Linear systems and the line in Year 2

End of September 2026: in Year 2 of the language-focused high school (liceo linguistico) the chapter on radicals is closing, and in November the next block starts, linear systems and the line in the Cartesian plane. The test falls six to nine weeks from now, so between early November and early December. With 3 hours of maths a week and a timetable dominated by languages, this is one of the few blocks of the year in which algebra and analytic geometry are studied together, in the same period: it is worth understanding from the start how they hold together, instead of treating them as two separate chapters that happen to sit close together on the calendar.

The idea that holds the two chapters together

A system of two first-degree equations in two unknowns is, in essence, a single question: where do these two lines meet? Every first-degree equation in $x$ and $y$, written in the form $ax+by+c=0$, represents a line in the Cartesian plane. Solving the system made of two of these equations means finding the point (or points, or no point) that belongs to both lines at once — that is, the point where the two graphs cross.

From this idea come, without having to learn them separately by heart, the three possible cases:

• one solution: the two lines are intersecting, they meet at a single point;
• no solution: the two lines are parallel and distinct, they never meet;
• infinitely many solutions: the two lines are really the same line, written twice in different ways (coincident).

The rest of this post follows the same order in which I cover it in lessons: first the three methods for solving a system algebraically, then the line in the Cartesian plane, then the explicit link between the two.

The three methods for solving a system

A linear system in two unknowns looks like this:

There are three classical methods. They should not be learned as equivalent alternatives to try at random: there is a practical criterion for choosing which to use, by looking at the coefficients before you start writing.

Substitution. You isolate an unknown from one of the two equations and substitute it into the other. It is best when an unknown already has coefficient 1 (or $-1$) in one of the two equations: isolating it needs no division, just moving terms.

Elimination. You multiply one or both equations by suitable numbers so that an unknown has equal or opposite coefficients, then you add or subtract the two equations, eliminating that unknown. It is best when the coefficients of an unknown are already equal, opposite, or become so by multiplying by a small number.

Equating (comparison). You isolate the same unknown in both equations, and equate the two expressions obtained. It is useful above all when both equations are already written (or are easily written) in explicit form with respect to the same letter — for example when the system arrives already as $y=\dots$ and $y=\dots$, that is as two lines in explicit form: in that case equating is almost automatic.

The practical criterion, in one line: look at the coefficients before choosing. If one of them is already 1, substitution. If two coefficients of the same unknown are equal or opposite (or become so with a simple multiplication), elimination. If the system arrives with both equations already isolated with respect to the same unknown, equating.

Typical mistake in the three methods

The most common problem is not choosing the wrong method — even the “less convenient” method still gets to the right result, just with more steps — but getting the signs wrong while using the chosen method: in elimination, multiplying an equation by a negative number and forgetting to change the sign of one of the two terms; in substitution, substituting an expression with a minus sign in front and forgetting to distribute it to all the terms inside the bracket. A second, more conceptual mistake is confusing “no solution” with “solution zero”: a system with no solution does not have $x=0,y=0$ as its result, it simply has no pair of numbers that satisfies both equations together.

The line in the Cartesian plane

The explicit form of the line is

$m$ is the slope: it says how much the line rises or falls, and in which direction. If $m>0$ the line rises from left to right, if $m<0$ it falls, if $m=0$ the line is horizontal. $q$ is the $y$-intercept: the value of $y$ at the point where the line meets the vertical axis, that is where $x=0$.

Given two points $A(x_1,y_1)$ and $B(x_2,y_2)$, the slope of the line joining them is calculated like this:

Once you have found $m$, you find $q$ by substituting the coordinates of one of the two points into the equation $y=mx+q$ and solving for $q$.

Parallelism. Two lines are parallel if and only if they have the same slope:

Perpendicularity. Two lines (neither of them horizontal or vertical) are perpendicular if and only if the product of their slopes is $-1$:

In practice, to find the slope of the perpendicular to a line of slope $m$, you calculate $-\frac{1}{m}$: you change the sign and take the reciprocal.

Typical mistake with the line

The most frequent is to get $m$ from the implicit form $ax+by+c=0$ with the wrong sign. Isolating $y$:

so $m=-\frac{a}{b}$, with the minus sign that many forget to write. For a line like $2x-3y+6=0$, the correct slope is $m=-\frac{2}{-3}=\frac{2}{3}$, not $-\frac{2}{3}$, which is what you get by reading the signs off the coefficients by eye without applying the formula.

The link: system as intersection

Putting together the equations of two lines in a system, and solving it, is the same as asking where those two lines meet. If the system has a solution $(x_0,y_0)$, that point lies on both lines: substituting it into each of the two equations, the equality must come out true in both — it is the same check you do to verify that a point belongs to a single line, applied twice. If the system is impossible, the two lines are parallel and distinct: same $m$, different $q$. If the system is indeterminate, the two equations describe the same line, just written with different coefficients (one is a multiple of the other).

