Topics

Unique, infinite and no solutions

The three possibilities

Every system of linear equations has exactly one of three outcomes. There is no fourth case, and "two solutions" is impossible.

Outcome	Algebraic signature	Geometry of three planes
Unique solution	elimination produces definite values	meet at a single point
Infinitely many	a true statement such as $0 = 0$	meet along a line, or coincide as a plane
No solution	a false statement such as $0 = 7$	have no common point

$0 = 0$ and $0 = 7$ look almost identical and mean opposite things. A true statement means every point on some line or plane works; a false one means nothing works.

Why two solutions is impossible

If two different points satisfy every equation, so does every point on the line joining them.
The reason is linearity: if $P_1$ and $P_2$ both satisfy $ax + by + cz = d$ , then so does any weighted average of them, since the left-hand side distributes across the combination.
So the solution set can be empty, a point, a line, or a plane — never two isolated points.

Recognising each case

Unique — every variable ends up with a definite value, and the check in all three equations balances.
Infinitely many — one equation turns out to be a combination of the others, adding no new information. Look for:
- one equation being the sum or difference of two others
- one equation being a multiple of another
- a row of zeros with a zero constant in the echelon form
None — two equations demand contradictory things. Look for:
- two equations with proportional coefficients but a non-proportional constant — parallel planes
- a row of zeros with a non-zero constant in the echelon form

Writing the general solution

"Infinitely many" is not an answer. The answer is the general solution: a description of every point in the solution set.
The procedure:
- Set the free variable equal to a parameter, usually $t$ .
- Express the other two variables in terms of $t$ , working up from the reduced system.
- State the parameter's range — usually $t \in \mathbb{R}$ .
- Check the general solution in all three original equations. Every $t$ term must cancel.
Worked through — from the reduced system $x + 2y + z = 8$ $x + 2 y + z = 8$ and $-y - 3z = -9$ $- y - 3 z = - 9$ :
- Let $z = t$
- Then $y = 9 - 3t$
- And $x = 8 - 2(9 - 3t) - t = 5t - 10$
- General solution: $(5t - 10,\ 9 - 3t,\ t)$ for $t \in \mathbb{R}$
Give a couple of particular solutions too. Substituting $t = 0$ and $t = 1$ produces two concrete points and demonstrates the family is real.

One parameter or two

One free variable → a line, described with one parameter.
Two free variables → a plane, described with two parameters — this happens when two of the three equations are multiples of the third, so only one independent equation remains.
Count the independent equations. Three unknowns minus the number of independent equations gives the number of parameters.

Describing an inconsistency

Name the contradiction. "Equations (1) and (3) require $2x + 4y - 2z$ to equal both 6 and 10, which is impossible."
Point to the geometry where you can: two parallel planes, or three planes forming a triangular prism with no common point.
In a context, say what it means: the requirements as stated cannot all be met, and which two conflict.

Tips and Tricks

A true statement means infinitely many; a false one means none. Getting these the wrong way round is the single most common error in this standard, and it costs the whole question.

Never write "infinitely many solutions" and stop. Give the general solution with a parameter — that is what the marks are for.

Check the general solution in all three equations. The $t$ terms must cancel completely; if any survives, the parametrisation is wrong.

Look for one equation being the sum of two others before starting a long elimination. It takes ten seconds and tells you the answer will be a line.

Two solutions is never the answer. If you find two, either they are the same point written differently, or one of them fails an equation you did not check.

In context, translate the outcome. "No solution" means the specification is contradictory; "infinitely many" means the information given was not enough to pin the answer down.

Worked ExampleClassifying and solving three systems

Determine the nature of the solutions of each system, and give the solution set in each case.

(a) $x + 2y + z = 8$ , $2x + 3y - z = 7$ , $3x + 5y = 15$ (b) $x + 2y + z = 8$ , $2x + 3y - z = 7$ , $3x + 5y = 20$ (c) $x + 2y + z = 8$ , $2x + 3y - z = 7$ , $x + y + z = 6$

Step 1 — Notice the structure common to all three

In every system the first two equations are the same. Adding them:

$(x + 2y + z) + (2x + 3y - z) = 8 + 7$

$3x + 5y = 15 \qquad (\star)$

Step 2 — System (a): the third equation IS the sum

The third equation is $3x + 5y = 15$ , exactly $(\star)$ .

