algebra 2?

Question

x+y+2z=1
x-y+z=0
3x+3y+6z=4

all the variables cancel out and i dont know whatthe answer is.

im trying to solve for x, y, & z
its linear equations in three variables

Answer 1

x+y+2z=1..........equation(1)
x-y+z=0 ..........equation(2)
3x+3y+6z=4 equation(3)
Subtractin (2) from (1) we get,
2y+z=1................(equation(4)
Multiplying (2) by 3 and subtracting it from (3),we get,
6y+3z=4...............equation(5)
Now multiplying equation no.4 by 3 and subtracting (5) from it,we get ALL VARIABLES ELLININATED.THE PRBLEM IS WRONG

Answer 2

If all the variables cancel out, then two of your equations are equivalent, and you will have a line, or, if you simplify down to something along the lines of 0 = 4, then there is no answer and x,y,z = Ø

Answer 3

x-y+z=0 SO y=x+z
PUT IT IN THE FIRST EQUATION
x+y+2z=1
x+(x+z)+2z=1
2x+3z=1
2x=1-3z
x=(1-3z)/2
AND THEN IN THE SECOND
3x+3y+6z=4
3x+3(x+z)+6z=4
3x+3x+3z+6z=4
6x+9z=4
6x=4-9z
x=(4-9z)/6
OK NOW SAY THAT X=X
x=(1-3z)/2 AND x=(4-9z)/6
(1-3z)/2 = (4-9z)/6
6(1-3z)=2(4-9z)
6-18z=8-18z
OUPS YOU'RE RIGHT!
so maybe you can't solve it?

Answer 4

what are you trying to do? solve for them?