Simultaneous Equations
Three methods — Substitution · Elimination · Graphical · Cambridge 0606 Ch.6
Which method do I use?
METHOD 1
Substitution
Use when one equation is linear and one is non-linear (quadratic, circle, etc.)
Most common in 0606 exams
Most common in 0606 exams
METHOD 2
Elimination
Use when both equations are linear. Add or subtract to remove one variable.
Quick for linear-linear systems
Quick for linear-linear systems
METHOD 3
Graphical
Plot both curves/lines. Solutions = intersection points.
Good for checking; required for some questions
Good for checking; required for some questions
Method 1 — Substitution (Linear & Non-Linear)
1
Rearrange the linear equation: make \(y\) (or \(x\)) the subject.
2
Substitute that expression into the non-linear equation.
3
Expand, simplify → form a quadratic \(ax^2+bx+c=0\). Solve by factorising or formula.
4
Substitute each \(x\) value back into the linear equation to find \(y\).
5
Write solutions as pairs: \((x_1, y_1)\) and \((x_2, y_2)\).
Key rule: Always sub back into the linear equation — never the quadratic (avoids sign errors).
Worked Example
Solve: \(y = x + 1\) and \(y = x^2 - 3x + 3\)
Sub
\(x + 1 = x^2 - 3x + 3\)
Rearr.
\(0 = x^2 - 4x + 2\)
Formula
\(x = \dfrac{4 \pm \sqrt{16-8}}{2} = 2 \pm \sqrt{2}\)
Find y
\(y = x+1\): \(y = 3\pm\sqrt{2}\)
Answers: \((2+\sqrt{2},\; 3+\sqrt{2})\) and \((2-\sqrt{2},\; 3-\sqrt{2})\)
Method 2 — Elimination (Linear & Linear)
1
Write both equations in the same form (line them up clearly).
2
Multiply one or both equations so a variable has the same coefficient.
3
Add or subtract the equations to eliminate that variable.
4
Solve for the remaining variable.
5
Substitute back into either original equation to find the other variable.
Add when signs are opposite — Subtract when signs are the same.
Worked Example
Solve: \(2x + 3y = 12\) and \(x - y = 1\)
Multiply
Eq2 ×3: \(3x - 3y = 3\)
Add
\((2x+3y)+(3x-3y) = 12+3\)
\(5x = 15 \Rightarrow x = 3\)
\(5x = 15 \Rightarrow x = 3\)
Find y
Sub into Eq2: \(3 - y = 1 \Rightarrow y = 2\)
Answer: \(x = 3,\; y = 2\)
Method 3 — Graphical Method
1
Draw both equations on the same axes (use a table of values).
2
Find where the curves/lines intersect — these are the solutions.
3
Read off the \((x, y)\) coordinates of each intersection point.
Graphical limit: You can only estimate solutions from a graph. For exact values, always use substitution.
Geometric interpretation of the discriminant:
• Line cuts curve at 2 points → \(\Delta > 0\)
• Line is tangent (1 point) → \(\Delta = 0\)
• Line misses curve → \(\Delta < 0\)
• Line cuts curve at 2 points → \(\Delta > 0\)
• Line is tangent (1 point) → \(\Delta = 0\)
• Line misses curve → \(\Delta < 0\)
Graph: \(y = x^2 - 2x - 4\) meets \(y = 2x + 1\)
— Curve \(y=x^2-2x-4\)
— Line \(y=2x+1\)
Intersection points: (−1, −1) and (5, 11)
Quadratic Functions & Curves
Forms · Completing the square · Max/min · Modulus Graphs \(y = |f(x)|\)
Forms of a quadratic function
A quadratic function represents a parabola (U or inverted U shape). It can be written in two main forms:
GENERAL FORM
\[f(x) = ax^2 + bx + c\]
Good for finding the y-intercept (\(c\)) and roots/solutions via quadratic formula.
VERTEX OR COMPLETED SQUARE FORM
\[f(x) = a(x-h)^2 + k\]
Good for finding the vertex (maximum/minimum point) at \((h, k)\) and sketching the curve.
Completing the Square — The Method
1
Factorise \(a\) out of the \(x^2\) and \(x\) terms ONLY. Ignore the constant \(c\).
2
Inside the bracket, take half of the new \(x\) coefficient. Square it. Add it and subtract it inside the bracket!
