Can I use substitution instead of elimination?

Yes, you can solve one equation for $x$ and substitute that expression into the other two. However, unless one variable already has a coefficient of 1, this often creates messy fractions early on. Elimination is usually cleaner for three variables.

What does it mean if all the variables cancel out and I get $0 = 0$?

Getting a true statement like $0 = 0$ means the system has infinitely many solutions. Geometrically, the planes intersect along a line, or they are the exact same plane.

What if I get a false statement like $0 = 5$?

A false statement means the system has no solution. The planes do not share a single common point; they might be parallel or form a triangular prism shape in space.

How do you solve a system with three variables?

To solve a system of three variables, you systematically eliminate one variable to create a simpler system of two variables. Think of it like taking apart a tall sandwich. You cannot eat the whole thing at once, so you remove one layer first to make it manageable. Once you have a two-variable system, you use the familiar methods of elimination or substitution to find those values. Finally, you substitute them back into an original equation to find the last missing piece.

What is a three-variable system?

A three-variable system consists of three linear equations containing three unknowns, usually $x$ , $y$ , and $z$ . Geometrically, each equation represents a flat plane in 3D space. When you solve the system, you are looking for the exact point $(x, y, z)$ where all three planes intersect. If they do not meet at a single point, the system might have no solution or infinitely many solutions.

The core strategy: eliminate one variable

The most reliable method is elimination. Pick one variable to eliminate, such as $z$ . Group your three equations into two different pairs. For each pair, multiply the equations by whatever numbers are necessary so that adding them together makes the $z$ terms cancel out. You now have two brand new equations that only contain $x$ and $y$ . You have successfully reduced the 3D problem into a standard 2D problem.

Common places students slip up

The biggest trap in these problems is organization. It is incredibly easy to drop a negative sign or forget to multiply a constant term on the right side of the equals sign. To avoid this, always label your equations (like Equation 1, Equation 2, and Equation 3). Write out every multiplication step explicitly, and draw clear separators between the different stages of your work.

Worked through

Solve the system: Eq 1: $x + y + z = 6$ , Eq 2: $2x - y + 3z = 14$ , Eq 3: $-x + 2y - z = -2$

Let us eliminate $y$ first. Add Eq 1 and Eq 2: $(x + y + z) + (2x - y + 3z) = 6 + 14$ , which simplifies to $3x + 4z = 20$ (we will call this Eq A). Next, we need another pair to eliminate $y$ . Multiply Eq 1 by -2 to get $-2x - 2y - 2z = -12$ . Add this to Eq 3: $(-2x - 2y - 2z) + (-x + 2y - z) = -12 + (-2)$ , giving $-3x - 3z = -14$ (we will call this Eq B). Now solve the 2D system of Eq A and Eq B: $3x + 4z = 20$ and $-3x - 3z = -14$ . Add them together: $(3x - 3x) + (4z - 3z) = 20 - 14$ , which gives $z = 6$ . Substitute $z = 6$ into Eq A: $3x + 4(6) = 20$ , so $3x = -4$ , meaning $x = -4/3$ . Finally, plug $x$ and $z$ into Eq 1: $-4/3 + y + 6 = 6$ , so $y = 4/3$ . The final solution point is $(-4/3, 4/3, 6)$ .

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Where this comes from: OpenStax College Algebra, Chapter 7 · Khan Academy: Systems of equations with three variables