Solve the system using any method.

2x-2y+4z=2,

x+3y+z=8

3y-z=5

x = 1, y = 2, z = 1.

Step-by-step explanation:

From the third equation:

3y = z + 5

Substitute for 3y in the second equation:

x + z + 5 + z = 8

x + 2z = 3 ... (1)

Multiply the first equation by 3 and the second by 2 and add, we get:

6x - 6y + 12 z = 6

2x + 6y + 2z = 16 Adding:

8x + 14z = 22

4x + 7z = 11 ... (2)

Now multiply equation (1) by - 4:

-4x - 8z = - 12 ... (3) Adding (2) and (3):

-z = - 1

z = 1.

Substituting for z in (2):

4x + 7 (1) = 11

4x = 4

x = 1.

Now substitute x = 1 and z = 1 in the original first equation to find y:

2 (1) - 2y + 4 (1) = 2

-2y = 2 - 2 - 4 = - 4

y = 2.

Step-by-step explanation:

let's solve it by eliminating method

first we will eliminate x by using first and second equation

multiply second equation by - 2

-2 * (x+3y+z) = 8 ➡ - 2x - 6y-2z = - 16

now add it up to first equation

2x-2y+4z - 2x-6y-2z = 2 - 16 ➡ 2z - 8y = - 14

now using thus new equation with 3rd one to get rid of z

multiply 3rd equation by 2

2 * (3y-z) = 5 ➡ 6y - 2z = 10

add this to the new equation we found

6y - 2z + 2z - 8y = 10 - 14 ➡

y = 2

now use y to find the value of z in the 3rd equation

3*2 - z = 5

z = 1

lastly

x + 3*2 + 1 = 8

x = 1