3
Factorise the perfect square part.
4
Expand the outer bracket and combine the constants.
COMMON EXAM MISTAKE ⚠
Students often forget to multiply the subtracted term by the \(a\) that was taken out in Step 1. Always expand correctly in the final step!
Students often forget to multiply the subtracted term by the \(a\) that was taken out in Step 1. Always expand correctly in the final step!
Worked Example: Complete the square for \(f(x) = 2x^2 - 12x + 7\)
1
Factorise 2 out: \(2[x^2 - 6x] + 7\)
2
Half of -6 is -3. Add and subtract \((-3)^2\):
\(2[x^2 - 6x + (-3)^2 - (-3)^2] + 7\)
\(2[x^2 - 6x + (-3)^2 - (-3)^2] + 7\)
3
Form the square:
\(2[(x-3)^2 - 9] + 7\)
\(2[(x-3)^2 - 9] + 7\)
4
Expand the 2 and simplify:
\(2(x-3)^2 - 18 + 7\)
Answer: \(2(x-3)^2 - 11\)
\(2(x-3)^2 - 18 + 7\)
Answer: \(2(x-3)^2 - 11\)
PRO TIP FOR VERTEX 🛠️’¡
From \(2(x-3)^2 - 11\):
The minimum value of the function is −11 (since \((x-3)^2 \geq 0\)).
This minimum occurs when the bracket is 0, so \(x-3=0 \Rightarrow x=3\).
The vertex coordinates are (3, −11).
From \(2(x-3)^2 - 11\):
The minimum value of the function is −11 (since \((x-3)^2 \geq 0\)).
This minimum occurs when the bracket is 0, so \(x-3=0 \Rightarrow x=3\).
The vertex coordinates are (3, −11).
WARNING: MAXIMUM OR MINIMUM?
If \(a > 0\) (positive), you have a U-shape → Minimum.
If \(a < 0\) (negative), you have an inverted U → Maximum.
If \(a > 0\) (positive), you have a U-shape → Minimum.
If \(a < 0\) (negative), you have an inverted U → Maximum.
The Discriminant & Intersections
Roots of a quadratic · Nature of roots · Line satisfying a curve
What is the discriminant?
The discriminant, denoted by \(\Delta\) (delta), tells you how many times a quadratic curve crosses the x-axis, or whether a quadratic equation has any real solutions.
\(\Delta = b^2 - 4ac\)
"REAL ROOTS" vs "DISTINCT ROOTS"
If a question says "has real roots", it could be distinct OR repeated. Therefore, you must use \(b^2 - 4ac \geq 0\). Forgetting the "\(\geq\)" is the #1 mistake.
If a question says "has real roots", it could be distinct OR repeated. Therefore, you must use \(b^2 - 4ac \geq 0\). Forgetting the "\(\geq\)" is the #1 mistake.
Nature of Roots
>
Two real and distinct roots
\(b^2 - 4ac > 0\) (crosses x-axis twice)
\(b^2 - 4ac > 0\) (crosses x-axis twice)
=
Two real and equal roots (repeated)
\(b^2 - 4ac = 0\) (touches the x-axis once at a tangent)
\(b^2 - 4ac = 0\) (touches the x-axis once at a tangent)
<
No real roots
\(b^2 - 4ac < 0\) (never crosses or touches the x-axis)
\(b^2 - 4ac < 0\) (never crosses or touches the x-axis)
Interactive Explorer — Watch the roots change
Move the graph up and down (adjust the \(c\) value) and observe the discriminant.
y = x² + 2x + c
c = -3
Discriminant (\(2^2 - 4(1)(c)\))
16
Two real and distinct roots
Intersection of a Line and a Curve
The exact same rules apply when a straight line meets a curve. Instead of solving \(y = 0\), you equate the line and the curve: \(L(x) = C(x)\). Then rearrange everything to one side to form a single quadratic equation \(ax^2 + bx + c = 0\), and apply the discriminant to IT.
- Line cuts curve twice: \(b^2 - 4ac > 0\)
- Line is a TANGENT to the curve: \(b^2 - 4ac = 0\)
- Line does not intersect curve: \(b^2 - 4ac < 0\)
- Line is a TANGENT to the curve: \(b^2 - 4ac = 0\)
- Line does not intersect curve: \(b^2 - 4ac < 0\)
TANGENT TRICK 🛠️’¡
In exam questions, if they ask you to show a line is a tangent to a curve, set them equal, bring everything to one side, and prove that \(b^2 - 4ac = 0\). This proves there is exactly 1 intersection point (repeated root), meaning the line just kisses the curve!
