Linear systems: the Year 2 test

Linear systems are the first serious test of Year 2: they almost always fall between the end of September and mid-October, right after the review of inequalities. It is a test on which you can get an excellent grade, because the calculation is almost all mechanical — and for exactly that reason it is also the test where it is easiest to throw away marks through distraction.

The typical problem is not “I can’t solve a system”. It is that the book presents four methods, the teacher wants all four, and nobody ever explains which one is best to use when. Result: you always choose the first, substitution, even when it is the longest, and you reach the end of the hour with two exercises left blank.

Here you will find what solving a system really means, the four methods with the criterion for choosing them, how to recognise the impossible and indeterminate cases, three worked exercises and the mistakes I see repeated in every class.

What you are really looking for

A system of two first-degree equations in two unknowns looks like this:

{2 x + 3 y = 12 x - y = 1

Solving it means finding the pair $(x, y)$ that satisfies both equations together. Not any two numbers: a pair, which must be written as a pair — $(3, 2)$ — because that is what the teacher expects to read at the end of the exercise.

The picture that makes the whole chapter simpler is this: every first-degree equation in two unknowns is a line in the Cartesian plane. Solving the system means finding the point where the two lines meet. From this follow, without having to learn them by heart, the three possible cases:

the lines meet at one point → determinate system, a single solution;
the lines are parallel and distinct → impossible system, no solution;
the lines coincide → indeterminate system, infinitely many solutions.

If your teacher asks for the graphical interpretation (in many classes it is a point in the test), it is all here.

The four methods, and when each is best

Method	Best when	The risk
Substitution	an unknown has coefficient $1$ or $- 1$	fractions everywhere if the coefficient is not 1
Equating	the same unknown is already isolated, or easily isolated, in both	copying the right-hand side wrongly
Elimination	the coefficients of one unknown are equal or opposite (or become so with an easy multiplication)	forgetting to multiply the whole equation
Cramer	“ugly” coefficients, or when you need to discuss a parameter	swapping the columns in the determinants

Substitution. You express one unknown from one equation and substitute it into the other. It always works, but if no coefficient is $1$ you end up working with fractions for the whole exercise.

Equating (the comparison method). You isolate the same unknown in both equations and set the two results equal. It is the quickest method when the system is already written in explicit form, like $y = 2 x - 1$ and $y = - x + 5$ .

Elimination. You add or subtract the two equations side by side to make one unknown disappear, after multiplying them if necessary. It is the quickest of the four in most tests, and the one students use least.

Cramer. Mechanical and panic-proof: you calculate three determinants and divide. It is the only one that requires no choices, and it is the one to use when the numbers are awkward or when there is a parameter to discuss.

Cramer’s formulas, for the system in standard form

{a x + b y = c a^{'} x + b^{'} y = c^{'}

are

D D_{x} D_{y} = a a^{'} b b^{'} = a b^{'} - a^{'} b = c c^{'} b b^{'} = c b^{'} - c^{'} b = a a^{'} c c^{'} = a c^{'} - a^{'} c

and the solution is

x = \frac{D _{x}}{D}, y = \frac{D _{y}}{D}

The way not to get the columns wrong: $D_{x}$ is obtained from $D$ by replacing the column of the $x$ terms with that of the constant terms; $D_{y}$ by replacing the column of the $y$ terms. One second of attention here is worth half an exercise.

Determinate, impossible, indeterminate

Cramer also gives the quickest criterion for classifying a system, and this is almost always a question in the test:

$D \neq = 0$ → determinate, a single solution;
$D = 0$ and at least one of $D_{x}$ and $D_{y}$ is non-zero → impossible;
$D = 0$ and $D_{x} = D_{y} = 0$ → indeterminate.

It is worth seeing it on two almost identical systems. The first:

{2 x - y = 3 4 x - 2 y = 1

Here $D = 2 \cdot (- 2) - 4 \cdot (- 1) = - 4 + 4 = 0$ and $D_{x} = 3 \cdot (- 2) - 1 \cdot (- 1) = - 6 + 1 = - 5 \neq = 0$ : impossible. Looking at the two equations you see why — the second is double the first on the left, but not on the right: two parallel lines that never touch.

The second:

{2 x - y = 3 4 x - 2 y = 6

Now $D = 0$ as before, but $D_{x} = 3 \cdot (- 2) - 6 \cdot (- 1) = - 6 + 6 = 0$ and in the same way $D_{y} = 0$ : indeterminate. The second equation is exactly the first multiplied by 2, that is, the same line written twice. The solutions are infinite: all the pairs with $y = 2 x - 3$ .

Three worked exercises

1. Substitution, when it really is the right method

{2 x + 3 y = 12 x - y = 1

In the second equation $x$ has coefficient 1: this is the case where substitution is the best choice. I express $x$ :

x = 1 + y

and substitute it into the first:

2 (1 + y) + 3 y = 12 \Rightarrow 2 + 2 y + 3 y = 12 \Rightarrow 5 y = 10 \Rightarrow y = 2

I go back to the expression for $x$ : $x = 1 + 2 = 3$ . The solution is the pair $(3, 2)$ .

The check costs ten seconds and is worth every one: $2 \cdot 3 + 3 \cdot 2 = 12$ and $3 - 2 = 1$ . Both work.

