Systems & Polynomial Equations (AI SL)
A system of equations is a set of equations sharing the same unknowns, solved together rather than one at a time - think three coffee prices linked by three days of sales totals. A polynomial equation asks for the values of a single unknown that make an expression equal zero. AI SL leans entirely on the GDC for both: you set the equations up from the context, then let the calculator's equation solver do the algebra.
What the syllabus says
This topic maps onto one point in the official IB Applications & Interpretation syllabus.
| Code | Syllabus content |
|---|---|
| SL1.8 | Use technology to solve systems of linear equations in up to 3 variables, and to solve polynomial equations. In examinations no specific method of solution is required, and a system of equations will always have a unique solution. Standard terminology such as "zeros" or "roots" should be understood. |
Links to quadratic models (SL2.5) - the same GDC solver handles a quadratic written as a polynomial equation.
Key terms
Five words worth knowing cold before you touch the formulas below - each with a worked example showing exactly what it means.
What is a system of equations?
A system of equations is a collection of two or more equations that share the same unknowns. Solving the system means finding the values of every unknown that satisfy all the equations at once, not just one of them.
e.g. \(x+y=7\) and \(x-y=1\) together give \(x=4,\ y=3\) - the only pair that fits both.
What is a polynomial equation?
A polynomial equation sets a polynomial expression (a sum of terms in powers of \(x\)) equal to zero. Solving it means finding every value of \(x\) - the roots or zeros - that makes the equation true.
e.g. \(x^2-5x+6=0\) has roots \(x=2\) and \(x=3\), since \((x-2)(x-3)=0\).
What does "unique solution" mean for a system?
A unique solution means exactly one set of values for the unknowns satisfies every equation in the system simultaneously. The IB guarantees any system you're asked to solve fully in an exam has exactly one such solution.
e.g. The system \(2x+y=8,\ x-y=1\) has the single solution \(x=3,\ y=2\) - no other pair works.
What are roots or zeros of an equation?
Roots (or zeros) are the input values that make an equation equal zero. For a polynomial \(f(x)=0\), the roots are exactly the \(x\)-values where the graph of \(y=f(x)\) crosses or touches the \(x\)-axis.
e.g. \(f(x)=x^2-4\) has roots \(x=2\) and \(x=-2\), since \(f(2)=0\) and \(f(-2)=0\).
What is an inconsistent system?
An inconsistent system is one with no solution at all - the equations describe lines or planes that never meet. It usually happens when two conditions contradict each other, such as demanding both a specific total and a specific ratio that can't both hold.
e.g. \(3x+y=7\) and \(6x+2y=10\) are parallel lines (same gradient \(-3\), different intercepts) - no pair \((x,y)\) satisfies both.
Key formulas
This topic isn't formula-driven so much as method-driven - the table below summarises the two GDC-based approaches, and the sections underneath walk through when to use each.
Formula reference
There's no formula booklet entry for "how to solve a system" - the syllabus deliberately leaves the method to technology. The table below shows the general forms you'll be setting up, not formulas to memorise.
| Form | Used for | Booklet? |
|---|---|---|
| \(a_1x+b_1y(+c_1z)=d_1\) etc. | General linear system, 2 or 3 unknowns | Not in booklet - solved by GDC |
| \(f(x)=0\) | General polynomial equation, any degree | Not in booklet - solved by GDC |
| \(y=mx+c\) | Line through two points (prior knowledge) | Not in booklet - prior knowledge |
Linear systems vs polynomial equations
Both are solved with a GDC solver, but they're set up differently - a system needs several equations entered together, a polynomial equation needs just one.
| Feature | System of equations | Polynomial equation |
|---|---|---|
| Unknowns | Multiple (\(x, y\), sometimes \(z\)) | One (\(x\)) |
| Equations needed | Same number as unknowns | One equation, one degree |
| GDC tool | Simultaneous equation solver | Equation/polynomial solver, or graph + zeros |
| What you get back | One value per unknown | Up to (degree) real roots |
| Example | \(2x+3y=12,\ x-y=1\) | \(x^2-x-6=0\) |
Setting up a system from context
The hardest part is usually translating words into equations - the GDC handles everything after that.
Count the unknowns first
Identify exactly how many quantities you don't know - that tells you how many independent equations you need to write down before reaching for the GDC.
One equation per condition
Each piece of given information (a total, a ratio, a comparison) should translate into exactly one equation. Missing a condition leaves you with too few equations to pin down a unique solution.
Enter coefficients carefully
Write every equation in the same order of variables (e.g. always \(x, y, z\)) before typing coefficients into the GDC's matrix or equation-solver screen - a swapped order gives a wrong but plausible-looking answer.
Reading the number of solutions
For a 2-variable system, the number of solutions depends on how the two lines relate - this matters even though full 3-variable exam systems are always guaranteed a unique solution.
One solution
The lines have different gradients, so they cross at exactly one point. This is the case the GDC solver is built to return directly.
No solution
The lines are parallel (same gradient, different intercept) - they never meet, so there's no pair of values satisfying both equations.
Infinitely many solutions
The two equations describe the same line (one is a multiple of the other) - every point on that line satisfies both, so there are infinitely many solutions.
Worked examples
Two full exam-style questions, marked exactly like the real thing. Try each one yourself before checking the worked solution.
\(4\) coffees and \(3\) teas cost \(\$23\); \(2\) coffees and \(5\) teas cost \(\$22.\)
(a)(i) Find the price of a coffee.
(a)(ii) Find the price of a tea.
(b) Find the cost of \(6\) coffees and \(4\) teas.
