The discussion of a system
Learning Objectives
- Classify a system as compatible determinate, compatible indeterminate, or incompatible.
- Recognise the three cases from the outcome of elimination: a value, \(0 = c\), or \(0 = 0\).
- Interpret the cases geometrically: intersecting, parallel, or coincident lines.
- Apply the ratio criterion to classify a system from its coefficients.
- Describe the solution set of an indeterminate system in parametric form.
When elimination misbehaves
Every system in the last lesson had exactly one solution, because its two lines had different slopes. But run elimination on this innocent-looking system:
\[\begin{cases} x + y = 2 \\ x + y = 5 \end{cases}\]
Subtracting the first equation from the second gives \(0 = 3\) — both unknowns vanished, leaving a false statement. We met this situation with single equations in the previous unit: it is the contradiction case of the discussion of \(ax + b = 0\). And just as there, it has a precise meaning: no pair \((x, y)\) satisfies both equations. The sum of two numbers cannot be \(2\) and \(5\) at once.
This lesson classifies everything that can happen.
Compatible and incompatible systems
A system of equations is:
compatible determinate if it has exactly one solution;
compatible indeterminate if it has infinitely many solutions;
incompatible if it has no solution.
"Compatible" records that the two conditions can hold together — the system has at least one solution. (Some English textbooks say consistent for compatible and inconsistent for incompatible; the meaning is identical.)
The three cases, worked
Exactly one solution. \(\begin{cases} x + y = 5 \\ x - y = 1 \end{cases}\) Adding eliminates \(y\): \(2x = 6\), so \(x = 3\), then \(y = 2\). One solution, \((3, 2)\): the system is compatible determinate.
No solution. \(\begin{cases} 2x + y = 3 \\ 4x + 2y = 10 \end{cases}\) Multiply the first equation by \(2\): \(4x + 2y = 6\). Subtracting this from the second gives \(0 = 4\) — false. No pair can satisfy both: incompatible.
Infinitely many solutions. \(\begin{cases} x - 2y = 1 \\ 3x - 6y = 3 \end{cases}\) The second equation is exactly \(3\) times the first — it adds no new information. Elimination gives \(0 = 0\), true for every pair satisfying the first equation. The solutions are all pairs on the line \(x - 2y = 1\); writing \(y = t\) for a free parameter,
\[\{\,(1 + 2t,\; t) \mid t \in \mathbb{R}\,\},\]
one solution for every real \(t\): compatible indeterminate.
Notice what elimination produced in each case: a genuine value, a false statement \(0 = c\) with \(c \neq 0\), or the identity \(0 = 0\) — precisely the three cases of the discussion of a single equation.
The picture: two lines in the plane
Each equation is a line, and the solutions of the system are the points common to both lines. Two lines in the plane can be positioned in exactly three ways — and they match the three cases one for one:
In the third figure the dashed line is drawn on top of the solid one: the two equations describe the same line, so every one of its points solves the system.
The ratio criterion
Consider the system
\[\begin{cases} a_1 x + b_1 y = c_1 \\ a_2 x + b_2 y = c_2 \end{cases}\]
with all coefficients of the second equation non-zero. Then the system is:
compatible determinate if \(\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}\) (the lines intersect);
incompatible if \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}\) (the lines are parallel and distinct);
compatible indeterminate if \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}\) (the lines coincide).
Assume also \(b_1 \neq 0\), so that each equation can be solved for \(y\) and read as a line in the familiar form:
\[y = -\frac{a_1}{b_1}x + \frac{c_1}{b_1}, \qquad y = -\frac{a_2}{b_2}x + \frac{c_2}{b_2}.\]
The slopes are equal exactly when \(\dfrac{a_1}{b_1} = \dfrac{a_2}{b_2}\), which after cross-multiplying and dividing is the condition \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}\).
If the slopes differ, the lines cross in exactly one point (Unit 1): one solution — determinate.
If the slopes are equal, the lines are parallel or coincident, and the intercepts decide which. The intercepts are equal exactly when \(\dfrac{c_1}{b_1} = \dfrac{c_2}{b_2}\), i.e. \(\dfrac{c_1}{c_2} = \dfrac{b_1}{b_2}\). Equal slope and equal intercept mean the same line — every point common, indeterminate. Equal slope and different intercepts mean parallel distinct lines — no point common, incompatible.
(If \(b_1 = 0\) or \(b_2 = 0\), one of the lines is vertical; the same three-way analysis goes through by comparing the vertical lines \(x = c/a\) directly.)
Classifying by ratios alone
Reading off the coefficients. For \(\begin{cases} 2x + 3y = 5 \\ 4x + 6y = 7 \end{cases}\): \(\dfrac{2}{4} = \dfrac{1}{2}\) and \(\dfrac{3}{6} = \dfrac{1}{2}\), but \(\dfrac{5}{7} \neq \dfrac{1}{2}\). Equal coefficient ratios, different constant ratio: incompatible — no need to run elimination at all.
A parameter. For which \(m\) is \(\begin{cases} x + 2y = 3 \\ 2x + my = 8 \end{cases}\) not determinate? The lines fail to intersect in a single point exactly when \(\dfrac{1}{2} = \dfrac{2}{m}\), i.e. \(m = 4\). And with \(m = 4\), the constant ratio is \(\dfrac{3}{8} \neq \dfrac{1}{2}\), so the system is incompatible: parallel lines. For every other \(m\) it is determinate.
Why never exactly two solutions?
A system of two linear equations can have \(0\), \(1\), or infinitely many solutions. Could some cleverly chosen system have exactly two? Or seventeen?
No — and the reason is geometric. Suppose a system had two different solutions, \(P\) and \(Q\). Both points lie on the first line, and both lie on the second. But through two distinct points there passes exactly one line — so the "two" lines are in fact the same line. And then every point of that line is a solution: infinitely many, not two.
So the moment a linear system acquires a second solution, it is forced to surrender infinitely many. The possible solution counts are exactly \(0\), \(1\), and \(\infty\) — nothing in between. This rigidity is special to linear equations, whose solution sets are lines; curves of other shapes can meet in \(2\), \(3\), or any finite number of points.
Exercises
Exercise 1
Run elimination (or compare ratios) and classify each system.
Exercise 2
Find the value of the parameter in each case.
Exercise 3
Decide whether each statement about systems is true or false.
Exercise 4
Classify each system as compatible (at least one solution) or incompatible (none).
Exercise 5
Each item describes the two lines of a system. How many solutions does the system have?
Summary
A system of two linear equations is compatible determinate (one solution), compatible indeterminate (infinitely many), or incompatible (none) — and these are the only possibilities.
Elimination reveals the case: a genuine value (determinate), a false statement \(0 = c\) (incompatible), or the identity \(0 = 0\) (indeterminate) — the same three outcomes as the discussion of a single first-degree equation.
Geometrically: intersecting, parallel, or coincident lines. The ratio criterion reads the case off the coefficients: \(\tfrac{a_1}{a_2} \neq \tfrac{b_1}{b_2}\) determinate; equal, with \(\tfrac{c_1}{c_2}\) different, incompatible; all three equal, indeterminate.
An indeterminate system's solutions form a line, written parametrically — e.g. \(\{(1 + 2t,\, t) \mid t \in \mathbb{R}\}\). Exactly two solutions is impossible: two shared points force the lines to coincide.