y+z=2....(2)
x+2y-Z=2...(3)

By inspection, if we add (1) and (3), we eliminate y and z at the same time:
3x-2y+z+x+2y-z = 2+2
4x=4
x=1

Substitute x=1 into (2) and (3) to get:
-y+z=1 ...(2a)
2y-z=1 ....(3a)

Add (2a) and (3a):
y +0 = 2, so
y=2
Substitute x=1, y=2 in (2) to get
1-(2)+z = 2
so
z=3

Now substitute x=1, y=2, z=3 into equations (1), (2) and (3) as a check.

To solve the linear system of equations, we can use the method of elimination or substitution. I will use the method of elimination in this case.

Step 1: Write down the three equations of the system:
1) 3x - 2y + z = 2
2) x - y + z = 2
3) 5x + 10y - 5z = 10

Step 2: Multiply equation 2 by -3 to make the x term cancel out when combined with equation 1:
-3(x - y + z) = -3(2)
-3x + 3y - 3z = -6

The new equation 2 is: -3x + 3y - 3z = -6

Step 3: Add equation 1 and the new equation 2 to eliminate the x term:
3x - 2y + z + (-3x + 3y - 3z) = 2 + (-6)
-2y - 2z = -4

The new equation 3 is: -2y - 2z = -4

Step 4: Multiply equation 2 by 5 to make the z term cancel out when combined with equation 3:
5(x - y + z) = 5(2)
5x - 5y + 5z = 10

The new equation 2 is: 5x - 5y + 5z = 10

Step 5: Add equation 3 and the new equation 2 to eliminate the z term:
-2y - 2z + (5x - 5y + 5z) = -4 + 10
5x - 7y + 3z = 6

The new equation 3 is: 5x - 7y + 3z = 6

Now we have the following simplified system of equations:
1) 3x - 2y + z = 2
2) 5x - 5y + 5z = 10
3) 5x - 7y + 3z = 6

Step 6: Solve the simplified system of equations. I will use the method of elimination again.

Multiply equation 1 by 5 and equation 2 by 3 to make the x term and z term cancel out:
5(3x - 2y + z) = 5(2)
3(5x - 5y + 5z) = 3(10)

The new equation 1 is: 15x - 10y + 5z = 10
The new equation 2 is: 15x - 15y + 15z = 30

Subtract the new equation 2 from the new equation 1:
(15x - 10y + 5z) - (15x - 15y + 15z) = 10 - 30
5y - 10z = -20

The new equation 3 is: 5y - 10z = -20

Step 7: Solve for y in terms of z from the new equation 3:
5y = 10z - 20
y = 2z - 4

Step 8: Substitute the value of y into equation 1:
3x - 2(2z - 4) + z = 2
3x - 4z + 8 + z = 2
3x - 3z = -6

The new equation 1 is: 3x - 3z = -6

Step 9: Solve for x in terms of z from the new equation 1:
3x = 3z - 6
x = z - 2

Therefore, the solution to the linear system of equations is:
x = z - 2
y = 2z - 4

Now you can substitute any value for z and find the corresponding values of x and y to form a solution.

What have you done at school, substitution, elimination or echelon form?

substitution