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Systems of Linear Equations

⚗ Dr. Möbius, from the lab

Two equations, two unknowns. Each one is a line; together they're either one intersection point, no intersection, or the exact same line. The geometry tells you the solution count before you touch the algebra. Pay attention to that picture — in the Matrices stratum, this exact game gets played in $n$ dimensions, and Gaussian elimination is the industrial-strength version of what you're about to learn.

THE BIG IDEA

Solving a system of two linear equations means finding the point(s) where two lines intersect; the algebra of substitution and elimination corresponds directly to the geometry.

What a system is

A system of linear equations is two (or more) equations that must be satisfied simultaneously. A solution is an ordered pair $(x, y)$ that makes every equation in the system true. Geometrically, each equation is a line; the solution is the intersection.

Three cases, each with its own algebra and its own picture:

One solution. Lines cross at exactly one point. Consistent and independent.
No solution. Lines are parallel — same slope, different intercepts. Inconsistent.
Infinitely many solutions. Lines are the same — coincident. Consistent and dependent.

You met the algebra of cases 2 and 3 in the Linear Equations lesson (identity vs contradiction). Now they have geometric explanations.

The substitution method

Rule: solve one equation for one variable, substitute into the other.

Solve: $y = 2x - 1$ and $3x + y = 9$ .

Step 1: substitute $y = 2x-1$ into the second: $3x + (2x-1) = 9 \implies 5x - 1 = 9 \implies 5x = 10 \implies x = 2.$

Step 2: back-substitute: $y = 2(2) - 1 = 3$ .

Solution: $(2, 3)$ .

Check both equations:

$y = 2x-1$ : $3 = 2(2)-1 = 3$ . Yes.
$3x + y = 9$ : $3(2)+3 = 9$ . Yes.

Substitution is cleanest when one equation already has a variable isolated.

The elimination method

Rule: add a multiple of one equation to the other to make one variable disappear.

Solve: $2x + 3y = 7$ and $4x - 3y = 5$ .

Add the equations (the $y$ -terms cancel immediately): $6x = 12 \implies x = 2.$

Substitute back: $2(2) + 3y = 7 \implies 4 + 3y = 7 \implies y = 1$ .

Solution: $(2, 1)$ .

More often you'll need to multiply first to align coefficients:

Solve: $3x + 2y = 11$ and $5x - 4y = 11$ .

Multiply equation 1 by 2: $6x + 4y = 22$ . Now add to equation 2: $6x + 4y + 5x - 4y = 22 + 11 \implies 11x = 33 \implies x = 3.$

Back-substitute: $3(3) + 2y = 11 \implies 2y = 2 \implies y = 1$ .

Solution: $(3, 1)$ .

Elimination is the star method. It grows up to become Gaussian elimination in the Matrices stratum, which handles hundreds of variables with the same basic idea.

Recognizing inconsistency and dependence

If elimination produces a false constant equation (like $0 = 5$ ), the system is inconsistent — the lines are parallel, no solution.

If elimination produces $0 = 0$ , the system is dependent — the lines are the same, infinitely many solutions (every point on the shared line).

Word problems: mixtures and rates

A system is needed whenever you have two unknown quantities with two independent conditions.

Example. A chemist mixes a 20% acid solution with a 50% acid solution to get 60 liters of a 30% acid solution. Find the volume of each.

Let $x$ = liters of 20% solution, $y$ = liters of 50% solution.

Condition 1 (total volume): $x + y = 60$ .

Condition 2 (total acid): $0.20x + 0.50y = 0.30(60) = 18$ .

Multiply equation 2 by 10: $2x + 5y = 180$ . From equation 1: $x = 60 - y$ . Substitute: $2(60-y) + 5y = 180 \implies 120 - 2y + 5y = 180 \implies 3y = 60 \implies y = 20.$

Then $x = 40$ . Check: $0.20(40) + 0.50(20) = 8 + 10 = 18 = 0.30(60)$ . Confirmed.

