2×2 Systems of Equations (AA HL)
Most 2×2 systems have exactly one pair \((x,y)\) that satisfies both equations, but the IB likes to test the two exceptions too: lines that never meet, and lines that are secretly the same line. This page is about spotting which of the three cases you're in and handling each correctly, alongside a standard elimination worked example. It's part of the broader Systems of Equations topic.
19 questions on this sub-topic.
Reading the three cases
Covered under IB syllabus reference AHL1.16 - solutions of systems of linear equations, including cases with a unique solution, an infinite number of solutions, or no solution, and finding a general solution when there are infinitely many.
Elimination / substitution
Scale one equation so a variable's coefficient matches the other equation's, then add or subtract to eliminate it and solve for what remains.
Not a formula-booklet result - it's a method, usable by hand or with your GDC's simultaneous-equation solver.
No solution vs. infinitely many
If scaling makes the \(x\)- and \(y\)-coefficients match but the constants don't, you get a contradiction: no solution (parallel lines). If the entire scaled equation matches, every point on the line works: infinitely many solutions, written parametrically as \((x,y)=(f(t),t)\).
Need the full syllabus wording and formula-booklet reference table? See Systems of Equations.
Worked examples
Show that the system has infinitely many solutions and give them in parametric form:
\[x - 3y = 2, \qquad 2x - 6y = 4.\]
Worked solution
the second equation is exactly twice the first, so they represent the same line. M1 A1 Hence infinitely many solutions. A1
let \(y = t\); then \(x = 2 + 3t.\) M1 Solution set: \((x, y) = (2 + 3t,\ t),\ t \in \mathbb{R}.\) A1
For the system
\[x + 2y = 3, \qquad 2x + ky = 6,\]
find the value of \(k\) for which there are infinitely many solutions, and state what happens for other values of \(k\).
Worked solution
if \(k = 4\), the second equation \(2x + 4y\) M1 \(= 6\) is exactly twice the first, giving the same line. A1 So infinitely many solutions. A1
for \(k \neq 4\), the lines have different gradients M1 and intersect once A1 - a unique solution. R1
Common mistakes
- Assuming every system has a unique solution. Before diving into elimination, glance at the coefficients - if one equation looks like a multiple of the other, you're likely dealing with a no-solution or infinite-solutions case, not a single pair \((x,y)\).
- Writing "no solution" without justifying it. You need to show the contradiction explicitly, e.g. that the scaled equations give \(3 \ne 5\), not just assert that the lines don't meet.
- Forgetting the parameter constraint \(t \in \mathbb{R}\) when giving an infinite solution set. The parametric answer \((x,y)=(2+3t,\ t)\) is incomplete without stating that \(t\) ranges over all real numbers.
Ready to practise properly?
18 two-variable-system questions, marked instantly like the real exam.
Quick answers
How do you know if a 2x2 system has no solution?
If, after scaling to match one variable's coefficients, the other coefficients also match but the constants differ, you get a contradiction - the lines are parallel and distinct, so there is no solution.
How do you know if a 2x2 system has infinitely many solutions?
If one equation is exactly a scalar multiple of the other, they describe the same line, so every point on it is a solution - write the solution set in parametric form using a free parameter \(t\).