In exam questions, if they ask you to show a line is a tangent to a curve, set them equal, bring everything to one side, and prove that \(b^2 - 4ac = 0\). This proves there is exactly 1 intersection point (repeated root), meaning the line just kisses the curve!
Quadratic Inequalities
Solving inequalities by sketching
The Golden Rule
NEVER cross-multiply an x across an inequality sign without knowing its sign!
ALWAYS solve quadratic inequalities by SKETCHING THE CURVE!
ALWAYS solve quadratic inequalities by SKETCHING THE CURVE!
Method: Solve \(x^2 - x - 6 > 0\)
1
Factorise: \((x - 3)(x + 2) > 0\)
2
Find critical values: \(x=3\) and \(x=-2\). These are where it crosses the x-axis.
3
Sketch a quick U-shape cutting at -2 and 3.
4
We want \(> 0\) (above the x-axis). Shade the regions above the x-axis.
5
Read off: \(x < -2\) OR \(x > 3\).
Writing the final answer
Between the roots (< 0)
If the question asks for the bottom part of a U-shape (e.g. \(f(x) < 0\)), the solution is a single continuous region.
Write as: \(-2 < x < 3\)
(Do NOT write \(x > -2, x < 3\) disjointedly)
Outside the roots (> 0)
If the question asks for the top branches of a U-shape (e.g. \(f(x) > 0\)), the solution splits into two divergent directions.
Write as: \(x < -2 \text{ or } x > 3\)
(Do NOT write \(-2 > x > 3\) — that mathematically makes no sense!)
Practice Questions
3 Easy · 3 Medium · 3 Hard — attempt first, then reveal the full worked solution
Easy 1
[3]
Express \(3x^2 - 12x + 7\) in the form \(a(x-h)^2 + k\). State the minimum point.
STEP-BY-STEP SOLUTION
Step 1
Take out \(a=3\) from first two terms only:
\(3[x^2 - 4x] + 7\)
\(3[x^2 - 4x] + 7\)
Step 2
Half of \(-4\) is \(-2\). \((-2)^2=4\). Add & subtract inside bracket:
\(3[(x-2)^2 - 4] + 7\)
\(3[(x-2)^2 - 4] + 7\)
Step 3
Expand the 3: \(\;3(x-2)^2 - 12 + 7\)
Answer
\(\boldsymbol{3(x-2)^2 - 5}\)
Minimum point: \(\mathbf{(2,\,-5)}\) since \(a = 3 > 0\) (U-shape).
Minimum point: \(\mathbf{(2,\,-5)}\) since \(a = 3 > 0\) (U-shape).
⚠ Common mistake: forgetting to multiply \(-4\) by the outer 3. Always expand \(3[\ldots - 4]\) carefully.
Easy 2
[3]
Solve by elimination: \(2x + 3y = 12\) and \(x - y = 1\).
STEP-BY-STEP — ELIMINATION
Step 1
Multiply Eq2 by 3: \(\quad 3x - 3y = 3\)
Step 2
Add to Eq1: \((2x+3y)+(3x-3y) = 12+3\)
\(5x = 15 \Rightarrow x = 3\)
\(5x = 15 \Rightarrow x = 3\)
Step 3
Sub \(x=3\) into \(x - y = 1\): \(\quad y = 2\)
Answer
\(\boldsymbol{x=3,\; y=2}\)
Check: \(2(3)+3(2)=12\) ✓ and \(3-2=1\) ✓
Easy 3
[2]
State the nature of roots of \(x^2 - 6x + 9 = 0\) with justification.
STEP-BY-STEP SOLUTION
Identify
\(a=1,\; b=-6,\; c=9\)
Discriminant
\(\Delta = (-6)^2 - 4(1)(9) = 36 - 36 = 0\)
Answer
\(\Delta = 0 \Rightarrow\) Two real, equal (repeated) roots.
The curve touches the x-axis once at \(x = 3\).
The curve touches the x-axis once at \(x = 3\).
Medium 1
[5]
Solve the simultaneous equations \(x + 2y = 7\) and \(x^2 - 4y^2 = 21\).