The trailer

In the Matrices stratum, a system of equations becomes a matrix equation $Ax = b$ , and the elimination process is formalized as row reduction. Every idea here — pivots, leading variables, free variables, consistent/inconsistent — reappears there in $n$ -dimensional form. This lesson is the warm-up.

🔬 SPECIMENS (worked examples)

Worked example 1 — substitution, clean case▸

Solve the system: $y = -x + 5$ and $2x + y = 8$ .

The first equation already isolates $y$ . Substitute into the second: $2x + (-x + 5) = 8 \implies x + 5 = 8 \implies x = 3.$

Back-substitute: $y = -3 + 5 = 2$ .

Solution: $(3, 2)$ .

Check both equations:

$y = -x+5$ : $2 = -3+5 = 2$ . Yes.
$2x+y=8$ : $6+2 = 8$ . Yes.

Geometrically, the lines $y = -x+5$ (slope $-1$ ) and $y = -2x+8$ (slope $-2$ ) cross at $(3,2)$ .

Worked example 2 — elimination, two-step▸

Solve: $3x - 2y = 4$ and $5x + 3y = 1$ .

To eliminate $y$ , get coefficients that are additive inverses. Multiply equation 1 by 3 and equation 2 by 2:

$9x - 6y = 12$ $10x + 6y = 2.$

Add: $19x = 14 \implies x = \frac{14}{19}.$

Back-substitute into equation 1: $3 \cdot \frac{14}{19} - 2y = 4 \implies \frac{42}{19} - 2y = 4 \implies -2y = 4 - \frac{42}{19} = \frac{76-42}{19} = \frac{34}{19} \implies y = -\frac{17}{19}$ .

Solution: $\left(\frac{14}{19}, -\frac{17}{19}\right)$ .

Check equation 2: $5 \cdot \frac{14}{19} + 3 \cdot (-\frac{17}{19}) = \frac{70}{19} - \frac{51}{19} = \frac{19}{19} = 1$ . Confirmed.

Worked example 3 — the inconsistent system trap▸

A student tries to solve: $4x - 2y = 6$ and $-6x + 3y = 4$ . They eliminate $y$ by tripling the first and doubling the second, then adding. What do they get, and what does it mean?

Multiply equation 1 by 3: $12x - 6y = 18$ .

Multiply equation 2 by 2: $-12x + 6y = 8$ .

Add: $0x + 0y = 26$ , i.e., $0 = 26$ .

This is a contradiction. There is no ordered pair satisfying both equations. The system is inconsistent.

Why geometrically? Rewrite equation 1: $y = 2x - 3$ . Rewrite equation 2: $y = 2x + \frac{4}{3}$ . Both have slope 2 but different intercepts — parallel lines that never meet. The algebra correctly reports "no solution."

☠ KNOWN HAZARDS

Checking only one equation after solving. The solution must satisfy EVERY equation. Always substitute into both (all) equations.
Misidentifying "no solution" vs "infinitely many". $0 = 5$ means no solution (contradiction). $0 = 0$ means infinitely many (identity). They look similar but are opposites.
Forgetting to multiply ALL terms when scaling an equation. If you multiply equation 1 by 3, the right-hand side also gets multiplied by 3. Missing a term creates a fake system.
In substitution: forgetting to substitute back. After finding $x$ , you must substitute into one of the original equations to find $y$ . "I found $x = 2$ " is half an answer.

TL;DR

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A system solution is a point satisfying all equations simultaneously — geometrically, the intersection of lines.
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Three cases: one solution (lines cross), no solution (parallel), infinitely many (same line). The algebra outcome matches: $x = k$ , $0 = 5$ , or $0 = 0$ .
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Substitution: isolate one variable, plug into the other. Best when a variable is already isolated.
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Elimination: multiply to align opposing coefficients, then add. This method scales to $n$ equations and becomes Gaussian elimination.
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Word problems: name the unknowns, write one equation per independent condition, solve, check.

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