2x+2y+3z=-1
4x+4y+3z=-1
4x+2y+2z=0
To solve this system of equations, we can use the method of elimination or substitution. Let's use the method of elimination:
First, let's eliminate the x variable.
To do that, we can multiply the first equation by -2 and the third equation by 2, then add them together:
-4x - 4y - 6z = 2
+ 4x + 2y + 2z = 0
---------------------
-2y - 4z = 2
Now, let's eliminate the y variable.
To do that, we can multiply the first equation by -1 and the third equation by 2, then add them together:
-2x - 2y - 3z = 1
+ 4x + 2y + 2z = 0
-----------------------
2x - z = 1
We now have two equations left:
-2y - 4z = 2
2x - z = 1
Let's solve the second equation for x:
2x = 1 + z
x = (1 + z)/2
Now let's substitute this expression for x into the first equation:
-2y - 4z = 2
-2y - 4z = 2(1 + z)/2
-2y - 4z = 1 + z
-2y - 5z = 1
Now, let's solve the second equation for y:
-2y = -5z - 1
y = (-5z - 1)/(-2)
y = (5z + 1)/2
Now, we have expressions for x and y in terms of z. Let's substitute these values into any of the original equations. For simplicity, let's use the third equation:
4x + 2y + 2z = 0
4((1 + z)/2) + 2((5z + 1)/2) + 2z = 0
2 + 2z + 5z + 1 + 2z = 0
9z + 3 = 0
9z = -3
z = -3/9
z = -1/3
Substituting this value for z into the expressions we found earlier for x and y:
x = (1 + (-1/3))/2
x = (1/3)/2
x = 1/6
y = (5(-1/3) + 1)/2
y = (-5/3 + 1)/2
y = (-5/3 + 3/3)/2
y = (-2/3)/2
y = -1/3
Therefore, the solution to the system of equations is x = 1/6, y = -1/3, and z = -1/3.