Systems of Three Linear Equations

A system of three linear equations involves three variables, typically $x$, $y$, and $z$. To find the solution, you need to find the specific values for all three variables that make all three equations true at the same time.

The Elimination Method

The most reliable way to solve a $3 \times 3$ system is to reduce it to a $2 \times 2$ system using the elimination method.

1. Pick two pairs of equations from the system.
2. Eliminate the exact same variable from both pairs. This leaves you with two new equations that only have two variables.
3. Solve the new $2 \times 2$ system using standard elimination or substitution.
4. Back-substitute the two values you just found into one of the original three-variable equations to find the final unknown.

Example 1: Solving Step-by-Step

Solve the system: $1)\; x + y + z = 6$ $2)\; 2x - y + z = 3$ $3)\; x + 2y - z = 2$

Step 1: Eliminate $z$ using two different pairs of equations. First, add Equation (1) and Equation (3): $(x + y + z) + (x + 2y - z) = 6 + 2$ $2x + 3y = 8 \quad \text{--- (Equation A)}$

Next, add Equation (2) and Equation (3): $(2x - y + z) + (x + 2y - z) = 3 + 2$ $3x + y = 5 \quad \text{--- (Equation B)}$

Step 2: Solve the new $2 \times 2$ system. We now have a system with just $x$ and $y$: $A)\; 2x + 3y = 8$ $B)\; 3x + y = 5$

From Equation B, we can express $y$ in terms of $x$: $y = 5 - 3x$

Substitute this into Equation A: $2x + 3(5 - 3x) = 8$ $2x + 15 - 9x = 8$ $-7x = -7 \implies x = 1$

Now, find $y$: $y = 5 - 3(1) = 2$

Step 3: Back-substitute to find $z$. Plug $x = 1$ and $y = 2$ into original Equation (1): $1 + 2 + z = 6$ $3 + z = 6 \implies z = 3$

Solution: $(x, y, z) = (1, 2, 3)$.

Example 2: Working with Multipliers

Solve the system: $1)\; 2x + y - z = 1$ $2)\; x - y + 2z = 5$ $3)\; 3x + 2y + z = 8$

Step 1: Eliminate $z$. Add Equation (1) and Equation (3): $(2x + y - z) + (3x + 2y + z) = 1 + 8$ $5x + 3y = 9 \quad \text{--- (Equation A)}$

To eliminate $z$ from Equations (1) and (2), multiply Equation (1) by $2$: $4x + 2y - 2z = 2$ Add this to Equation (2): $(4x + 2y - 2z) + (x - y + 2z) = 2 + 5$ $5x + y = 7 \quad \text{--- (Equation B)}$

Step 2: Solve the $2 \times 2$ system. $A)\; 5x + 3y = 9$ $B)\; 5x + y = 7 \implies y = 7 - 5x$

Substitute into Equation A: $5x + 3(7 - 5x) = 9$ $5x + 21 - 15x = 9$ $-10x = -12 \implies x = 1.2$

Find $y$: $y = 7 - 5(1.2) = 7 - 6 = 1$

Step 3: Back-substitute to find $z$. Plug $x = 1.2$ and $y = 1$ into Equation (1): $2(1.2) + 1 - z = 1$ $2.4 + 1 - z = 1$ $3.4 - z = 1 \implies z = 2.4$

Solution: $(x, y, z) = (1.2, 1, 2.4)$.