$\textbf{Infinitely many solutions — the planes meet in a LINE.}$

Find the general solution. Let $z = t$ .

From the first equation: $x + 2y = 8 - t$ From the second: $2x + 3y = 7 + t$

Eliminate $x$ : multiply the first by 2 and subtract from the second… setting it out:

$2(x + 2y) = 2(8 - t) \quad \Rightarrow \quad 2x + 4y = 16 - 2t$

$(2x + 4y) - (2x + 3y) = (16 - 2t) - (7 + t)$

$y = 9 - 3t$

Then:

$x = 8 - t - 2(9 - 3t) = 8 - t - 18 + 6t = 5t - 10$

$\boxed{(x,\ y,\ z) = (5t - 10,\ 9 - 3t,\ t), \quad t \in \mathbb{R}}$

Check in all three equations:

(1): $(5t-10) + 2(9-3t) + t = 5t - 10 + 18 - 6t + t = 8$ ✓
(2): $2(5t-10) + 3(9-3t) - t = 10t - 20 + 27 - 9t - t = 7$ ✓
(3): $3(5t-10) + 5(9-3t) = 15t - 30 + 45 - 15t = 15$ ✓

Every $t$ cancels, as a correct general solution requires.

Two particular solutions: $t = 2$ gives $(0, 3, 2)$ ; $t = 3$ gives $(5, 0, 3)$ . Both satisfy all three equations ✓

Step 3 — System (b): the third equation CONTRADICTS the sum

The third equation is $3x + 5y = 20$ , but $(\star)$ says the first two force $3x + 5y = 15$ .

Subtracting:

$0 = 20 - 15 \quad \Rightarrow \quad 0 = 5$

$\textbf{A FALSE statement — NO SOLUTION.}$

Naming the inconsistency: the first two equations together require $3x + 5y$ to be 15, while the third requires it to be 20. No triple can satisfy both demands.

Geometrically, the third plane is parallel to the line of intersection of the first two, and offset from it — the three planes form a triangular prism, with each pair meeting in a line but no point lying on all three.

Step 4 — System (c): an independent third equation

The third equation is $x + y + z = 6$ , which is not a multiple of, or a combination of, the other two — its coefficient pattern $(1,1,1)$ cannot be produced from $(1,2,1)$ and $(2,3,-1)$ with a constant $z$ coefficient of 1 unless… test it directly.

Any combination $\alpha(1,2,1) + \beta(2,3,-1)$ giving $(1,1,1)$ would need

$\alpha + 2\beta = 1, \qquad 2\alpha + 3\beta = 1, \qquad \alpha - \beta = 1$

From the first and third: $\alpha = 1 + \beta$ , so $1 + 3\beta = 1$ , giving $\beta = 0$ and $\alpha = 1$ . But then the second requires $2(1) + 0 = 1$ , which is false. No combination works, so the third equation is independent.

Solve the system.

$(2) - 2(1)$ : $(2x + 3y - z) - (2x + 4y + 2z) = 7 - 16$

$-y - 3z = -9 \quad \Rightarrow \quad y + 3z = 9 \qquad (4)$

$(3) - (1)$ : $(x + y + z) - (x + 2y + z) = 6 - 8$

$-y = -2 \quad \Rightarrow \quad y = 2$

From (4): $2 + 3z = 9$ , so $z = \frac{7}{3}$ .

From (1): $x = 8 - 2(2) - \frac{7}{3} = 4 - \frac{7}{3} = \frac{5}{3}$ .

$\textbf{UNIQUE solution: } \left(\frac{5}{3},\ 2,\ \frac{7}{3}\right)$

Check in all three:

(1): $\frac{5}{3} + 4 + \frac{7}{3} = \frac{5 + 12 + 7}{3} = \frac{24}{3} = 8$ ✓
(2): $\frac{10}{3} + 6 - \frac{7}{3} = \frac{10 - 7}{3} + 6 = 1 + 6 = 7$ ✓
(3): $\frac{5}{3} + 2 + \frac{7}{3} = \frac{12}{3} + 2 = 4 + 2 = 6$ ✓

Step 5 — Compare the three

System	Third equation	Signature	Solutions	Geometry
(a)	equals the sum of the first two	$0 = 0$	infinitely many, one parameter	planes meet in a line
(b)	contradicts the sum	$0 = 5$	none	triangular prism
(c)	independent	definite values	unique	meet at a point

Unique, infinite and no solutions

Covers AS91587

The three possibilities

Every system of linear equations has exactly one of three outcomes. There is no fourth case, and "two solutions" is impossible.