(c) If the tea price increased by \(\$0.50,\) find the new cost of \(2\) coffees and \(5\) teas.
Worked solution
(a)(i) \(4c+3t=23,\ 2c+5t=22\Rightarrow c=3.5,\) M1
(a)(ii) \(t=3.\) A1
(b) \(6(3.5)+4(3)=21+12=$33.\) M1
\($33.\) A1
(c) New tea price \(=$3.50;\ 2(3.5)+5(3.5)=7+17.5=$24.50.\) M1
\($24.50.\) A1
A café sells small, medium and large coffees at prices \(\$s\), \(\$m\) and \(\$l.\) Sales over three days give:
| Day | Small | Medium | Large | Total \(\$\) |
|---|---|---|---|---|
| Mon | 10 | 8 | 5 | 64 |
| Tue | 6 | 10 | 8 | 74 |
| Wed | 12 | 6 | 4 | 58 |
(a) Write down three equations in \(s\), \(m\) and \(l.\)
(b)(i) Find the price of a small coffee.
(b)(ii) Find the price of a medium coffee.
(b)(iii) Find the price of a large coffee.
(c) Find the cost of 3 small, 2 medium and 1 large coffee.
Worked solution
(a) \(10s+8m+5l=64,\ 6s+10m+8l=74,\ 12s+6m+4l=58.\) M1
The three equations are \(10s+8m+5l=64,\ 6s+10m+8l=74,\ 12s+6m+4l=58.\) A1
(b)(i) Solving the system with technology: M1
\(s=$2.\) A1
(b)(ii) \(m=$3.\) A1
(b)(iii) \(l=$4.\) A1
(c) \(3(2)+2(3)+1(4)\) M1
\(=$16.\) A1
Common mistakes
The four slip-ups that account for most of the marks lost on this topic - worth reading before you start practising, not just after you get one wrong.
- Missing an equation. Three unknowns need three independent equations. Forgetting to translate one of the given conditions (a ratio, a comparison) into an equation leaves the system unsolvable or gives an infinite family of answers.
- Inconsistent variable order. Entering coefficients for \(x, y, z\) in one equation and \(y, x, z\) in the next produces a solvable-looking system with the wrong answer - always keep the same order across every equation.
- Treating "no solution" as an error. If the GDC reports no solution or infinitely many, that's often the correct mathematical answer (parallel or identical lines), not a sign you mistyped something - check the equations before assuming a mistake.
- Rounding mid-calculation. Rounding an intermediate coffee price or root before using it in the next part compounds errors. Keep full GDC precision until the final answer, then round.
Using your GDC
Every step below is a real button sequence, not a vague "use your calculator" hint - covering the TI-84 Plus, TI-Nspire, and Casio fx-9860/fx-CG50. Pick your model to filter down to just the steps that apply to you.
Solve two or three simultaneous equations (linear systems) without elimination by hand.
- Write each equation in the form \(ax+by(+cz)=d.\)
- APPS → PlySmlt2 → Simultaneous Eqn Solver; set the number of equations/unknowns and enter the coefficients.TI-84
- menu → Algebra → Solve System of Equations, or use linSolve.Nspire
- Main menu → Equation → Simultaneous, set the number of unknowns, enter the coefficients, SOLVE.Casio
Tip: No solution or infinitely many? The calculator will flag it - that means the lines/planes are parallel or coincident.
Faster and safer than algebra for messy equations - and essential in AI, where many equations can't be solved by hand. The trick is getting all solutions, not just one.
- Graph \(f(x)\) first so you can see how many solutions exist and roughly where they are.
- Rearrange so everything is on one side: \(f(x)=0\) - or graph both sides as separate functions and find intersections.
- MATH → Solver: enter the expression, type a starting guess close to one root, press ALPHA + ENTER. Move the guess to near a different root and repeat for each solution.TI-84
- Type nSolve(f(x)=0, x, guess) - include a guess or interval e.g. nSolve(f(x)=0, x, 2) or nSolve(f(x)=0, x, {1,5}) to target a specific root.Nspire
- Run-Matrix → SolveN(f(x), x) returns all real roots at once; or use the Equation app for a visual approach.Casio
- Always verify each solution by substituting back into the original equation.
Tip: The solver finds ONE root near your starting guess - change the guess to find others. The graph shows you how many to expect.
See the full GDC guide for more calculator models and topics.
Ready to practise properly?
Systems & polynomial equations questions, marked instantly like the real exam.
Quick answers
The questions students on this topic ask most often.
Do I have to solve systems of equations by hand?
No. The AI SL syllabus explicitly says technology should be used to solve systems of linear equations and polynomial equations - no specific by-hand method is required in examinations, and every exam question in this topic assumes a GDC.
How many variables can a system have in the exam?
Up to 3 variables (a 3x3 system). The syllabus guarantees a unique solution in examinations, so you don't need to worry about handling infinite or no-solution cases for a full system - though you may be asked to recognise those cases for 2-variable systems.
What does it mean if my GDC says a system has no solution?
It means the equations represent parallel, non-intersecting lines (or planes) - there's no point that satisfies all of them at once. This can happen for 2-variable systems you're asked to analyse, even though full 3-variable systems in exams always have a unique solution.
What's the difference between a system of equations and a polynomial equation here?
A system of equations has several unknowns linked by several equations, solved simultaneously. A polynomial equation has one unknown raised to various powers, solved by finding its roots. Both are solved using your GDC's equation-solving tools rather than by hand.
Sub-topics
Systems & Polynomial Equations broken down into its individual skills, each with its own focused page.
Related topics
More Number & Algebra topics from the same AI SL syllabus unit, in case you want to keep going.