2. Elimination, the quickest method when the coefficients are opposite

{3 x + 2 y = 7 5 x - 2 y = 1

The coefficients of $y$ are already opposite, $+ 2$ and $- 2$ : I add the two equations side by side and $y$ disappears.

(3 x + 5 x) + (2 y - 2 y) = 7 + 1 \Rightarrow 8 x = 8 \Rightarrow x = 1

I substitute into either of the two, for example the first:

3 \cdot 1 + 2 y = 7 \Rightarrow 2 y = 4 \Rightarrow y = 2

Solution $(1, 2)$ . With substitution the same exercise would have required fractions and at least three more steps.

3. Cramer, when the numbers are awkward

{4 x - 3 y = 6 2 x + 5 y = 16

No coefficient is 1, no pair is opposite: this is the case for Cramer.

D = 4 \cdot 5 - 2 \cdot (- 3) = 20 + 6 = 26

D_{x} D_{y} = 6 \cdot 5 - 16 \cdot (- 3) = 30 + 48 = 78 = 4 \cdot 16 - 2 \cdot 6 = 64 - 12 = 52

x = \frac{78}{26} = 3, y = \frac{52}{26} = 2

Solution $(3, 2)$ . Check: $4 \cdot 3 - 3 \cdot 2 = 6$ and $2 \cdot 3 + 5 \cdot 2 = 16$ . Both work.

Problems to translate into a system

In the test there is almost always a problem in words to turn into a system, and it is worth more than the other exercises. The method is always the same, in three steps.

State the unknowns in full, with the unit of measurement: “let $x$ be the base in cm, let $y$ be the height in cm”. It seems formal, but it is the step that stops you confusing them halfway through the exercise, and in marking schemes it is worth a mark.
Write one equation for each numerical piece of information in the text. Two pieces of information, two equations.
Check that the solution makes sense in the problem, not just in the system: a negative length or a number of people with a decimal point must be discarded, and that is what the teacher is checking.

A typical example. A rectangle has a perimeter of 34 cm and the base exceeds the height by 5 cm: find the dimensions.

{2 (x + y) = 34 x - y = 5 \Rightarrow {x + y = 17 x - y = 5

Adding side by side: $2 x = 22$ , so $x = 11$ and $y = 17 - 11 = 6$ . Base 11 cm, height 6 cm — both positive, so acceptable. The perimeter works: $2 \cdot 17 = 34$ .

The mistakes that cost the most marks

1. Multiplying only half the equation. In the elimination method, if you multiply by 3 you must multiply all the terms, constant term included. It is mistake number one of the chapter, and it is invisible: the calculation carries on smoothly and the result is wrong.

2. Not reducing to standard form before Cramer. Determinants are calculated on a system written as $a x + b y = c$ , with the unknowns in the same order in both rows and the constant terms on the right. Applying Cramer to a messy system gives meaningless numbers.

3. Answering with a single number. The solution is a pair. Finding $x = 3$ and stopping there is half an exercise.

4. Wrong signs in the determinants. $D_{y} = a c^{'} - a^{'} c$ : when the coefficients are negative, $- 3 \cdot (- 2)$ is $+ 6$ . One second of slowness here is worth more than ten extra exercises.

5. Not checking. Substituting the pair found into the two starting equations costs ten seconds and finds practically every calculation error. In a one-hour test it is the best possible investment.

The two weeks before the test

Do five systems with each method, in separate blocks. Not alternating: five of substitution in a row, then five of elimination, and so on. It serves to build the automatic response.
Then do ten mixed systems and, before solving them, write only which method you would use and why. Without solving them. Five minutes in all, and it is the exercise that gains you the most time in the test.
Add two systems with a parameter and two problems to translate, which are the exercises worth the most in the test.
Always check. Get into the habit now, not on the day of the test.

If revising is not the problem and what freezes is your head in the first minutes of the test, I have written about it in how to manage test anxiety in maths and physics.

How we work on this in lessons

On systems the work that pays off most is not doing twenty more exercises: it is making twenty choices of method without solving anything. I look at the system with the student, ask which method they would use and why, and we move on to the next. In half an hour we cover what solving them all would take three lessons, and in the test the difference shows in the time.

The other point I insist on is translating problems: stating the unknowns aloud, with the unit of measurement, before writing any equation. Lessons are online and one-to-one: you will find up-to-date rates on the pricing page — the 10-hour package is good value if Year 2 has started uphill and it is worth putting last year’s inequalities in order too. If radicals appear in the coefficients, rationalisation is in radicals and rationalisation; and for systems that arise from two lines, with the intersection and the parameter, linear systems and the line in Year 2.

The quickest way to find out where you stand is to try the simulated test on linear systems: fifteen questions in fifty minutes, with a breakdown of weak points by topic at the end. The correct answer to every question is free and with no registration; the step-by-step guided solution is unlocked with your first lesson. If after the test you see that the calculation holds but the classification of systems does not (or vice versa), write to me for a free 30-minute introductory call via the contact page: we look at the syllabus of your test and I tell you honestly how many lessons you need. The full picture of how I set up the course is on the maths tutoring page.

Linear systems in Year 2: four methods and when to use them