Outcome	Algebraic signature	Geometry of three planes
Unique solution	elimination produces definite values	meet at a single point
Infinitely many	a true statement such as $0 = 0$	meet along a line, or coincide as a plane
No solution	a false statement such as $0 = 7$	have no common point

$0 = 0$ and $0 = 7$ look almost identical and mean opposite things. A true statement means every point on some line or plane works; a false one means nothing works.

Why two solutions is impossible

If two different points satisfy every equation, so does every point on the line joining them.
The reason is linearity: if $P_1$ and $P_2$ both satisfy $ax + by + cz = d$ , then so does any weighted average of them, since the left-hand side distributes across the combination.
So the solution set can be empty, a point, a line, or a plane — never two isolated points.

Recognising each case

Unique — every variable ends up with a definite value, and the check in all three equations balances.
Infinitely many — one equation turns out to be a combination of the others, adding no new information. Look for:
- one equation being the sum or difference of two others
- one equation being a multiple of another
- a row of zeros with a zero constant in the echelon form
None — two equations demand contradictory things. Look for:
- two equations with proportional coefficients but a non-proportional constant — parallel planes
- a row of zeros with a non-zero constant in the echelon form

Writing the general solution

"Infinitely many" is not an answer. The answer is the general solution: a description of every point in the solution set.
The procedure:
- Set the free variable equal to a parameter, usually $t$ .
- Express the other two variables in terms of $t$ , working up from the reduced system.
- State the parameter's range — usually $t \in \mathbb{R}$ .
- Check the general solution in all three original equations. Every $t$ term must cancel.
Worked through — from the reduced system $x + 2y + z = 8$ $x + 2 y + z = 8$ and $-y - 3z = -9$ $- y - 3 z = - 9$ :
- Let $z = t$
- Then $y = 9 - 3t$
- And $x = 8 - 2(9 - 3t) - t = 5t - 10$
- General solution: $(5t - 10,\ 9 - 3t,\ t)$ for $t \in \mathbb{R}$
Give a couple of particular solutions too. Substituting $t = 0$ and $t = 1$ produces two concrete points and demonstrates the family is real.

One parameter or two

One free variable → a line, described with one parameter.
Two free variables → a plane, described with two parameters — this happens when two of the three equations are multiples of the third, so only one independent equation remains.
Count the independent equations. Three unknowns minus the number of independent equations gives the number of parameters.

Describing an inconsistency

Name the contradiction. "Equations (1) and (3) require $2x + 4y - 2z$ to equal both 6 and 10, which is impossible."
Point to the geometry where you can: two parallel planes, or three planes forming a triangular prism with no common point.
In a context, say what it means: the requirements as stated cannot all be met, and which two conflict.

Tips and Tricks

A true statement means infinitely many; a false one means none. Getting these the wrong way round is the single most common error in this standard, and it costs the whole question.

Never write "infinitely many solutions" and stop. Give the general solution with a parameter — that is what the marks are for.

Check the general solution in all three equations. The $t$ terms must cancel completely; if any survives, the parametrisation is wrong.

Look for one equation being the sum of two others before starting a long elimination. It takes ten seconds and tells you the answer will be a line.

Two solutions is never the answer. If you find two, either they are the same point written differently, or one of them fails an equation you did not check.

In context, translate the outcome. "No solution" means the specification is contradictory; "infinitely many" means the information given was not enough to pin the answer down.

Worked ExampleClassifying and solving three systems

Determine the nature of the solutions of each system, and give the solution set in each case.

