Pair of Linear Equations

⚡ Key Takeaways

A pair of linear equations in two variables is a system of two equations of the form $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ , whose common solution is the point (or points) satisfying both simultaneously. Key result: geometrically each equation is a straight line, so the system has exactly one solution (intersecting lines), no solution (parallel lines), or infinitely many solutions (coincident lines), determined entirely by the ratios of coefficients. Where it applies: mixture problems, geometry, economics, age/cost word problems. Who/when: rooted in ancient Babylonian and Chinese algebra (200 BCE), formalised by Cramer, Gauss, and others in the 18th–19th centuries.

💡 The Core Insight

Two straight lines in a plane can relate in only three ways: they cross at one point, they never meet (parallel), or they lie exactly on top of each other (coincident). Solving a pair of linear equations is nothing more than identifying which of these three cases occurs and locating the common point(s). The entire theory — substitution, elimination, graphing, determinants — is just different machinery for answering this one geometric question.

DefinitionPair of Linear Equations in Two Variables

A pair (or system) of linear equations in two variables $x$ and $y$ is written as

1

\quad a_1x + b_1y + c_1 = 0

Derivation hint: This is the general form of any straight line; "linear" means each variable appears to the first power only, with no products like $xy$ .

2

\quad a_2x + b_2y + c_2 = 0

Derivation hint: A second independent line constraint; together [1] and [2] form the system. where $a_1, b_1, c_1, a_2, b_2, c_2$ are real numbers, with $a_1, b_1$ not both zero and $a_2, b_2$ not both zero. A solution is an ordered pair $(x_0, y_0)$ that makes both equations true simultaneously. The symbols mean:

$x, y$ — the unknown variables we wish to find,
$a_1, a_2$ — coefficients of $x$ ,
$b_1, b_2$ — coefficients of $y$ ,
$c_1, c_2$ — constant terms.

TheoremConsistency and the Coefficient-Ratio Criterion

Given the system [1]–[2] with $a_2, b_2, c_2 \neq 0$ , exactly one of the following holds:

3

\quad \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \;\Longrightarrow\; \text{unique solution (lines intersect)}

Derivation hint: Different slope ratios force the lines to have different directions, so they must cross at exactly one point.

4

\quad \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \;\Longrightarrow\; \text{no solution (lines parallel)}

Derivation hint: Equal coefficient ratios mean equal slopes; unequal constant ratio means different intercepts — parallel, never meeting.

5

\quad \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \;\Longrightarrow\; \text{infinitely many solutions (lines coincident)}

Derivation hint: All three ratios equal means equation [2] is just a scalar multiple of [1] — the same line written twice. A system with at least one solution is consistent; with no solution it is inconsistent. A consistent system with a unique solution is independent; with infinitely many it is dependent.

ProofElimination Yields the Solution and Reveals the Cases

We solve the general system [1]–[2] by elimination.

Step 1. Multiply [1] by $b_2$ and [2] by $b_1$ :

6

\quad a_1b_2\,x + b_1b_2\,y + c_1b_2 = 0, \qquad a_2b_1\,x + b_1b_2\,y + c_2b_1 = 0

Derivation hint: Scaling each equation makes the $y$ -coefficients identical ( $b_1b_2$ ), preparing them for subtraction.

Step 2. Subtract the second from the first to eliminate $y$ :

7

\quad (a_1b_2 - a_2b_1)\,x + (c_1b_2 - c_2b_1) = 0

Derivation hint: The $b_1b_2\,y$ terms cancel, leaving a single equation in $x$ .

Step 3. Solve for $x$ , provided $a_1b_2 - a_2b_1 \neq 0$ :

8

\quad x = \frac{c_2b_1 - c_1b_2}{a_1b_2 - a_2b_1}, \qquad y = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1}

Derivation hint: Rearrange [7] for $x$ ; repeat the elimination on $x$ to obtain $y$ symmetrically — this is Cramer's Rule in disguise.

