1.1 Equations and solution sets

When a system of linear equations is written down, the real object of interest is not the list of equations itself. It is the set of all tuples of numbers that make every equation true at the same time.

That solution set can be a single point, no point at all, or an infinite family of points. Solving the system means determining which tuples belong to that set.

Which equations are linear?

Fix an ordered list of unknowns $x_1,\ldots,x_n$, all taking real values. A linear equation in those unknowns has the form

where the coefficients $a_1,\ldots,a_n$ and right-hand side $b$ are fixed real numbers. A linear system is a finite collection of such equations using the same list of unknowns. An unknown may have coefficient zero in some equations; its position in the list is still retained. The coefficient belongs to the problem data, whereas the unknown is what we are allowed to choose.

For example, $3x-2y=7$ is linear in $x,y$, but $xy=7$ and $x^2+y=7$ are not: the first contains a product of unknowns and the second contains a square of an unknown. An equation such as $ax+y=1$ is linear in $x,y$ when $a$ is a fixed parameter. If $a$ were itself included among the unknowns, the product $ax$ would violate this definition. Thus “linear” is always read relative to the specified unknowns, not merely by looking for a familiar symbol.

Allowing zero coefficients is essential for elimination. The equation $0x+0y=0$ is satisfied by every pair in $\mathbb R^2$, while $0x+0y=1$ is satisfied by none. Neither is an instruction to divide by zero. Each remains a condition to be interpreted, and together they explain why a row of zeros and a contradiction row have completely different meanings.

What a solution set records

Definition 1.1.1

(Solution set)

A solution set is the collection of every number or vector that satisfies the whole system.

If the system has unknowns $x_1, x_2, \ldots, x_n$, then a solution is an ordered $n$-tuple $(s_1, s_2, \ldots, s_n)$ such that every equation becomes a true statement after we substitute $x_i = s_i$ for each $i$.

That order matters. The tuple $(2, 3)$ is not the same solution as $(3, 2)$. Likewise, a tuple is not a set. We do not write $\{2,3\}$ when the intended solution is the ordered pair $(2, 3)$.

The system itself can be read as an intersection of conditions:

• the first equation cuts out the tuples that satisfy it;
• the second equation cuts out the tuples that satisfy it;
• the solution set is the intersection of all those individual solution sets.

This is why even one failed equation disqualifies a tuple.

Concept lensStructural

Equations describe a set of admissible tuples

Keep the ambient space $\mathbb R^n$ fixed. An equation selects some of its tuples, and a system keeps only tuples selected by every equation. This changes what it means to check an answer: finding one acceptable tuple establishes existence, while solving the system requires a description of the entire set.

Adding an equation intersects the old solution set with one more condition. It can leave that set unchanged or make it smaller; it cannot introduce a new solution. For instance, adding $2x+2y=8$ to $x+y=4$ leaves the set unchanged, whereas adding $x=1$ keeps only $(1,3)$. Adding $x+y=5$ removes every tuple. These are three different effects of adding one equation, so counting equations alone cannot determine the outcome.

Conversely, deleting an equation can enlarge the solution set. To justify a deletion without changing the answer, show that the remaining equations already force the deleted one. The word “redundant” expresses precisely this logical relationship. It is stronger than noticing that two equations look similar.

Consistent or inconsistent

Definition 1.1.2

(Consistent and inconsistent systems)

A linear system is consistent if it has at least one solution.

Otherwise, it is inconsistent.

• A consistent system may have exactly one solution.
• A consistent system may have infinitely many solutions.
• An inconsistent system has no solutions, so its solution set is empty.

Later row-reduction arguments prove that these are the only possibilities.

Theorem 1.1.3

(A linear system has only three possible kinds of solution set)

For a linear system, the solution set is either:

1. a singleton, so the system has a unique solution;
2. the empty set, so the system is inconsistent;
3. an infinite set, so the system has infinitely many solutions.

Row reduction will prove this classification. In a consistent system, a free variable gives infinitely many choices; a contradiction rules out every solution.

A system with one solution

Worked example 1.1.4

(A tiny system with one solution)

Solve

$$x + y = 4, \qquad x - y = 0.$$

The second equation says $x = y$. Substitute that into the first equation:

$$y + y = 4.$$

So $2y = 4$, hence $y = 2$, and therefore $x = 2$.

The solution set is

$$\{(2, 2)\}.$$

This is the simplest possible example of a consistent system: there is exactly one ordered pair that works.

A system with no solution

Worked example 1.1.5

(A contradiction gives an empty solution set)

Consider

$$x + y = 1, \qquad x + y = 3.$$

If a pair $(x, y)$ satisfied both equations, then the same left-hand side would have to equal two different numbers. Subtracting the first equation from the second gives

$$0 = 2,$$

which is impossible.

So the system is inconsistent, and its solution set is empty:

$$\varnothing.$$

The important subtlety is that an inconsistent system is not a system with "many" solutions. It has none.

Common mistake

(Inconsistent does not mean more than one solution)

Students sometimes read "inconsistent" as "too complicated" or "overdetermined." That is not the definition. Inconsistent means that there is no tuple of values that satisfies all equations simultaneously.