First worked exercise: substitution and check with elimination

Solve the system:

With substitution. In the first equation the unknown $x$ already has coefficient 1: I isolate it.

I substitute into the second equation:

Substituting $y=3$ into $x=8-2y$: $x=8-6=2$. Solution: $(2,3)$.

Check with elimination. I restart from the original system and multiply the second equation by 2, so that the coefficient of $y$ becomes opposite to that of the first:

I add side by side: $7x=14$, so $x=2$. Substituting into the first equation: $2+2y=8\Rightarrow y=3$. The same solution, $(2,3)$: the two methods check each other.

Final check in the starting system: $2+2\cdot 3=8$, correct; $3\cdot 2-3=3$, correct.

To practise other systems with these three methods you will find exercises organised by topic in the maths test — linear systems: the correct answer is free, the step-by-step guided solution is unlocked with your first lesson.

Second worked exercise: impossible and indeterminate systems

Impossible case. Solve:

With elimination, I multiply the first equation by 2:

Subtracting side by side I get $0=7$, a false equality: the system is impossible, no pair $(x,y)$ satisfies it. Graphically, the two lines have the same slope (isolating $y$ in both you find $m=-2$ for each) but different $q$: they are parallel and distinct, they never meet.

Indeterminate case. Solve:

The second equation is the first multiplied by 2: dividing the second by 2 you get exactly the first. With elimination, multiplying the first by $-2$ and adding it to the second gives $0=0$, an always-true equality: the system is indeterminate, it has infinitely many solutions, because the two equations describe the same line.

Third worked exercise: line through two points, parallel and perpendicular

Given $A(2,1)$ and $B(6,9)$, find the equation of the line $r$ through $A$ and $B$; then the line $s$ parallel to $r$ through $C(0,4)$; then the line $t$ perpendicular to $r$ through the same point $C$.

Line $r$. Slope:

I use the point $A$ to find $q$: $y-1=2(x-2)\Rightarrow y=2x-3$.

Check with $B$: $2\cdot 6-3=9$, correct.

Line $s$, parallel to $r$ through $C(0,4)$. Same slope, $m_s=2$. Since $C$ has $x=0$, $q$ can be read directly: $q=4$. So $s:y=2x+4$.

Line $t$, perpendicular to $r$ through the same $C(0,4)$. We need $m_t$ such that $2\cdot m_t=-1$, so $m_t=-\frac{1}{2}$. With $q=4$ read in the same way: $t:y=-\frac{1}{2}x+4$.

Fourth worked exercise: a problem translated into a system

In a stationery shop, 3 notebooks and 2 pens cost €12.50 together; 1 notebook and 4 pens cost €11 together. How much does a notebook cost and how much a pen?

I call $x$ the price of a notebook and $y$ the price of a pen:

In the second equation $x$ has coefficient 1: I isolate it with substitution.

I substitute into the first: $3(11-4y)+2y=12.5\Rightarrow 33-12y+2y=12.5\Rightarrow -10y=-20.5\Rightarrow y=2.05$.

Then $x=11-4\cdot 2.05=11-8.2=2.8$. A notebook costs €2.80, a pen €2.05.

Check: $3\cdot 2.8+2\cdot 2.05=8.4+4.1=12.5$, correct; $2.8+4\cdot 2.05=2.8+8.2=11$, correct.

How we work on this block

At the language high school maths has fewer hours than on other tracks, and it is often the subject on which a student bets the least — and that is perfectly legitimate. That does not make the time spent understanding it well worth any less: a block like systems and the line can be closed in a few targeted lessons, with no need to reinvent yourself as “good at maths”. In lessons I always start from a couple of systems to see whether the weak point is the calculation (signs, distribution) or the method (which to choose, and why), and I work only there.

If the previous chapter, radicals, is not yet fully solid, it is best to close it first: the precision required in the calculations is the same. On the parallel block seen from another track, same time of year but a different approach, there is factorising and the line: the first Year 2 test. For those who reach Year 3 at the same language high school, the next chapter on this algebraic framework is told in polynomial division and Ruffini’s rule in Year 3. For the case in which the line depends on a parameter, and the intersection is sought for the whole family of lines, there is pencils of lines: proper and improper.

To practise linear systems with immediate marking, the dedicated test is at /simulazioni/verifiche-liceo/seconda/sistemi-lineari/: the correct answer is free, the guided solution is unlocked with your first lesson.

The full maths programme for the track is on the maths page. If you want a hand on this block before the November test, write to me through the contact page.