To solve the system of equations:
- 2x + 3y + z = 1
- 3x + y + 2z = 12
- x + 2y + 3 = 5
We will use the method of substitution or elimination. Let’s first rewrite the equations in a more standard format:
- Equation 1: 2x + 3y + z = 1
- Equation 2: 3x + y + 2z = 12
- Equation 3: x + 2y = 2
Next, we’ll express z in terms of x and y from Equation 1:
z = 1 – 2x – 3y
Now, we can substitute this expression for z into Equations 2:
- 3x + y + 2(1 – 2x – 3y) = 12
- 3x + y + 2 – 4x – 6y = 12
- -x – 5y + 2 = 12
- -x – 5y = 10
- x + 5y = -10
Now we substitute z into Equation 3 as well:
- x + 2y = 2
We now have a new system of equations to solve:
- Equation 4: x + 5y = -10
- Equation 5: x + 2y = 2
We can subtract Equation 5 from Equation 4:
- (x + 5y) – (x + 2y) = -10 – 2
- 3y = -12
- y = -4
Now that we have y, we can substitute -4 back into Equation 5 to find x:
- x + 2(-4) = 2
- x – 8 = 2
- x = 10
Finally, we substitute x = 10 and y = -4 into the expression for z:
- z = 1 – 2(10) – 3(-4)
- z = 1 – 20 + 12
- z = -7
So the solution to the system of equations is:
- x = 10, y = -4, z = -7