No Solution or Infinite Solutions in 3-Variable Systems

When you solve a system of linear equations with three variables, you are usually looking for a single point \((x, y, z)\) that satisfies all three equations at once. But sometimes that point does not exist, or there are infinitely many points that work. These two special outcomes come up constantly in searches for "no solution one solution infinite solution" style problems, and the good news is that the same elimination steps you already know reveal both cases automatically.

This builds directly on the idea of determining the number of solutions to linear equations for two variables. With three variables the logic is identical, you just track one extra unknown.

To solve a 3-variable system, you eliminate one variable at a time until you are left with equations in fewer unknowns. If you have not seen this yet, it is worth reviewing the elimination method for two variables first, and the full 3-variable substitution method as an alternative approach. The difference in this lesson is what happens when the elimination steps do not lead to a clean, unique answer.

A system has no solution when, at some point during elimination, you end up with a statement that is never true, such as \(0 = 5\) or \(8 = 9\). This happens because two of the equations describe conditions that directly contradict each other, no matter what values you choose for the variables.

Example 1

\(x + y + z = 4\)
\(2x + 2y + 2z = 9\)
\(x - y + z = 1\)

Multiply the first equation by \(2\):

\(2x + 2y + 2z = 8\)

Compare this to the second equation, \(2x + 2y + 2z = 9\). The left sides are identical, but the right sides are not, which gives \(8 = 9\). This is false for every possible \(x\), \(y\), and \(z\), so the system has no solution. Geometrically, this means the three planes never share a common point, at least two of them are parallel and never meet.

A system has infinite solutions when elimination leads to a statement that is always true, such as \(0 = 0\). This means one equation carries no new information, it is really just a multiple or combination of another equation in the system. Once that redundancy is removed, you are left with fewer independent equations than variables, so an entire line (or plane) of points satisfies the system.

Example 2

\(x + y + z = 3\)
\(2x + 2y + 2z = 6\)
\(x - y + 2z = 4\)

Multiply the first equation by \(2\):

\(2x + 2y + 2z = 6\)

This is exactly the second equation, so it adds nothing new, \(0 = 0\). Only two independent equations remain for three unknowns, so infinitely many solutions exist. To describe them, solve the first equation for \(x\): \(x = 3 - y - z\), then substitute into the third equation:

\((3 - y - z) - y + 2z = 4\)
\(3 - 2y + z = 4\)
\(z = 1 + 2y\)

Let \(y = t\) be a free parameter. Then \(z = 1 + 2t\) and \(x = 3 - t - (1 + 2t) = 2 - 3t\). The solution set is every triple of the form \((2 - 3t,\ t,\ 1 + 2t)\), one solution for every value of \(t\).

Each linear equation in three variables represents a flat plane. A unique solution is a single point where all three planes cross. No solution means the planes never all meet at once, often because two are parallel. Infinite solutions means the planes collapse so that they all share an entire line (or the same plane), giving endlessly many shared points.

Whenever you eliminate variables in a 3-variable system, keep an eye on what the leftover equation says. A contradiction rules out any solution, an always-true statement opens the door to infinitely many, and a specific numeric result for each variable gives you the single point you were looking for.

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