IB Analysis and Approaches HL · Systems of linear equations · Section 1 of 3

Linear equations and systems

3 worked examples · By Álvaro López

Key idea

A system of linear equations is a set of equations that must all hold at the same time. With three unknowns it can have a unique solution, infinitely many or none, and recognising which of the three cases you have is the basis of the whole topic. In the IB exam you need it to set up problems, express solutions with parameters and justify what happens when the system has no unique solution.

The linear equation in three unknowns

An equation is linear when each unknown appears on its own, raised to the power 1 and not multiplied by another unknown.

For example, $4 x + y - 2 z = 9$ is linear. In contrast, $x y + z = 1$ (a product of unknowns) or $x^{2} + y + z = 0$ (an unknown squared) are not.

A solution is a triple of numbers $(x, y, z)$ that makes the equality true when substituted. Solving the equation means finding all its solutions. With one equation and three unknowns there are infinitely many: you choose two values freely and the third is then determined. For example, in $4 x + y - 2 z = 9$ , with $x = 1$ and $y = 1$ we get $z = - 2$ , and $(1, 1, - 2)$ is a solution, since $4 + 1 + 4 = 9$ .

To describe all of them at once, we use the general solution: call $t$ and $s$ the unknowns you choose freely (the parameters), and solve for the third.

For $4 x + y - 2 z = 9$ : with $y = t$ and $z = s$ , $x = \frac{9 - t + 2 s}{4}$ , and the general solution is $(\frac{9 - t + 2 s}{4}, t, s)$ with $t, s \in R$ . Giving values to $t$ and $s$ produces each of the specific solutions.

Systems of three equations

When the three constant terms are zero, the system is homogeneous.

The reason is immediate: substituting $x = y = z = 0$ , each left-hand side equals $0$ , which is the constant term of each equation.

Equivalent systems

Two systems are equivalent when they have exactly the same solutions. This idea is what allows you to simplify a system without changing its answer, and it is what the Gaussian elimination method uses.

How many solutions it can have

According to the number of solutions, a system is classified as follows:

Type	Abbreviation	Solutions
Consistent, unique solution	SCD	Exactly one
Consistent, infinitely many solutions	SCI	Infinitely many (they depend on one or more parameters)
Inconsistent	SI	None

"Consistent" means that it has at least one solution; "unique" means that the solution is the only one. A system never has, for example, exactly two solutions: it has 0, 1 or infinitely many.

Two simple recognition criteria, which the Gaussian elimination method makes systematic:

If combining equations produces an impossible equality such as $0 = 3$ , the system is inconsistent.
If one equation is redundant (it is a combination of the others) and some unknown is free, there are infinitely many solutions, and they are expressed with parameters.

Worked examples

Example 1 Basic

Solutions of a linear equation

Consider the equation $2 x - y + 3 z = 7$ .

a) Find the value of $z$ so that $(2, - 3, z)$ is a solution.

b) Write the general solution, taking $y = t$ and $z = s$ as parameters.

Solution

a) Substitute $x = 2$ and $y = - 3$ :

$2 \cdot 2 - (- 3) + 3 z = 7 \Rightarrow 7 + 3 z = 7 \Rightarrow z = 0$

So $(2, - 3, 0)$ is a solution.

b) With $y = t$ and $z = s$ , solve for $x$ :

$2 x = 7 + t - 3 s \Rightarrow x = \frac{7 + t - 3 s}{2}$

The general solution is $(\frac{7 + t - 3 s}{2}, t, s)$ with $t, s \in R$ . Check with $t = - 3$ , $s = 0$ : $x = 2$ , which agrees with part a).

