Simultaneous Equations (AA SL)
Where a straight line meets a parabola is a question about two equations holding true at once. This page covers the substitution method for solving one linear and one quadratic equation together, with worked examples and the mistakes that trip students up when reporting the final coordinates. It's part of the broader Quadratics & Equations topic.
16 questions on this sub-topic.
The substitution method
Covered under IB syllabus reference SL2.7: solution of quadratic equations, including those arising when a linear and a quadratic relationship are solved together, using factorisation, completing the square, or the quadratic formula.
Substitute to eliminate a variable
Rearrange the linear equation for \(y\) (if it isn't already), then substitute that expression into the quadratic. This leaves a single equation in \(x\) alone.
Quadratic formula
\(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\)
Given in the formula booklet. Use it once substitution leaves you with a quadratic in \(x\) that won't factorise nicely.
Need the full syllabus wording, formula-booklet table, or calculator methods? See Quadratics & Equations (calculator steps under Using your GDC).
Worked examples
Find the coordinates of the points of intersection of \(y=3x+2\) and \(y=x^2+4x-4\).
Worked solution
Equate the two expressions for \(y\): \(x^2+4x-4=3x+2\Rightarrow x^2+x-6=0.\) M1
Factorise and solve: \((x+3)(x-2)=0\Rightarrow x=-3\) or \(x=2.\) A1
Substitute each \(x\)-value back into \(y=3x+2\): \(x=-3\Rightarrow y=-7\); \(x=2\Rightarrow y=8.\) M1
The points are \((-3,-7)\) A1 and \((2,8)\). A1
Find the coordinates of the points of intersection of \(y = x^2\) and \(y = 2x + 3\).
(a)(i) State the coordinates of one point of intersection.
(a)(ii) State the coordinates of the other point of intersection.
Worked solution
Set the expressions equal: \(x^2=2x+3\Rightarrow x^2-2x-3=0.\) M1
Factorise: \((x-3)(x+1)=0\Rightarrow x=3\) or \(x=-1.\) A1 A1
Substitute each \(x\)-value into \(y=2x+3\): \(x=-1\Rightarrow y=1\); \(x=3\Rightarrow y=9.\) M1
(a)(i) \((-1,1)\) A1 (a)(ii) \((3,9)\) A1
Common mistakes
- Stopping at the \(x\)-values. Solving the quadratic gives \(x\), but the question asks for coordinates - always substitute back into the linear equation to find the matching \(y\)-value for each root.
- Reporting only one intersection point. A line meeting a parabola almost always gives two points; if the quadratic has two distinct roots, there should be two coordinate pairs in the final answer, not one.
- Substituting into the wrong equation. Once \(x\) is known, plug it back into the simpler (linear) equation to find \(y\) - substituting into the quadratic instead is more error-prone and wastes time.
Ready to practise properly?
16 simultaneous-equation questions, marked instantly like the real exam.
Quick answers
How do you solve a linear and a quadratic equation simultaneously?
Substitute the linear equation into the quadratic to eliminate one variable, solve the resulting quadratic for \(x\), then substitute each \(x\)-value back into the linear equation to find the matching \(y\)-value.
How many solutions does a line and a parabola have?
Up to two, matching the two roots of the resulting quadratic - a line can cross a parabola at two points, touch it at one (tangent), or miss it completely.