Step 4 (case analysis). The quantity $D = a_1b_2 - a_2b_1$ governs everything:

If $D \neq 0$ , formulas [8] give a single $(x,y)$ — the unique solution, equivalent to condition [3].
If $D = 0$ but the numerators are nonzero, [7] becomes $\text{(nonzero)} = 0$ , a contradiction — no solution, condition [4].
If $D = 0$ and the numerators also vanish, [7] becomes $0 = 0$ , an identity — infinitely many solutions, condition [5].

This completes the proof, tying the algebraic determinant $D$ to the three geometric cases. $\blacksquare$

∎

mathematician · Gabriel Cramer | 1704–1752

The Swiss mathematician Gabriel Cramer published, in his 1750 Introduction à l'analyse des lignes courbes algébriques, the rule that now bears his name. Cramer's Rule expresses the solution of a linear system as ratios of determinants — precisely formulas [8] above. Though Cramer worked with curves rather than pure linear algebra, his determinant-based method became the standard textbook approach to solving small systems and laid groundwork for the broader theory of matrices and determinants developed later by Cayley and Sylvester.

🧠 Intuition

Imagine each equation as a constraint — a rule that pins down a relationship between $x$ and $y$ . One equation alone leaves infinitely many possibilities (a whole line of points). The job of the second equation is to narrow those possibilities. If the second line genuinely points in a different direction, it slices through the first at exactly one spot — your answer. If it points in the same direction but sits elsewhere, the two constraints contradict each other and no point can satisfy both. If the second line is secretly the first one in disguise (just scaled up), it adds no new information at all, and every point on the line works. The three ratios $\frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2}$ are simply a quick way of asking: are these lines tilted the same way? And if so, are they in the same place?

Figure

The three configurations of a pair of linear equations

ExampleUnique Solution by Elimination

Given: Solve

2x + 3y = 13 \qquad\text{and}\qquad 4x - y = 5.

Step 1 — Check the case. Here $\frac{a_1}{a_2} = \frac{2}{4} = \frac12$ and $\frac{b_1}{b_2} = \frac{3}{-1} = -3$ . Since $\frac12 \neq -3$ , by [3] there is a unique solution.

Step 2 — Eliminate $y$ . Multiply the second equation by 3: $12x - 3y = 15$ . Add to the first:

2x + 3y + 12x - 3y = 13 + 15 \;\Rightarrow\; 14x = 28 \;\Rightarrow\; x = 2.

Step 3 — Back-substitute. From $4x - y = 5$ : $\;4(2) - y = 5 \Rightarrow y = 3.$

Step 4 — Verify. $2(2)+3(3)=4+9=13$ ✓ and $4(2)-3=5$ ✓.

\boxed{x = 2,\quad y = 3}

ExampleA Mixture Word Problem

Given: A shopkeeper sells 5 pens and 3 notebooks for ₹120, and 2 pens and 4 notebooks for ₹100. Find the price of one pen ( $p$ ) and one notebook ( $n$ ).

Step 1 — Set up equations.

5p + 3n = 120, \qquad 2p + 4n = 100.

Step 2 — Eliminate $p$ . Multiply the first by 2 and the second by 5:

10p + 6n = 240, \qquad 10p + 20n = 500.

Subtract: $\;14n = 260 \Rightarrow n = \dfrac{260}{14} = \dfrac{130}{7} \approx 18.57.$ Let us instead use cleaner numbers: solve directly. Subtracting gives $14n = 260$ , so $n = 18.57$ . Back-substitute into $2p + 4n = 100$ :

2p + 4(18.57) = 100 \Rightarrow 2p = 100 - 74.29 = 25.71 \Rightarrow p = 12.86.

Step 3 — Verify. $5(12.86)+3(18.57)=64.3+55.7 = 120$ ✓.

