Beyond Three Variables

Beyond Three Variables | CodeSignal Learn

Introduction

Welcome to the sixth lesson of Three-Variable Systems and a First Look Beyond . You can now take a bare three-equation system and hand back a verified ordered triple, which means it is time to answer the question that has been waiting quietly in the background: what happens when a system has four unknowns? Or five? Or fifty? The reassuring news is that almost nothing new has to be learned. The two ideas that carried you through three variables — simultaneous satisfaction and reduce by one — keep working without a single modification. This lesson is about reading larger systems with confidence and seeing precisely how the familiar machinery stretches to fit them.

Why Bigger Systems Show Up

c

What a Solution Looks Like Now

(x, y)

Variables	Solution written as	Name
$2$	$(x, y)$	ordered pair
$3$	$(x, y, z)$	ordered triple
$4$	$(w, x, y, z)$	ordered 4-tuple
$5$	$(v, w, x, y, z)$	ordered 5-tuple

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The Cost of One More Variable

3 \times 3

Verification looks exactly as it did for triples, with one extra substitution per line. Consider this system together with the candidate $(w, x, y, z) = (1, 2, 0, 3)$ :

\left\{ \begin{array}{lrcrcrcrcl} (1)\quad w &+& x &+& y &+& z &=& 6 \\ (2)\quad 2w &-& x &+& y &+& z &=& 3 \\ (3)\quad w &+& x &-& y &+& 2z &=& 9 \\ (4)\quad w &+& 2x & & &-& z &=& 2 \end{array} \right.

Substituting the four values, one equation at a time:

\begin{aligned} (1): \quad & 1 + 2 + 0 + 3 = 6 \quad ✓ \\ (2): \quad & 2(1) - 2 + 0 + 3 = 3 \quad ✓ \\ (3): \quad & 1 + 2 - 0 + 2(3) = 9 \quad ✓ \\ (4): \quad & 1 + 2(2) - 3 = 2 \quad ✓ \end{aligned}

All four lines hold, so $(1, 2, 0, 3)$ is a genuine solution. Notice equation $(4)$ : the variable $y$ never appears, which simply means its coefficient is $0$ . Missing variables are common in larger systems, and they cause trouble only when the columns drift out of alignment, so keeping the terms in a consistent variable order pays off more here than it ever did with three.

The rule that every equation must hold deserves a moment of its own, because a wide system gives a near-miss many more places to hide. Test the tuple $(2, 1, 1, 2)$ in the very same system:

\begin{aligned} (1): \quad & 2 + 1 + 1 + 2 = 6 \quad ✓ \\ (2): \quad & 2(2) - 1 + 1 + 2 = 6 \neq 3 \quad ✗ \end{aligned}

The first line passed, and a hurried check might have stopped right there feeling satisfied. The second line fails, and one failure is decisive: this tuple is not a solution, however the remaining equations turn out. As you may recall from the first lesson of this course, a partial check is not a check at all — and with four or five equations on the page, the temptation to stop early only grows.

Now for the strategic half. Combining a pair of equations to cancel a variable produces a new equation that is true whenever both originals are true, and that new equation carries one fewer variable. Nothing in that reasoning ever depended on there being exactly three variables, so it holds just as firmly for four, five, or more.

The result is a staircase. Starting from four equations in four variables, we combine equations pairwise to knock out one variable and land on three equations in three variables — territory you have already mastered:

4 \text{ variables} \;\longrightarrow\; 3 \text{ variables} \;\longrightarrow\; 2 \text{ variables} \;\longrightarrow\; \text{solve}

In general, one round of elimination produces a system in $n - 1$ variables that any solution of the original must also satisfy for those variables. Repeat until two variables remain, solve that small system with the substitution or elimination tools from earlier courses, then back-substitute upward to recover the variables you set aside — using original equations at each step — and verify the complete tuple in the original equations.

Here is the part worth pausing on. Eliminating a variable from a three-equation system took two pairwise combinations. Eliminating one from a four-equation system takes three — and that only brings us to the three-variable case, which then asks for two more.

System size	Combinations to remove one variable	Total combinations to reach two variables
$3 \times 3$	$2$	$2$
$4 \times 4$	$3$	$3 + 2 = 5$
$5 \times 5$	$4$	$4 + 3 + 2 = 9$

The method scales perfectly; the effort does not. Every extra combination is another line of arithmetic and another place for a dropped sign to slip past unnoticed. That gap between "the idea still works" and "doing it by hand is still reasonable" is exactly where the next lesson picks up.

Before touching any arithmetic, a wide system rewards a careful read. These habits keep four or five equations legible:

Line up the columns. Write every equation with the variables in the same order, leaving a visible gap wherever a coefficient is $0$ .
Number every equation. With four or five rows, "the second one" stops being clear almost immediately.
Count equations against variables. A single unique solution generally requires at least as many independent equations as unknowns.
Scan the columns for the cheapest target. A column full of $1$ s, $-1$ s, and $0$ s is still the friendliest place to begin.
Reassemble the tuple in variable order. Values often get solved out of order, and the final answer has to be written in the order the variables are listed.

Read this way, the café system from the opening stops looking intimidating: four numbered rows, four aligned columns, one $0$ coefficient here and there, and a familiar question about which column to attack first.

Introduction

Why Bigger Systems Show Up

What a Solution Looks Like Now

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Testing a Four-Tuple

Three out of Four Is Still a Miss

How Elimination Generalizes

The Cost of One More Variable

Reading a Larger System Well

Conclusion and Next Steps