4.4 Solve Systems of Equations with Three Variables

Solving Systems of Equations with Three Variables

Systems with three variables extend what you already know about two-variable systems. Instead of finding where two lines cross, you're now finding where three planes intersect in space. The logic is the same: reduce the system step by step until you can solve for each variable.

Verification of Ordered Triples

An ordered triple $(x, y, z)$ is a potential solution to a system of three equations. To verify whether it actually works, plug the values into every equation and check.

Steps to verify:

1. Substitute the values of $x$, $y$, and $z$ into the first equation. Simplify and check whether both sides are equal.
2. Repeat for the second equation.
3. Repeat for the third equation.
4. If all three equations produce true statements, the ordered triple is a solution. If even one equation fails, it's not.

Example: Suppose you're checking whether $(2, -1, 3)$ solves the system:

$x + y + z = 4$ $2x - y + z = 8$

$x + 2y - z = -3$

• Equation 1: $2 + (-1) + 3 = 4$ ✓
• Equation 2: $2(2) - (-1) + 3 = 4 + 1 + 3 = 8$ ✓
• Equation 3: $2 + 2(-1) - 3 = 2 - 2 - 3 = -3$ ✓

All three check out, so $(2, -1, 3)$ is a solution.

Methods for Three-Variable Systems

Elimination Method

The idea is to eliminate one variable at a time until you reduce the system to two equations with two unknowns, which you already know how to solve.

1. Pick a variable to eliminate first. Look for a variable that's easy to cancel (small or matching coefficients help).
2. Pair up two equations and multiply one or both by constants so the coefficients of your target variable are opposites. Add the equations to eliminate that variable. This gives you one new equation with two unknowns.
3. Pair up a different combination of two equations from the original system and eliminate the same variable. This gives you a second equation with two unknowns.
4. Solve the resulting two-equation, two-variable system using elimination or substitution.
5. Back-substitute the two values you found into any original equation to find the third variable.

The most common mistake here is eliminating a different variable in step 3 than in step 2. You must eliminate the same variable both times, or you won't get a clean two-variable system.

Substitution Method

This works best when one equation already has a variable isolated (or nearly isolated).

1. Solve one equation for one variable in terms of the other two. For example, $x = 2y + 3z$.
2. Substitute that expression into the other two equations. You now have two equations with two unknowns.
3. Solve that two-variable system using elimination or substitution.
4. Plug the two values back into the expression from step 1 to find the third variable.

Gaussian Elimination

This is a more systematic approach that uses an augmented matrix to organize the work.

1. Write the system as an augmented matrix $[A|b]$, where each row represents one equation.

2. Use elementary row operations to transform the matrix into row echelon form (upper triangular):

   • Swap two rows
   • Multiply a row by a nonzero constant
   • Add a multiple of one row to another row

3. The goal is to get zeros below each leading entry (pivot), creating a staircase pattern.

4. Use back-substitution starting from the bottom row to solve for each variable.

At the intermediate algebra level, you'll mostly use elimination and substitution. Gaussian elimination is worth understanding because it's the same logic in a more organized format, and you'll use it heavily in later courses.

Real-World Applications

Word problems with three unknowns follow a predictable setup process:

1. Identify three unknown quantities and assign a variable to each. For example: $x$ = number of apples, $y$ = number of bananas, $z$ = number of oranges.
2. Translate three distinct pieces of information from the problem into three equations. Each equation should capture a different relationship (total cost, total quantity, a ratio or comparison).
3. Solve the system using elimination, substitution, or Gaussian elimination.
4. Interpret and check your answer in context. Ask yourself: do the values make sense? Are quantities non-negative? Are they integers if the problem requires whole numbers?

A few things to watch for:

• If the system has no solution, the problem contains conflicting information (an inconsistent system).
• If the system has infinitely many solutions, there isn't enough independent information to pin down a unique answer (a dependent system). This usually means two of the equations are really saying the same thing.