Example 2 Medium

Three systems, three behaviours

Classify each system (SCD, SCI or SI) and, when it has a solution, solve it.

a) $⎩ ⎨ ⎧ x + y + z 2 x - y + z x + 2 y - z = 6 = 3 = 2$

b) $⎩ ⎨ ⎧ x + y + z x - y + z 2 x + 2 z = 0 = 0 = 0$

c) $⎩ ⎨ ⎧ x + y + z 2 x + 2 y + 2 z x - y = 1 = 5 = 0$

Solution

a) Adding the first and third equations eliminates $z$ : $2 x + 3 y = 8$ . Adding the second and third also: $3 x + y = 5$ . From the second, $y = 5 - 3 x$ ; substituting into the first, $2 x + 15 - 9 x = 8$ , so $x = 1$ and $y = 2$ . From the first equation of the system, $z = 6 - 1 - 2 = 3$ . It is checked in the three equations. The solution is unique: SCD, $(1, 2, 3)$ .

b) It is homogeneous, so it has the trivial solution. Subtracting the second equation from the first: $2 y = 0$ , so $y = 0$ . The first and second equations become $x + z = 0$ , and the third, $2 x + 2 z = 0$ , says the same. One equation is redundant and $z$ is free: with $z = t$ , $x = - t$ . There are infinitely many solutions: SCI, $(- t, 0, t)$ with $t \in R$ .

c) If you subtract twice the first equation from the second, you get $0 = 3$ , which is impossible. No triple satisfies both: SI.

Example 3 Hard

A system that comes from a problem

A shop buys $x$ T-shirts, $y$ sweatshirts and $z$ caps. It buys 20 items in total and pays 270 euros. Each T-shirt costs 10 euros, each sweatshirt 25 euros and each cap 5 euros.

a) Set up the system that describes the situation with these two conditions, and express it with $y = t$ as a parameter. What type of system is it?

b) Given that $x$ , $y$ , $z$ are non-negative integers, how many different purchases are possible?

c) It is also known that there are 2 caps fewer than sweatshirts. Find $x$ , $y$ and $z$ .

Solution

a) The two conditions give:

${x + y + z 10 x + 25 y + 5 z = 20 = 270$

Dividing the second by 5: $2 x + 5 y + z = 54$ . Subtracting the first: $x + 4 y = 34$ , so $x = 34 - 4 y$ . Substituting into the first, $z = 20 - x - y = - 14 + 3 y$ . With $y = t$ :

$(x, y, z) = (34 - 4 t, t, 3 t - 14)$

There are two equations and three unknowns, and no contradiction: the system is consistent with infinitely many solutions.

b) We need $x \geq 0$ and $z \geq 0$ : $34 - 4 t \geq 0$ gives $t \leq 8.5$ , and $3 t - 14 \geq 0$ gives $t \geq 4.67 \dots$ . As $t$ is an integer, $t \in {5, 6, 7, 8}$ : there are 4 possible purchases.

c) The new condition is $y - z = 2$ , that is, $t - (3 t - 14) = 2$ , which gives $- 2 t + 14 = 2$ and $t = 6$ . Then $x = 34 - 24 = 10$ , $y = 6$ , $z = 4$ . Check: $10 + 6 + 4 = 20$ items and $100 + 150 + 20 = 270$ euros.

In the exam: set up the system by stating what each unknown represents and, if the solution depends on a parameter, name it and say what type of system it is. After obtaining the general solution, check the constraints of the context (integers, non-negative values): these usually earn marks of their own.

Common mistakes

Summary

Linear equation in three unknowns: $a_{1} x + a_{2} y + a_{3} z = b$ . Its solutions are triples $(x, y, z)$ and there are infinitely many; they are described with two parameters $t$ and $s$ .
$3 \times 3$ system: three linear equations with coefficients $a_{ij}$ and constant terms $b_{i}$ . A solution satisfies all three at once.
Homogeneous system ( $b_{1} = b_{2} = b_{3} = 0$ ): always consistent, with the trivial solution $(0, 0, 0)$ .
Classification: SCD (one solution), SCI (infinitely many, with parameters) and SI (none). There is never a finite number of solutions greater than one.
An impossible equality such as $0 = 3$ signals an inconsistent system; a redundant equation with free unknowns signals a system with infinitely many solutions.
Geometrically, each equation is a plane, and the classification describes how the planes meet (developed in topic 8).
With the GDC (Paper 2) the solution is only obtained when it is unique; in the other cases you must reason by hand, as in Paper 1.

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