How To Solve System Of Equations 3 Variables

A system of equations with three variables represents the intersection of three planes in three-dimensional space. Finding the solution means identifying the single point $(x, y, z)$ where all three planes meet, determining if they intersect along a line (infinite solutions), or realizing they never share a common point (no solution). Mastering how to solve system of equations 3 variables is a fundamental skill in linear algebra, essential for fields ranging from engineering and physics to economics and computer graphics. While the addition of a third variable increases complexity, the core logic remains an extension of the two-variable methods you already know: elimination and substitution.

Understanding the Geometry and Possible Outcomes

Before diving into the algebra, visualizing the geometry helps build intuition. In real terms, each linear equation in three variables ($Ax + By + Cz = D$) graphs as a flat plane extending infinitely in 3D space. When you have a system of three such equations, you are essentially asking: "How do these three planes relate to one another?

There are three distinct possibilities for the solution set:

1. One Unique Solution (Consistent & Independent): The three planes intersect at a single, specific point. This is the most common scenario in textbook problems. The coordinates of that point $(x, y, z)$ satisfy all three equations simultaneously.
2. Infinitely Many Solutions (Consistent & Dependent): The planes intersect along a common line, or all three planes are coincident (the exact same plane). In this case, the equations are not independent; one can be derived from the others. The solution is expressed parametrically (e.g., $z = t, y = 2t+1, x = 3t-4$).
3. No Solution (Inconsistent): The planes do not share a single common point. This happens if two planes are parallel, or if they form a triangular prism shape where each pair intersects in a line, but those three lines are parallel to each other (never meeting at one spot). Algebraically, this reveals a contradiction, such as $0 = 5$.

Recognizing which outcome you are heading toward early in the process can save significant calculation time.

Method 1: Gaussian Elimination (The Standard Algorithm)

Gaussian Elimination is the most systematic, algorithmic approach. It transforms the system into an equivalent one that is easier to solve, typically an upper triangular form (Row Echelon Form), allowing for back-substitution. This method is the basis for how computers solve large systems Practical, not theoretical..

Step-by-Step Procedure

Consider the generic system: $ \begin{cases} a_1x + b_1y + c_1z = d_1 \ a_2x + b_2y + c_2z = d_2 \ a_3x + b_3y + c_3z = d_3 \end{cases} $

1. Write the Augmented Matrix Strip away the variables and operators. Write only the coefficients and constants in a matrix bracket. $ \left[\begin{array}{ccc|c} a_1 & b_1 & c_1 & d_1 \ a_2 & b_2 & c_2 & d_2 \ a_3 & b_3 & c_3 & d_3 \end{array}\right] $

2. Perform Row Operations to Achieve Row Echelon Form You are allowed three operations that do not change the solution set:

• Swap Rows: $R_i \leftrightarrow R_j$ (Useful to get a '1' or non-zero number in the top-left pivot position).
• Multiply a Row by a Non-zero Scalar: $kR_i \rightarrow R_i$ (Useful to create a leading 1, called a pivot).
• Add a Multiple of One Row to Another: $R_i + kR_j \rightarrow R_i$ (The primary tool for eliminating variables).

Goal: Create zeros below the main diagonal (positions $(2,1)$, $(3,1)$, and $(3,2)$) Easy to understand, harder to ignore. Still holds up..

• Target Column 1: Use Row 1 to zero out the first element in Row 2 and Row 3.
• Target Column 2: Use the new Row 2 to zero out the second element in Row 3.

3. Back-Substitution Once the matrix looks like this: $ \left[\begin{array}{ccc|c} 1 & * & * & * \ 0 & 1 & * & * \ 0 & 0 & 1 & * \end{array}\right] $ You have a triangular system: $ \begin{cases} x + \dots = \dots \ y + \dots = \dots \ z = \text{value} \end{cases} $ Solve the bottom equation for $z$, plug that into the middle equation to find $y$, and plug both into the top equation to find $x$.

Worked Example: Elimination in Action

Solve: $ \begin{cases} x + 2y - z = 4 \quad (Eq 1) \ 2x - y + 3z = 9 \quad (Eq 2) \ 3x + y - 2z = 0 \quad (Eq 3) \end{cases} $

Augmented Matrix: $ \left[\begin{array}{ccc|c} 1 & 2 & -1 & 4 \ 2 & -1 & 3 & 9 \ 3 & 1 & -2 & 0 \end{array}\right] $

Eliminate $x$ from Row 2 and Row 3:

• $R_2 - 2R_1 \rightarrow R_2$: $(2-2), (-1-4), (3-(-2)), (9-8) \rightarrow [0, -5, 5, 1]$
• $R_3 - 3R_1 \rightarrow R_3$: $(3-3), (1-6), (-2-(-3)), (0-12) \rightarrow [0, -5, 1, -12]$

Matrix now: $ \left[\begin{array}{ccc|c} 1 & 2 & -1 & 4 \ 0 & -5 & 5 & 1 \ 0 & -5 & 1 & -12 \end{array}\right] $

Simplify Row 2 (Optional but helpful): Divide Row 2 by -5: $R_2 / -5 \rightarrow [0, 1, -1, -0.2]$ (or keep as fractions: $1/5$). Let's keep integers for now to avoid decimals.

Eliminate $y$ from Row 3:

• $R_3 - R_2 \rightarrow R_3$: $(0-0), (-5-(-5)), (1-5), (-12-1) \rightarrow [0, 0, -4, -13]$

Matrix now (Row Echelon Form): $ \left[\begin{array}{ccc|c} 1 & 2 & -1 & 4 \ 0 & -5 & 5 & 1 \ 0 & 0 & -4 & -13 \end{array}\right] $

Back-Substitution:

1. Row 3: $-4z = -13 \implies z = \frac{13}{4} = 3.25$
2. Row 2: $-5y + 5(\frac{13}{4}) = 1 \implies -5y = 1 - \frac{65}{4} = -\frac{61}{4} \implies y = \frac{61}{20} = 3.05$
3. Row 1: $x + 2(\frac{61}{20}) - \frac{13}{4} = 4 \implies x + \frac{61}{1