11.2 Systems of Linear Equations: Three Variables

Solving Systems of Linear Equations with Three Variables

Solving three-variable linear systems

A system of three linear equations with three variables looks like three equations, each containing some combination of $x$, $y$, and $z$. Your goal is to find the values of all three variables that satisfy every equation at once. Two main methods get you there: elimination and substitution.

Elimination method

The core idea is to strategically combine equations so that variables cancel out, reducing the system step by step until you can solve for a single variable.

1. Pick a variable to eliminate first (say $z$). Multiply equations by constants so that when you add or subtract two equations, $z$ drops out. Do this with two different pairs of equations.
2. You now have two equations with only $x$ and $y$. Use elimination again to remove one of those variables.
3. Solve the resulting single-variable equation.
4. Substitute that value back into one of your two-variable equations to find the second variable.
5. Substitute both known values into any original equation to find the third variable.

Substitution method

1. Pick one equation and solve it for one variable in terms of the other two (for example, express $x$ in terms of $y$ and $z$).
2. Substitute that expression into the other two equations. You now have a two-variable system.
3. Solve the two-variable system using either substitution or elimination.
4. Plug the two known values back into your expression from step 1 to find the third variable.

Both methods give the same answer. Elimination tends to be cleaner when coefficients line up nicely; substitution works well when one variable is already isolated or has a coefficient of 1.

Consistency of three-equation systems

Not every system has a neat single answer. Before you solve, it helps to know what kinds of outcomes are possible.

• Consistent and independent: The system has exactly one solution, an ordered triple $(x, y, z)$. Geometrically, the three planes intersect at a single point.
• Consistent and dependent: The system has infinitely many solutions. The three planes intersect along a line (or all three are the same plane). You'll notice during solving that one equation is a combination of the others, and at least one variable is "free," meaning it can take any value.
• Inconsistent: The system has no solution. At some point during elimination, you'll get a contradiction like $0 = 5$. Geometrically, the planes never all meet at a common point. This can happen when two or more planes are parallel, or when the three planes form a triangular "prism" shape with no shared intersection.

Interpreting solutions for three-equation systems

One solution (consistent and independent)

The solution is an ordered triple $(x, y, z)$. Always verify by plugging the values back into all three original equations.

No solution (inconsistent)

If your algebra produces a false statement (like $0 = 3$), the system is inconsistent. There's no point that lies on all three planes.

Infinitely many solutions (consistent and dependent)

When you end up with an identity like $0 = 0$ instead of a value for a variable, the system is dependent. To express the solution set:

• Designate one variable as a free variable (often called $t$). This variable can be any real number.
• Express the other two variables in terms of $t$.
• Write the solution in parametric form: $x = a + bt$, $y = c + dt$, $z = t$ (or whichever variable you chose as free).

For example, if solving gives you $x = 2 - 3z$ and $y = 1 + z$ with $z$ free, you'd let $z = t$ and write the solution as $(2 - 3t,\; 1 + t,\; t)$.

Matrix Representation and Row Reduction

You can organize a three-variable system into an augmented matrix, where each row holds the coefficients of one equation and the constant on the right side. Then you use row operations to solve it without rewriting variables over and over.

The goal is to reach row echelon form:

• Each leading coefficient (called a pivot) is 1.
• Each pivot sits to the right of the pivot in the row above.
• Rows of all zeros go to the bottom.

From there, you can back-substitute to find each variable. For an even cleaner result, continue to reduced row echelon form (RREF), where every column with a pivot has zeros in all other entries. In RREF, you can read the solution directly from the matrix.

If a row reduces to something like $[0\; 0\; 0\; |\; 5]$, the system is inconsistent. If you get a row of all zeros $[0\; 0\; 0\; |\; 0]$, the system is dependent and you'll have a free variable.

Applying Systems of Linear Equations with Three Variables

Solving real-world problems with three-variable linear systems

Three-variable systems show up whenever a problem involves three unknown quantities connected by three relationships. Setting these up correctly is half the battle.

1. Define your variables clearly. For example: let $x$ = number of adult tickets, $y$ = number of student tickets, $z$ = number of child tickets.
2. Translate each piece of information into an equation. If the problem says "200 total tickets were sold," that gives you $x + y + z = 200$. Each distinct relationship becomes one equation.
3. Solve the system using elimination, substitution, or row reduction.
4. Check that your answer makes sense in context. If you're counting physical objects, negative or fractional answers signal an error. Always substitute back into the original equations to verify.

A common mistake is writing equations that are actually restatements of each other, which gives you a dependent system instead of a unique answer. Make sure each equation captures genuinely different information from the problem.