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2009-09-29T18:50:25+02:00
¹ x-y-z=2
² x+y-z=4
³ x+y+z=6
z 1 i 2 wyznaczamy x i przyrównujemy
x=2+y+z
x=4-y+z
2+y+z=4-y+z
2+y=4-y
2y=2
y=1
podstawiam do 1-go
x=2+1+z
x=3+z
podstawiam do 3-go
3+z+1+z=6
2z=2
z=1
x=4
zad 2
x=1-y-z
x=3+z-y
1-y-z=3+z-y
1-z=3+z
-2=2z
z=-1
x=1-y+1
x=2-y
2-y+2y-3=0
y=1
x=1
1 5 1
2009-09-29T19:14:09+02:00
Zad.2
{x-y-z =2 x= 2+y+z
{x+y-z =4
{x+y+z =6

(2+y+z) + y + z = 6
2+2y+2z= 6 / -2
2y+2z = 4 / - 2z
2y = 4 - 2z / :2
y = 2-z
x = 2 + ( 2-z) +z = 4
4 + (2-z) - z = 4
6 - 2z = 4 / - 6
-2z = -2 / : -2
z = 1 y = 2 -1 = 1
z=1 y = 1 x=4

zad.3
x+2y+3z=0 x= -2y - 3z
x+y+z=1
x+y-z=3

(-2y -3z) +y +z = 1
-y - 2z = 1
y = -2z - 1
x = -2( -2z -1) - 3z = 4z +2 -3z = 2 + z
2+z -2z - 1 -z = 3
1 -2z = 3
-2z = 2 / : -2
z=1 x=2+1=3 y=-2-1=-3
z=1 x=3 y=-3