STEP-BY-STEP — SUBSTITUTION (spot the difference of squares)
Spot it
\(x^2-4y^2 = (x+2y)(x-2y)\). We know \(x+2y=7\), so:
\(7(x-2y)=21 \Rightarrow x-2y=3\)
\(7(x-2y)=21 \Rightarrow x-2y=3\)
Solve
Add \((x+2y=7)\) and \((x-2y=3)\):
\(2x=10 \Rightarrow x=5,\quad y=1\)
\(2x=10 \Rightarrow x=5,\quad y=1\)
Answer
\(\boldsymbol{x=5,\; y=1}\) — only one solution (line is tangent to curve).
Medium 2
[4]
Find the set of values of \(k\) for which \(x^2 + kx + 9 = 0\) has no real roots.
STEP-BY-STEP SOLUTION
Condition
No real roots \(\Rightarrow \Delta < 0\)
Set up
\(k^2 - 4(1)(9) < 0 \Rightarrow k^2 - 36 < 0\)
Critical values: \(k = \pm 6\)
Critical values: \(k = \pm 6\)
Sketch
Draw \(y=k^2-36\) (U-shape cutting at \(\pm6\)). We want BELOW the k-axis: between the roots.
Answer
\(\boldsymbol{-6 < k < 6}\)
⚠ If question asked for "real roots" (not "no real"): use \(\Delta \geq 0\) giving \(k\leq -6\) or \(k\geq 6\).
Medium 3
[5]
Solve simultaneously: \(2x - y = 3\) and \(x^2 + xy = 10\).
STEP-BY-STEP — SUBSTITUTION
Step 1
From linear: \(y = 2x - 3\)
Step 2
Substitute: \(x^2 + x(2x-3)=10 \Rightarrow 3x^2-3x-10=0\)
Step 3
\(x = \dfrac{3\pm\sqrt{9+120}}{6} = \dfrac{3\pm\sqrt{129}}{6}\)
Answer
Two solution pairs with \(y=2x-3\) substituted back.
\(\boldsymbol{x = \frac{3\pm\sqrt{129}}{6}}\)
\(\boldsymbol{x = \frac{3\pm\sqrt{129}}{6}}\)
Hard 1
[6]
The line \(y = mx + 2\) is a tangent to the curve \(y = x^2 - 4x + 6\). Find the exact values of \(m\).
STEP-BY-STEP — TANGENT CONDITION \(\Delta = 0\)
Key idea
Tangent touches curve at exactly 1 point ⇒ repeated root ⇒ \(\Delta = 0\)
Equate
\(mx+2=x^2-4x+6 \Rightarrow x^2-(4+m)x+4=0\)
Apply \(\Delta=0\)
\((4+m)^2-16=0 \Rightarrow (4+m)=\pm4\)
Answer
\(\boldsymbol{m=0}\) or \(\boldsymbol{m=-8}\)
Verify \(m=0\): line \(y=2\) touches minimum of \((x-2)^2+2\) at \((2,2)\). ✓
Hard 2
[5]
Find the set of values of \(p\) such that \(px^2 - 4x + p = 0\) has real roots, where \(p \neq 0\).
STEP-BY-STEP SOLUTION
Condition
Real roots \(\Rightarrow \Delta \geq 0\)
Set up
\(\Delta = (-4)^2 - 4(p)(p) \geq 0\)
\(16 - 4p^2 \geq 0 \Rightarrow p^2 \leq 4\)
\(16 - 4p^2 \geq 0 \Rightarrow p^2 \leq 4\)
Solve
Critical values: \(p = \pm 2\)
Between the roots of U-shape: \(-2 \leq p \leq 2\)
Between the roots of U-shape: \(-2 \leq p \leq 2\)
Answer
Excluding \(p=0\): \(\boldsymbol{-2 \leq p \leq 2,\; p\neq 0}\)
Hard 3
[6]
Sketch \(y = |x^2 - 5x + 6|\). Mark all axis intercepts and turning points.
