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24 November, 16:37

Solve the system using any method.

2x-2y+4z=2,

x+3y+z=8

3y-z=5

+1
Answers (2)
  1. 24 November, 16:44
    0
    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.
  2. 24 November, 16:45
    0
    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
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