Equations in Two Unknowns

One equation in one unknown is too easy; the hard part of algebra starts when you have two unknowns at once. Two equations in two unknowns is the first place where you can no longer just isolate the variable and read off the answer. The geometric truth makes everything click: each equation is a line in the $xy$-plane, and the solution is wherever the two lines cross.

The standard shape

A system of two linear equations in two unknowns looks like

where $a_1, b_1, c_1, a_2, b_2, c_2$ are given numbers and $x, y$ are the unknowns. The job is to find every pair $(x, y)$ that satisfies both equations simultaneously.

Each equation is a line

The set of points $(x, y)$ that satisfy $a_1 x + b_1 y = c_1$ is a straight line. So our system is really two lines drawn on the same plane, and the solution set is their intersection. Three things can happen:

1. The lines cross at one point — the system has a unique solution.

2. The lines are parallel and distinct — the system has no solution.

3. The two equations describe the same line — the system has infinitely many solutions.

(Drag the coefficient sliders in the widget. Try $a_1 = 1, b_1 = 1, c_1 = 4$ and $a_2 = 1, b_2 = 1, c_2 = 6$. The lines should be parallel and you should see "no solution." Now change $c_2$ to $4$: the lines snap together and you get infinitely many.)

Two algebraic methods to find the crossing

Elimination. Multiply one or both equations so that one variable has matching coefficients (or coefficients that cancel). Add or subtract to eliminate that variable. Solve for the remaining unknown, then back-substitute. This is the workhorse and the foundation of Gaussian elimination later in the book.

Substitution. Solve one equation for one variable, e.g., $x = c_1/a_1 - (b_1/a_1) y$. Plug that expression into the other equation. Now you have one equation in one unknown — solve, then back-substitute.

The two methods always agree on the answer. Pick whichever is faster for the specific system.

Try it

• Predict first: where do the lines $2x + 3y = 12$ and $x - y = 1$ cross? Set those coefficients in the widget to verify.
• By elimination: solve $x + y = 7$ and $x - y = 3$. (Hint: add the two equations to kill $y$ instantly.)
• By substitution: solve $3x - 2y = 7$ and $x + y = 4$.
• Modify the widget sliders until you produce all three outcomes (unique, none, infinitely many). What relationship between the coefficients triggers each?

Pause: in the widget, the "determinant" readout switches sign as you drag a slider. What is the geometric meaning of determinant = 0? (We will name this the "singular case" in Chapter 17.)

Try it in code

A trap to watch for

When elimination produces a contradiction like $0 = 7$, students sometimes write down $x = 7$ or panic. Neither is right. A contradiction means the lines are parallel and the system has no solution — the correct answer is "no solution." When elimination produces $0 = 0$ (always true), the lines coincide and there are infinitely many solutions parameterised by one free variable. The fix: read the final equation literally before deciding what it says about $x$ and $y$.

What you now know

You can solve any 2x2 linear system either by elimination or substitution, and you can read off the geometric story (one crossing, parallel, coincident) directly from the algebra. The next section extends both ideas to three unknowns, where each equation is a plane in 3D space.

Quick check

References

• Lang, S. (1971). Basic Mathematics. Springer. Chapter 2, §1 — the geometric view of two equations in two unknowns.
• Strang, G. (2016). Introduction to Linear Algebra, 5th ed. Wellesley-Cambridge. Chapter 1 — row picture vs column picture of linear systems.
• Hefferon, J. (2020). Linear Algebra, 4th ed. (open access). Chapter 1 — Gauss-Jordan elimination introduced as systematic 2x2 elimination.