\boxed{p \approx ₹12.86 \text{ per pen},\quad n \approx ₹18.57 \text{ per notebook}}

⚠️ Common mistake

"More equations always means a solution exists." False. Two equations may contradict each other (parallel lines, case [4]), giving no solution despite having "enough" equations. Conversely, two equations may be redundant (coincident lines, case [5]), giving infinitely many. Counting equations is not the same as counting independent constraints. Always check the ratios first.

"You can divide by a coefficient without checking it is nonzero." When applying [3]–[5], if some coefficient like $a_2$ or $c_2$ equals zero, the ratio criterion needs careful handling — return to elimination via the determinant $D = a_1b_2 - a_2b_1$ , which is always valid.

Counterexample

Consider the apparently reasonable pair

3x + 2y = 6 \qquad\text{and}\qquad 6x + 4y = 7.

A student might expect a unique answer. But $\frac{a_1}{a_2} = \frac{3}{6} = \frac12$ , $\frac{b_1}{b_2} = \frac{2}{4} = \frac12$ , yet $\frac{c_1}{c_2} = \frac{6}{7} \neq \frac12$ . By criterion [4] the lines are parallel — there is no solution at all. Elimination confirms it: doubling the first gives $6x+4y=12$ , contradicting $6x+4y=7$ . The system is inconsistent.

connection

A pair of linear equations is the two-variable case of the general theory of Systems of Linear Equations, which extends to any number of variables and is handled by Matrices and Determinants and Gaussian Elimination. The determinant $D$ in formula [8] is precisely the $2\times 2$ determinant studied in Cramer's Rule. Geometrically the subject belongs to Coordinate Geometry, since each equation is a line analysed via slope and intercept; the slope condition for parallelism links directly to Equation of a Straight Line. When inequalities replace equalities, the topic generalises to Linear Programming, the foundation of optimisation.

🔬 Where this appears

⏳ History

- c. 1800 BCE — Babylonia. Clay tablets show solutions to simultaneous linear (and quadratic) problems using rhetorical, word-based methods.

- c. 200 BCE — China. The Nine Chapters on the Mathematical Art presents the fangcheng method, an array-based elimination essentially identical to modern Gaussian elimination, for solving simultaneous linear equations.

- 3rd century CE — Diophantus of Alexandria develops symbolic techniques for indeterminate and determinate equations in his Arithmetica.

- 9th century CE — al-Khwārizmī systematises algebra (al-jabr), giving general procedures that underlie linear equation solving.

- 1545 — Gerolamo Cardano and Renaissance algebraists advance symbolic manipulation, enabling cleaner treatment of systems.

- 1683 — Seki Takakazu (Japan) and 1693 — Gottfried Leibniz (Europe) independently introduce determinants for solving linear systems.

- 1750 — Gabriel Cramer publishes Cramer's Rule, expressing solutions as determinant ratios (formulas [8]).

- 1809 — Carl Friedrich Gauss formalises systematic elimination while computing the orbit of Ceres, giving "Gaussian elimination" its name.

- 1850s — Cayley and Sylvester create matrix theory, embedding the humble pair of equations within a vast general framework.

open-problem

The two-variable case is completely understood, but it opens onto deeper questions. Numerical conditioning: when the determinant $D$ is very small but nonzero (lines almost parallel), tiny measurement errors in the coefficients produce wildly different solutions — a phenomenon of ill-conditioning that dominates real-world computational linear algebra. Higher dimensions: with $n$ variables and $m$ equations, the geometry becomes intersections of hyperplanes, and questions of existence and uniqueness require the rank of matrices; deciding solvability efficiently for huge sparse systems remains an active engineering challenge. Integer constraints: if $x$ and $y$ must be whole numbers, the problem becomes a Diophantine system, and its higher-dimensional generalisation (integer linear programming) is NP-hard — among the hardest computational problems known. Thus a topic that looks "finished" at high school becomes, with small twists, a frontier of computational mathematics.

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also