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حل نظام معادلات بثلاث متغيرات

Quiz by amal

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5 questions
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  • Q1
    ما هو الحل للمعادلات التالية: \( x + y + z = 6 \)، \( 2x - y + 3z = 14 \)، \( -x + 4y + z = 2 \)؟
    (x = 1, y = 2, z = 3)
    (x = 0, y = 1, z = -1)
    (x = 2, y = 3, z = 4)
    (x = 5, y = -1, z = 2)
    30s
  • Q2
    إذا كانت المعادلات \( 3x - 2y + z = 7 \)، \( x + 4y - z = 1 \)، \( 2x - y + 3z = 12 \)، فما هي قيمة \( z \)؟
    z = 5
    z = 2
    z = 4
    z = 3
    30s
  • Q3
    إذا كان لدينا نظام المعادلات \( x - 2y + 3z = 0 \)، \( 2x + y - z = 1 \)، \( x + 3y + 2z = 9 \)، فما هو الحل الصحيح لهذا النظام؟
    (x = 3, y = 2, z = 0)
    (x = 2, y = -1, z = 1)
    (x = 0, y = 1, z = 3)
    (x = 1, y = 2, z = -1)
    30s
  • Q4
    أي من الخيارات التالية يمثل حلاً للنظام \( 2x + y - z = 4 \)، \( x - 3y + z = -2 \)، \( 3x + 2y + 2z = 10 \)؟
    (x = 0, y = 2, z = -1)
    (x = 1, y = -2, z = 3)
    (x = 2, y = 1, z = 0)
    (x = -1, y = 1, z = 2)
    30s
  • Q5
    ما هو المتغير الذي يمكن التخلص منه بسهولة ضمن نظام المعادلات \( x + y + z = 6 \)، \( 3x - y + 2z = 7 \)، \( 2x - y - z = 1 \)؟
    z
    y
    لا يمكن التخلص من أي متغير بسهولة
    x
    30s

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