A system with infinitely many solutions

Worked example 1.1.6

(The same line can be written in more than one way)

Consider

$$x + y = 4, \qquad 2x + 2y = 8.$$

The second equation is just $2$ times the first, so it adds no new condition. Every pair on the line $x + y = 4$ satisfies both equations.

If we let $x = t$, then $y = 4 - t$, where $t \in R$. So the solution set is

$$\{(t, 4 - t) \mid t \in R\}.$$

There are infinitely many solutions because every real value of $t$ gives a different ordered pair.

The same phenomenon appears in larger systems. The notation gets longer, but the logic is the same: each free variable introduces one independent choice.

Worked example 1.1.7

(A four-variable system written as a full solution set)

Solve

$$x_1 - 2x_2 - x_3 + x_4 = 1, \qquad x_2 + x_3 - x_4 = 2, \qquad x_3 + 2x_4 = 3.$$

Start from the last equation:

$$x_3 = 3 - 2x_4.$$

Substitute this into the second equation:

$$x_2 + (3 - 2x_4) - x_4 = 2,$$

so

$$x_2 = -1 + 3x_4.$$

Now substitute both expressions into the first equation:

$$x_1 - 2(-1 + 3x_4) - (3 - 2x_4) + x_4 = 1.$$

Simplifying gives

$$x_1 = 2 + 3x_4.$$

Let $x_4 = t$. Then every solution has the form

$$(x_1, x_2, x_3, x_4) = (2 + 3t, -1 + 3t, 3 - 2t, t),$$

so the solution set is

$$\{(2 + 3t, -1 + 3t, 3 - 2t, t) \mid t \in R\}.$$

Each real value of $t$ specifies a solution; describing the entire family also requires checking that every solution has this form.

Verify a parameterization in both directions

The four-variable calculation has shown that any solution must have the stated form. This is the exhaustiveness direction: start with an arbitrary solution, call its fourth coordinate $t$, and use the equations to recover the other three coordinates. No solution can lie outside the displayed family.

There is a second obligation. For every real $t$, the proposed tuple must actually satisfy the original equations. Substitution gives, in their original order,

$$\begin{aligned} (2+3t)-2(-1+3t)-(3-2t)+t&=1,\\ (-1+3t)+(3-2t)-t&=2,\\ (3-2t)+2t&=3. \end{aligned}$$

These identities hold for every real parameter, so every member of the family is a solution. Finally, different values of $t$ give different fourth coordinates. Thus the family contains infinitely many distinct tuples, not merely infinitely many names for one tuple.

This two-direction check is useful whenever algebra produces a proposed answer set. Substitution establishes that there are no extraneous answers; exhaustiveness establishes that no answers were lost. Checking only a few sample values proves neither the universal substitution claim nor completeness. A parameter can also be renamed without changing the set: for $x+y=4$, the families $(t,4-t)$ and $(4-s,s)$ agree because setting $s=4-t$ converts one to the other, and the substitution is reversible. Equality of solution sets does not require identical parameter letters or identical-looking formulas.

Why equivalent systems matter

Definition 1.1.8

(Equivalent systems)

Two linear systems are equivalent if and only if they have exactly the same solution set.

Common mistake

(Same number of equations does not mean equivalent)

Two systems can have the same number of equations but different solution sets. They can also have different numbers of equations and still be equivalent.

There are three elementary equation operations:

1. swap two equations;
2. multiply one equation by a nonzero scalar;
3. add a multiple of one equation to another equation.

These are the equation-level version of the row operations used later on augmented matrices.

Theorem 1.1.9

(Elementary equation operations preserve the solution set)

If one system is obtained from another by a finite sequence of the three elementary equation operations, then the two systems are equivalent.

Proof

(Why the three elementary equation operations are safe)

Each operation is reversible. In row replacement, the source and target equations are distinct, so the source equation remains available to undo the operation.

• Swapping two equations only changes the order in which the conditions are listed.
• Multiplying an equation by a nonzero scalar produces an equivalent equation, because we can reverse the move by multiplying by its reciprocal.
• Replacing equation $j$ by $\alpha \times$ equation $i$ plus equation $j$ is reversible by subtracting $\alpha \times$ equation $i$ from the new equation $j$.

Since each step is reversible, no step changes the solution set.

This is the formal reason elimination is allowed. We are not changing the problem; we are rewriting it in a more readable form.

Two variables: geometry gives a quick picture

When there are two unknowns, an equation $ax+by=c$ describes a line provided $a$ and $b$ are not both zero. For two such equations, the solution set is the intersection of their lines.

• If the lines meet at one point, the system has a unique solution.
• If the lines are parallel and distinct, the system is inconsistent.
• If the lines coincide, the system has infinitely many solutions.

The zero-coefficient cases must be interpreted separately: $0x+0y=0$ selects the whole plane and $0x+0y=c$ with nonzero $c$ selects the empty set. Neither selects a line. With three or more equations, having pairwise intersections also does not guarantee a common solution. The lines $x=0$, $y=0$, and $x+y=1$ meet pairwise, but no point belongs to all three. The definition of a solution requires simultaneous satisfaction of every equation.

