Math
posted by Linda .
Solve the system using any algebraic method.
2x  2y + z = 3
5y  z = 31
x + 3y + 2z = 21
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The answer I got for this is ( 26, 6,67) but I know its wrong and I don't know why. Help please >< my work is at the bottom
I used substitution method changed 5yz=31 to z=5y+31 then I:
Plugged it in the first equation=
2x2y+5y+31=3 > 2x+3y+31=3 > 2x+3y=28 is my new 1st equation...
Plugged in z=5y+31 to equation 3 =
x+3y+2(5y+31)=21 > x+3y+10y+62=21 > x+13y+62=21 > x+13y=83 is my new 2nd equation.
2(=13y=82)+3y=28 > 26166+3y=28 > 23y=138  > y=6
2x+3(6)=28 > 2x+18=28 > 2x=46 > x=23
2(26)2*6)+z=3 > 5212+z=3 > 64+z=3 > z=67

5y  z = 31
z = 5y + 31
2x  2y + z = 3
( Now put z = 5y + 31 in equation )
2x  2y + 5y + 31 = 3
2x + 3y = 3  31
2x + 3y =  28
2x =  28  3y Divide both sides with 2
x =  14  ( 3 / 2 ) y
x + 3y + 2z = 21
( Now put x =  14  ( 3 / 2 ) y and z = 5y + 31 in equation )
 14  ( 3 / 2 ) y + 3y + 2 ( 5y + 31 ) =  21
 14  ( 3 / 2 ) y + 3y + 10y + 62 =  21
 ( 3 / 2 ) y + 13y =  21 + 14  62
 ( 3 / 2 ) y + ( 26 / 2 ) y =  69
( 23 / 2 ) y =  69 Multiply both sides with 2
23y =  138 Divide both sides with 23
y =  6
z = 5y + 31
z = 5 ( 6 ) + 31
z =  30 + 31
z = 1
x =  14  ( 3 / 2 ) y
x =  14  ( 3 / 2 ) 6
x =  14 + 18 / 2
x =  14 + 9
x =  5
So solutions are :
x =  5
y =  6
z = 1