(a) $x + 2y + z = 8$ , $2x + 3y - z = 7$ , $3x + 5y = 15$ (b) $x + 2y + z = 8$ , $2x + 3y - z = 7$ , $3x + 5y = 20$ (c) $x + 2y + z = 8$ , $2x + 3y - z = 7$ , $x + y + z = 6$

Step 1 — Notice the structure common to all three

In every system the first two equations are the same. Adding them:

$(x + 2y + z) + (2x + 3y - z) = 8 + 7$

$3x + 5y = 15 \qquad (\star)$

Step 2 — System (a): the third equation IS the sum

The third equation is $3x + 5y = 15$ , exactly $(\star)$ .

$\textbf{Infinitely many solutions — the planes meet in a LINE.}$

Find the general solution. Let $z = t$ .

From the first equation: $x + 2y = 8 - t$ From the second: $2x + 3y = 7 + t$

Eliminate $x$ : multiply the first by 2 and subtract from the second… setting it out:

$2(x + 2y) = 2(8 - t) \quad \Rightarrow \quad 2x + 4y = 16 - 2t$

$(2x + 4y) - (2x + 3y) = (16 - 2t) - (7 + t)$

$y = 9 - 3t$

Then:

$x = 8 - t - 2(9 - 3t) = 8 - t - 18 + 6t = 5t - 10$

$\boxed{(x,\ y,\ z) = (5t - 10,\ 9 - 3t,\ t), \quad t \in \mathbb{R}}$

Check in all three equations:

(1): $(5t-10) + 2(9-3t) + t = 5t - 10 + 18 - 6t + t = 8$ ✓
(2): $2(5t-10) + 3(9-3t) - t = 10t - 20 + 27 - 9t - t = 7$ ✓
(3): $3(5t-10) + 5(9-3t) = 15t - 30 + 45 - 15t = 15$ ✓

Every $t$ cancels, as a correct general solution requires.

Two particular solutions: $t = 2$ gives $(0, 3, 2)$ ; $t = 3$ gives $(5, 0, 3)$ . Both satisfy all three equations ✓

Step 3 — System (b): the third equation CONTRADICTS the sum

The third equation is $3x + 5y = 20$ , but $(\star)$ says the first two force $3x + 5y = 15$ .

Subtracting:

$0 = 20 - 15 \quad \Rightarrow \quad 0 = 5$

$\textbf{A FALSE statement — NO SOLUTION.}$

Naming the inconsistency: the first two equations together require $3x + 5y$ to be 15, while the third requires it to be 20. No triple can satisfy both demands.

Step 4 — System (c): an independent third equation

Any combination $\alpha(1,2,1) + \beta(2,3,-1)$ giving $(1,1,1)$ would need

$\alpha + 2\beta = 1, \qquad 2\alpha + 3\beta = 1, \qquad \alpha - \beta = 1$

Solve the system.

$(2) - 2(1)$ : $(2x + 3y - z) - (2x + 4y + 2z) = 7 - 16$

$-y - 3z = -9 \quad \Rightarrow \quad y + 3z = 9 \qquad (4)$

$(3) - (1)$ : $(x + y + z) - (x + 2y + z) = 6 - 8$

$-y = -2 \quad \Rightarrow \quad y = 2$

From (4): $2 + 3z = 9$ , so $z = \frac{7}{3}$ .

From (1): $x = 8 - 2(2) - \frac{7}{3} = 4 - \frac{7}{3} = \frac{5}{3}$ .

$\textbf{UNIQUE solution: } \left(\frac{5}{3},\ 2,\ \frac{7}{3}\right)$

Check in all three:

(1): $\frac{5}{3} + 4 + \frac{7}{3} = \frac{5 + 12 + 7}{3} = \frac{24}{3} = 8$ ✓
(2): $\frac{10}{3} + 6 - \frac{7}{3} = \frac{10 - 7}{3} + 6 = 1 + 6 = 7$ ✓
(3): $\frac{5}{3} + 2 + \frac{7}{3} = \frac{12}{3} + 2 = 4 + 2 = 6$ ✓

Step 5 — Compare the three

System	Third equation	Signature	Solutions	Geometry
(a)	equals the sum of the first two	$0 = 0$	infinitely many, one parameter	planes meet in a line
(b)	contradicts the sum	$0 = 5$	none	triangular prism
(c)	independent	definite values	unique	meet at a point