Interactive · Compare casesone solution

Three geometric possibilities for two-variable linear systems

Choose a case and compare the two lines with their common solution set.

Relative position

one solutionno solutioninfinitely many

Selected: one solution

intersection point

no solution

parallel distinct lines

infinitely many

same line

A common point exists

Yes

Exactly one common point

Yes

The lines coincide

No

Solution set

one solution

Holds: The two lines have a common solution

Two distinct intersecting lines share exactly one point. If they shared two distinct points they would be the same line.

NoticeThese are schematic drawings, not specified equations or mathematical coordinates. Each equation here describes a genuine line; degenerate equations such as 0 = 0 or 0 = 1 need a separate check.

← → Switch while a case button has focus.

This geometric picture is useful because it makes the three possibilities feel inevitable instead of arbitrary.

From equations to a matrix record

In an equation operation, the coefficients and constants change while the ordered list of unknowns stays fixed. A matrix records those changing numbers in fixed positions. The explorer below places each equation beside its matrix row; the same three equation operations then become row operations. Their purpose remains to preserve the whole solution set while making it easier to describe.

Common mistakes and subtle points

Common mistake

(A solution is an ordered tuple, not a bag of numbers)

$(2, 3)$ solves a system in two variables, but $\{2,3\}$ does not mean the same thing. Order matters because the first number belongs to $x_1$ and the second belongs to $x_2$.

Common mistake

(A contradiction row means no solution)

If elimination produces a row like $0 = 1$, the system is inconsistent. Do not continue trying to solve it as if it were a valid equation.

Common mistake

(A free variable is part of the answer)

When a solution set is written with parameters, the parameter is not a missing answer. It is the correct way to describe the whole family of solutions.

Quick checks

Quick check

(Which ordered pair solves both equations $x + y = 4$ and $x - y = 0$?)

Test the pair against both equations.

Solution · Answer

$(2, 2)$.

Quick check

(Is the system $x + y = 1$, $x + y = 3$ consistent?)

Use the definition of consistency, not the number of equations.

Solution · Answer

No. It is inconsistent because the two equations contradict each other.

Quick check

(If two systems have the same solution set, what do we call them?)

This is the course definition.

Solution · Answer

Equivalent systems.

Quick check

(Why does swapping two equations not change the solution set?)

Think about what the word "solution" means.

Solution · Guided solution

A solution must satisfy every equation in the system. The order of the equations does not matter, so swapping them does not change which tuples satisfy all of them.

Exercises

Quick check

(Write the solution set of $x_1 - 5x_4 = 1$, $x_2 + x_4 = 2$, $x_3 + 3x_4 = 3$ in parametric form.)

Use one free parameter.

Solution · Guided solution

Let $x_4 = t$. Then

$$x_1 = 1 + 5t, \qquad x_2 = 2 - t, \qquad x_3 = 3 - 3t.$$

So the solution set is

$$\{(1 + 5t, 2 - t, 3 - 3t, t) \mid t \in R\}.$$

Quick check

(Find $c$ so that the system $x + 2y - 5z = 6$, $2x + 3y - 2z = 7$, $x + cy + z = 0$ has no solution.)

You may eliminate $x$ first.

Solution · Guided solution

Eliminate $x$ from the second equation by subtracting $2$ times the first:

$$-y + 8z = -5.$$

Eliminate $x$ from the third equation by subtracting the first:

$$(c - 2)y + 6z = -6.$$

For inconsistency, the two resulting equations in $y$ and $z$ must become parallel but different. That happens when their coefficient rows are multiples. So solve

$$\frac{c - 2}{-1} = \frac{6}{8},$$

which gives

$$c = \frac{5}{4}.$$

With $c = 5/4$, the two equations are parallel but not the same, so the system has no solution.

Check the exceptional value and all other values

The coefficient comparison above finds a candidate exceptional value; a complete argument should also verify that it is the only one. From $-y+8z=-5$ we obtain $y=5+8z$. Substituting this in the remaining equation gives

$$(8c-10)z=4-5c.$$

At $c=5/4$ this reads $0=-9/4$, so inconsistency is explicit. For every other value of $c$, the coefficient $8c-10$ is nonzero, and the last equation determines exactly one $z$. It then determines $y$, and the first original equation determines $x=6-2y+5z$. Each substitution was reversible, so this triple solves the original system. This checks both the exceptional case and the claim that no further values were missed.

The same example distinguishes a parameter from a free variable. Here $c$ is part of the given system: changing it changes the problem. Once $c$ is fixed away from the exceptional value, none of $x,y,z$ is free. In the earlier four-variable example, by contrast, $t$ varies within the solutions of one fixed system. Always identify whether a letter indexes different problems or different solutions to the same problem before introducing cases.

Related notes

This note is the starting point for the matrix treatment of systems. The next useful pages are:

• 2.2 Augmented matrices and row operations
• 2.3 Gaussian elimination and RREF
• 4.1 Homogeneous systems and null space