Ask Question
13 April, 12:01

Solve the system by elimination.

-2x+2y+3z=0

-2x-y+z=-3

2x+3y+3z=5

+5
Answers (1)
  1. 13 April, 15:06
    0
    -2x + 2y + 3z = 0 → 2x - 2y - 3z = 0 → 2x - 2y - 3z = 0

    -2x - 1y + 1z = - 3 → 2x + 1y - 1z = 3 → 2x + 1y - 1z = 3

    2x + 3y + 3z = 5 → 2x + 3y + 3z = 5 - 3y - 2z = - 3

    -2x + 2y + 3z = 0 → 2x - 2y - 3z = 0

    -2x - 1y + 1z = - 3 → 2x + 1y - 1z = 3 → 2x + 1y - 1z = 3

    2x + 3y + 3z = 5 → 2x + 3y + 3z = 5 → 2x + 3y + 3z = 5

    -2y - 4z = - 2

    -3y - 2z = - 3 → - 6y - 4z = - 6

    -2y - 4z = - 2 → - 2y - 4z = - 2

    -4y = - 4

    -4 - 4

    y = 1

    -3y - 2z = - 3

    -3 (1) - 2z = - 3

    -3 - 2z = - 3

    + 3 + 3

    -2z = 0

    -2 - 2

    z = 0

    -2x + 2y + 3z = 0

    -2x + 2 (1) + 3 (0) = 0

    -2x + 2 + 0 = 0

    -2x + 2 = 0

    - 2 - 2

    -2x = - 2

    -2 - 2

    x = 1

    (x, y, z) = (1, 1, 0)
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “Solve the system by elimination. -2x+2y+3z=0 -2x-y+z=-3 2x+3y+3z=5 ...” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers