Quadratic systems: methods and worked exercises

It is early October 2026: in Year 3 at the Italian vocational schools (istituti professionali), quadratic equations have just been closed or are almost closed, and the next step is quadratic systems and the meeting of a line and a parabola. The test on this block usually falls five to eight weeks from now, that is in early December, when the first term closes and the school report is prepared.

This post is for those who have understood quadratic equations but get stuck when the unknowns become two. It is a jump that many experience as abrupt: a single equation is solved with a procedure you learn by heart; with a system you have to choose the order of the moves. In class the approach at vocational schools is applied, so here you will find few concepts, numbered steps and complete calculations. If the quadratic formula with the discriminant is not yet clear to you, start from quadratic equations in Year 3, vocational school and come back here afterwards.

What a quadratic system is

A quadratic system is a set of two equations in the unknowns $x$ and $y$, of which one is quadratic and the other is first-degree. The solution is not a number but a pair $(x,y)$ that makes both equations true together.

Example of a system:

The first equation describes a parabola, the second a line. Looking for the solutions means looking for the points where the line meets the parabola. This picture helps a lot, and we use it shortly to predict how many solutions to expect.

The substitution method, step by step

The method is always the same. Memorise it as a ladder, you need it identical in every exercise.

1. Isolate a variable in the first-degree equation (usually the most convenient one: $y$, or $x$ if it has coefficient 1).
2. Substitute that expression into the quadratic equation. Use brackets, always.
3. Expand and bring it to standard form $a{x}^2+bx+c=0$.
4. Calculate the discriminant $\Delta =b^2-4ac$ and decide how many solutions there are.
5. Solve the equation with the formula $x=\frac{-b\pm \sqrt{\Delta }}{2a}$.
6. Go back: for each value of $x$ found, calculate $y$ in the first-degree equation.
7. Check by substituting into the starting system, for at least one pair.

Step 6 is the one most often forgotten, and I talk about it in the typical mistakes later on.

The discriminant decides how many solutions there are

After the substitution you get a quadratic equation in $x$ alone. Its discriminant tells you straight away what to expect:

You don’t need to draw the graph to solve the exercise: the discriminant is enough. But knowing there is a drawing behind it lets you check the result by common sense. If a line passes above a parabola that lies entirely below, “two solutions” sounds suspicious.

Exercise 1: the line cuts the parabola

Here $y$ is already isolated in both equations, so you can equate them directly (it is a substitution, just faster):

I bring everything to the left, changing the sign of the terms that cross the equals sign:

I calculate the discriminant, with $a=1$, $b=-5$, $c=4$:

The discriminant is positive: let’s expect two pairs. I apply the formula:

Now the step that must not be skipped: I calculate $y$ for each $x$ using the line $y=x-1$.

• If $x=4$, then $y=4-1=3$.
• If $x=1$, then $y=1-1=0$.

Check on the parabola: for $x=4$ it gives $16-16+3=3$ ✓; for $x=1$ it gives $1-4+3=0$ ✓.

Solutions: $(4,3)$ and $(1,0)$.

After a first exercise like this, the useful thing is to do other similar systems straight away, against the clock, until the ladder becomes a reflex. On the site there is the category of tests for the liceo: there is no dedicated simulator for quadratic systems, but the maths tests train you on equations and calculation with the discriminant. The correct answer is free, the step-by-step guided solution is unlocked with your first lesson.

Exercise 2: the line is tangent

I equate the two expressions of $y$:

I bring everything to the left:

The discriminant, with $a=1$, $b=-2$, $c=1$:

Zero discriminant: a single solution. The equation is also the square of a binomial, $(x-1{)}^2=0$, and this confirms the result:

I calculate $y$ with the line: $y=4\cdot 1+2=6$. Check on the parabola: $1+2+3=6$ ✓.

Solution: the single pair $(1,6)$. It is not a mistake that only one comes out: it means the line touches the parabola at a single point, and is tangent. Many students see a single solution and think they have done something wrong. In fact it is the most “orderly” of the three cases.

Exercise 3: no real solution

I equate:

In standard form:

Discriminant, with $a=1$, $b=-2$, $c=4$:

The discriminant is negative, so the equation has no real solutions and the system is impossible. You write it like this, in full, in the answer: “the system has no real solutions, since $\Delta <0$”. You don’t leave the exercise without a conclusion.

By common sense: the parabola $y=x^2+1$ has its vertex at $(0,1)$ and always lies above height 1, while the line $y=2x-3$ near the origin lies lower. They never meet, consistent with the discriminant.

Exercise 4: two quadratic equations in a different form

Not all systems arrive with $y$ already isolated. This is the kind of system you often find in tests:

The first equation describes a circle, but to solve it you don’t need to know that: the method is enough. I isolate $y$ in the second equation, which is the first-degree one:

I substitute into the first, with brackets:

Here is the point where most mistakes happen: expanding the square of the binomial. The rule is $(a-b{)}^2=a^2-2ab+b^2$, so:

I put everything back into the equation:

I divide everything by 2 to simplify:

Discriminant, with $a=1$, $b=-7$, $c=12$:

I go back to $y=7-x$:

• If $x=4$, then $y=3$.
• If $x=3$, then $y=4$.

Check on the first equation: $16+9=25$ ✓ for $(4,3)$, and $9+16=25$ ✓ for $(3,4)$.

Solutions: $(4,3)$ and $(3,4)$.

Symmetric systems: sum and product

There is a shortcut for systems in which you know the sum and the product of two numbers:

In this case $x$ and $y$ are the two solutions of one and the same quadratic equation:

In words: ”$t$ squared, minus the sum times $t$, plus the product”. Solve that equation and you have the two numbers. Since the system does not distinguish $x$ from $y$, the solutions are written in both orders.

Example, with sum 10 and product 21:

Solutions: $(7,3)$ and $(3,7)$. Check: $7+3=10$ ✓ and $7\cdot 3=21$ ✓.

The same route also works with the substitution method, but the shortcut avoids expanding a square and therefore most sign errors.

The word problem: the rectangle

Word problems are the part where at vocational school you win back marks most easily, because the reasoning is concrete. The method is to translate sentences into equations.

A rectangle has a perimeter of 34 m and an area of 60 m². Find its dimensions.

I call $x$ and $y$ the two dimensions, base and height.

1. The perimeter is $2(x+y)=34$, so $x+y=17$ (semi-perimeter).
2. The area is $x\cdot y=60$.

The system is

The dimensions are 12 m and 5 m. Check against the text: perimeter $2(12+5)=34$ m ✓; area $12\cdot 5=60$ m² ✓. Here the two pairs $(12,5)$ and $(5,12)$ describe the same rectangle (base and height swapped), so in the answer it is enough to say that the sides measure 12 m and 5 m.

One more check on word problems: the solutions must make physical sense. A negative length must be discarded. In this case both are positive and are fine.

The mistakes I see most often

1. Stopping at the $x$ values. You have solved the quadratic equation, you have two values and feel you have finished. But the solution of a system is a pair. For each $x$ you must calculate $y$. If in a test you write only $x_1$ and $x_2$, you lose a substantial part of the marks.

2. Substituting badly, or only halfway. When you substitute $y=7-x$ into $x^2+y^2=25$ you must put the bracket: $(7-x{)}^2$, not $7-x^2$. Without brackets the square acts only on $x$ and everything else goes to pieces.

3. The square of a binomial without the double product. $(7-x{)}^2$ is not $49+x^2$. It is $49-14x+x^2$. The middle term, the double product, is forgotten in half the tests I mark. A good check: if the square of a binomial comes out with only two terms, there is a mistake.

4. Losing solutions. If you divide both sides by an expression containing $x$ you risk losing a result. In a system this happens when you cancel a common factor instead of taking it out. With numerical coefficients (like the 2 in exercise 4) the division is safe instead.

5. Getting the sign wrong when bringing everything to the left. Going from $x^2+2x+3=4x+2$ to $x^2-2x+1=0$, each term that changes side changes sign. Write the intermediate step, don’t do it in your head: $x^2+2x-4x+3-2=0$.

6. Not commenting on a negative discriminant. If $\Delta <0$ you don’t write “impossible” at random: you say that the system has no real solutions, and you cite the discriminant.

How to prepare over the next few weeks

With five to eight weeks ahead, the plan that works is simple:

• First week: redo quadratic equations until you solve them without looking at the formula. The discriminant should come out in a few seconds.
• Second and third week: line–parabola systems, three a day, always writing all the steps of the ladder. Then check with the test.
• From the fourth: add symmetric systems and one word problem every two days.
• Last week before the test: a complete timed test, as in class, and then correct the typical mistakes from the list above.

To understand better the link between the line and the parabola seen as a curve, read the parabola in Year 3: tangents and pencils: the tangent case of exercise 2 has its full space there. If instead you feel unsure about the substitution method itself, the post on linear systems reviews substitution, equating and elimination with the same logic, and for the first-degree systems and inequalities of a year earlier there is equations, inequalities and systems in Year 2, vocational school. The rest of the maths of the first two and last three years is collected in maths.

For the December test, the liceo tests category offers timed maths exercises on which to train equations and algebraic calculation: remember that the correct answer is free, while the step-by-step guided solution is unlocked with your first lesson.

If you want to understand where you lose marks

In lessons I always start from a real test of the student’s: I look at where it goes wrong, whether in expanding squares, in the signs or in the step from $x$ to $y$, and I build the exercises on that mistake. If you have a test on quadratic systems in the next few weeks and want to understand what to work on, you can ask for a free 30-minute introductory call via the contact page: we talk about it, I look at one of your tests and tell you what is best to do first.