To solve the system of equations, we have the following three equations:
- 4x + 2y + z = 5
- x + 3y + 2z = 3
- 3x + y + 2z = 5
We can solve this system using the method of substitution or elimination. Here, we will use the elimination method for clarity.
First, let’s label our equations for easy reference:
- Equation (1): 4x + 2y + z = 5
- Equation (2): x + 3y + 2z = 3
- Equation (3): 3x + y + 2z = 5
From Equation (1), we can express z in terms of x and y:
z = 5 – 4x – 2y
Now, we can substitute this expression for z into Equations (2) and (3).
Substituting into Equation (2):
x + 3y + 2(5 - 4x - 2y) = 3
x + 3y + 10 - 8x - 4y = 3
-7x - y + 10 = 3
-7x - y = -7
7x + y = 7 (Equation 4)
Next, substituting into Equation (3):
3x + y + 2(5 - 4x - 2y) = 5
3x + y + 10 - 8x - 4y = 5
-5x - 3y + 10 = 5
-5x - 3y = -5
5x + 3y = 5 (Equation 5)
Now we have a new system of two equations:
- Equation (4): 7x + y = 7
- Equation (5): 5x + 3y = 5
Next, we can express y from Equation (4):
y = 7 – 7x
Now substitute y into Equation (5):
5x + 3(7 - 7x) = 5
5x + 21 - 21x = 5
-16x + 21 = 5
-16x = -16
x = 1
Now that we found x, we can substitute it back into Equation (4) to find y:
7(1) + y = 7
7 + y = 7
y = 0
Finally, substitute x and y back into the expression for z:
z = 5 - 4(1) - 2(0)
z = 5 - 4 = 1
In conclusion, the solution to the system of equations is:
- x = 1
- y = 0
- z = 1
Thus, the solution is (x, y, z) = (1, 0, 1).