Solving the System of Equations for X, Y, Z

We are given the following system of three linear equations:

1. $X + 2Y + 4Z = 12$
2. $2X + 2Y + 3Z = 10$
3. $3X + 2Y + Z = 8$

The goal is to find the values of $X$, $Y$, and Z that satisfy all three equations simultaneously.

Step-by-Step Solution using Elimination

Notice that the coefficient of $Y$ is the same ($2$) in all three equations. This makes the elimination method particularly effective.

1. Eliminate Y using Equation 1 and Equation 2: Subtract Equation 1 from Equation 2.

   $(2X + 2Y + 3Z) - (X + 2Y + 4Z) = 10 - 12$

   $2X - X + 2Y - 2Y + 3Z - 4Z = -2$

   This simplifies to a new equation:

   $X - Z = -2$ (Equation 4)

2. Eliminate Y using Equation 2 and Equation 3: Subtract Equation 2 from Equation 3.

   $(3X + 2Y + Z) - (2X + 2Y + 3Z) = 8 - 10$

   $3X - 2X + 2Y - 2Y + Z - 3Z = -2$

   This results in another new equation:

   $X - 2Z = -2$ (Equation 5)

3. Solve the new system (Equation 4 and Equation 5): Now we have a system of two equations with two variables ($X$ and $Z$):

   • $X - Z = -2$ (Equation 4)
   • $X - 2Z = -2$ (Equation 5)

   Subtract Equation 5 from Equation 4 to solve for $Z$:

   $(X - Z) - (X - 2Z) = -2 - (-2)$

   $X - X - Z + 2Z = -2 + 2$

   $Z = 0$

4. Find X: Substitute the value $Z = 0$ back into Equation 4:

   $X - (0) = -2$

   $X = -2$

5. Find Y: Substitute the values $X = -2$ and $Z = 0$ back into the original Equation 1:

   $(-2) + 2Y + 4(0) = 12$

   $-2 + 2Y + 0 = 12$

   $2Y = 12 + 2$

   $2Y = 14$

   $Y = 14 / 2$

   $Y = 7$

The solution to the system of equations is X = -2, Y = 7, Z = 0.