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. We begin by describing those three possibilities. Substitution, elimination, matrices, and null spaces will then provide increasingly useful ways to determine the whole 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

$$a_1x_1+\cdots+a_nx_n=b,$$

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

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

Consistent and inconsistent systems

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

Otherwise, it is inconsistent.

This is a very small definition, but it carries a lot of meaning.

• 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

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.

The theorem is not a guess. Later sections justify it by elimination and row-reduction, but the point is already visible here: once a free variable appears, there are infinitely many choices; once a contradiction appears, there are none.

A system with one solution

Worked example

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

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

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

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

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

This is the kind of answer the course wants: a complete description of all solutions, not just one sample solution.

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

Two systems are equivalent if they have the same solution set.

Definition

Equivalent systems

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

This definition is stronger than "they look similar" and stronger than "they have the same number of equations." Equivalence is about solutions only.

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

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.

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

Use line-intersection examples to see solution sets as intersections, then connect reversible equation rewrites to the augmented-matrix explorer.

1. Conditions intersect

   A system is a stack of simultaneous conditions. A tuple belongs to the solution set only if it passes every equation.

2. One point

   For x+y=5 and 2x-3y=-5, the two lines meet at (2,3), so the solution set is the singleton {(2,3)}.

3. No point

   Parallel distinct lines give an empty intersection. In the two-variable picture, that is an inconsistent system.

4. Whole line

   Coincident lines produce infinitely many solutions. A parameter records the whole family rather than one sample point.

5. Equivalent rewrites

   The three elementary equation operations are safe because each has an inverse operation. They rewrite the system while preserving the solution set.

6. Matrix bridge

   The augmented matrix is only a compact way to package the same conditions: each row is still one equation.

The visible equations may be rewritten, but the object we protect is the full solution set: all ordered tuples that satisfy every equation at the same time.

Why the course starts with solution sets

The later matrix language does not replace this section. It formalizes it.

Once we introduce coefficient matrices and augmented matrices, the same system can be encoded more compactly. Once we introduce row operations, we can transform one equation system into an equivalent one. Once we reach row-echelon or reduced row-echelon form, the solution set becomes easier to read.

So if the solution set is the object, then every later technique is just a different lens.

Try the interactive preview below once you are comfortable reading a small system as a list of conditions.

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

Checkpoint

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

Test the pair against both equations.

Solution · Answer

$(2, 2)$.

Checkpoint

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.

Checkpoint

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

This is the course definition.

Solution · Answer

Equivalent systems.

Checkpoint

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

Checkpoint

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

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

Checkpoint

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.

Read this first

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