Resolver el siguiente Sistema de ecuaciones y compruebe su solución: 3x+2y-z=2 x-4y+2z=1 2x-3z=4
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metodo de sustitucion
3x+2y-x=2
x-4y+2z=1
2x-3z=4
3 x+2 y-z = 2
x-4 y+2 z = 1
2 x-3 z = 4
3 x+2 y-z = 2
x = 1+4 y-2 z
2 x-3 z = 4
2 y+3 (1+4 y-2 z)-z = 2
x = 1+4 y-2 z
2 (1+4 y-2 z)-3 z = 4
2 y+3 (1+4 y-2 z)-z = 2 y+(3+12 y-6 z)-z = 3+14 y-7 z:
3+14 y-7 z = 2
x = 1+4 y-2 z
2 (1+4 y-2 z)-3 z = 4
2 (1+4 y-2 z)-3 z = (2+8 y-4 z)-3 z = 2+8 y-7 z:
3+14 y-7 z = 2
x = 1+4 y-2 z
2+8 y-7 z = 4
3+14 y-7 z = 2
x = 1+4 y-2 z
2+8 y-7 z = 4
14 y = 7 z-1
x = 1+4 y-2 z
2+8 y-7 z = 4
y = z/2-1/14
x = 1+4 y-2 z
2+8 y-7 z = 4
y = z/2-1/14
x = 1+4 y-2 z
2+8 (z/2-1/14)-7 z = 4
2+8 (z/2-1/14)-7 z = (4 z-4/7)-7 z+2 = 10/7-3 z:
y = z/2-1/14
x = 1+4 y-2 z
10/7-3 z = 4
y = z/2-1/14
x = 1+4 y-2 z
10/7-3 z = 4
y = z/2-1/14
x = 1+4 y-2 z
-3 z = 18/7
y = z/2-1/14
x = 1+4 y-2 z
z = -6/7
y = (-1)/2
x = 4 y+19/7
z = -6/7
y = -1/2
x = 5/7
z = -6/7
x = 5/7
y = -1/2
z = -6/7
3x+2y-x=2
x-4y+2z=1
2x-3z=4
3 x+2 y-z = 2
x-4 y+2 z = 1
2 x-3 z = 4
3 x+2 y-z = 2
x = 1+4 y-2 z
2 x-3 z = 4
2 y+3 (1+4 y-2 z)-z = 2
x = 1+4 y-2 z
2 (1+4 y-2 z)-3 z = 4
2 y+3 (1+4 y-2 z)-z = 2 y+(3+12 y-6 z)-z = 3+14 y-7 z:
3+14 y-7 z = 2
x = 1+4 y-2 z
2 (1+4 y-2 z)-3 z = 4
2 (1+4 y-2 z)-3 z = (2+8 y-4 z)-3 z = 2+8 y-7 z:
3+14 y-7 z = 2
x = 1+4 y-2 z
2+8 y-7 z = 4
3+14 y-7 z = 2
x = 1+4 y-2 z
2+8 y-7 z = 4
14 y = 7 z-1
x = 1+4 y-2 z
2+8 y-7 z = 4
y = z/2-1/14
x = 1+4 y-2 z
2+8 y-7 z = 4
y = z/2-1/14
x = 1+4 y-2 z
2+8 (z/2-1/14)-7 z = 4
2+8 (z/2-1/14)-7 z = (4 z-4/7)-7 z+2 = 10/7-3 z:
y = z/2-1/14
x = 1+4 y-2 z
10/7-3 z = 4
y = z/2-1/14
x = 1+4 y-2 z
10/7-3 z = 4
y = z/2-1/14
x = 1+4 y-2 z
-3 z = 18/7
y = z/2-1/14
x = 1+4 y-2 z
z = -6/7
y = (-1)/2
x = 4 y+19/7
z = -6/7
y = -1/2
x = 5/7
z = -6/7
x = 5/7
y = -1/2
z = -6/7
jesaca122785:
si cualquier método o sistema de ecuación
b) Inecuaciones
c) Valor Absoluto
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