STEP-BY-STEP — MODULUS GRAPH
Step 1
Find roots
Find roots
\((x-2)(x-3)=0 \Rightarrow\) x-intercepts at \(\mathbf{(2,0)}\) and \(\mathbf{(3,0)}\)
Step 2
y-intercept
y-intercept
At \(x=0\): \(|6|=6\) ⇒ \(\mathbf{(0,6)}\)
Step 3
Vertex
Vertex
Axis of symmetry: \(x=2.5\)
Inner value: \((2.5)^2-5(2.5)+6 = -0.25\)
After modulus: local max \(\mathbf{(2.5,\;0.25)}\)
Inner value: \((2.5)^2-5(2.5)+6 = -0.25\)
After modulus: local max \(\mathbf{(2.5,\;0.25)}\)
Shape
Draw parent parabola → reflect the dip between \(x=2\) and \(x=3\) upward. Result: W-shape with small hill at \((2.5, 0.25)\).
The modulus curve is always \(\geq 0\). Every part that dips below the x-axis gets flipped upward.
Past Year Paper Questions
Cambridge IGCSE 0606 style — multi-part, show all working
0606 Style
Simultaneous & Discriminant
[10]
(a) Find the values of \(k\) for which \(kx^2 - 2kx + 3 = 0\) has real and distinct roots. [3]
(b) The line \(y = k - 2x\) intersects the curve \(y = x^2 - 3x + 5\) at two distinct points. Find the range of values of \(k\). [4]
(c) Hence find the value of \(k\) for which the line is a tangent to the curve, and state the point of tangency. [3]
(b) The line \(y = k - 2x\) intersects the curve \(y = x^2 - 3x + 5\) at two distinct points. Find the range of values of \(k\). [4]
(c) Hence find the value of \(k\) for which the line is a tangent to the curve, and state the point of tangency. [3]
FULL WORKED SOLUTION
(a) Setup
Real distinct roots ⇒ \(\Delta > 0\)
\((-2k)^2 - 4(k)(3) > 0\)
\(4k^2 - 12k > 0 \Rightarrow 4k(k-3) > 0\)
\((-2k)^2 - 4(k)(3) > 0\)
\(4k^2 - 12k > 0 \Rightarrow 4k(k-3) > 0\)
(a) Answer
U-shape positive outside roots: \(\boldsymbol{k < 0 \text{ or } k > 3}\)
(b) Equate
\(k-2x = x^2-3x+5\)
\(x^2 - x + (5-k) = 0\)
\(x^2 - x + (5-k) = 0\)
(b) Two points
\(\Delta > 0\): \(1 - 4(5-k) > 0 \Rightarrow 4k > 19\)
\(\boldsymbol{k > \tfrac{19}{4}}\)
\(\boldsymbol{k > \tfrac{19}{4}}\)
(c) Tangent
\(\Delta = 0 \Rightarrow k = \tfrac{19}{4}\)
Then \(x^2-x+\tfrac{1}{4}=0 \Rightarrow (x-\tfrac{1}{2})^2=0 \Rightarrow x=\tfrac{1}{2}\)
\(y = \tfrac{19}{4}-2(\tfrac{1}{2}) = \tfrac{15}{4}\)
\(\boldsymbol{k=\tfrac{19}{4},\text{ point } \left(\tfrac{1}{2},\tfrac{15}{4}\right)}\)
Then \(x^2-x+\tfrac{1}{4}=0 \Rightarrow (x-\tfrac{1}{2})^2=0 \Rightarrow x=\tfrac{1}{2}\)
\(y = \tfrac{19}{4}-2(\tfrac{1}{2}) = \tfrac{15}{4}\)
\(\boldsymbol{k=\tfrac{19}{4},\text{ point } \left(\tfrac{1}{2},\tfrac{15}{4}\right)}\)
0606 Style
Completing the Square & Inequalities
[9]
(a) Express \(f(x)=2x^2-12x+23\) in the form \(a(x-h)^2+k\). [3]
(b) State the minimum value of \(f(x)\) and the value of \(x\) at which it occurs. [2]
(c) State the range of \(f\) for \(x \geq 2\). [2]
(d) Solve \(f(x) \leq 29\). [2]
(b) State the minimum value of \(f(x)\) and the value of \(x\) at which it occurs. [2]
(c) State the range of \(f\) for \(x \geq 2\). [2]
(d) Solve \(f(x) \leq 29\). [2]
FULL WORKED SOLUTION
(a)
\(2[x^2-6x]+23 = 2[(x-3)^2-9]+23 = \boldsymbol{2(x-3)^2+5}\)
(b)
Min value = \(\boldsymbol{5}\) at \(\boldsymbol{x=3}\)
(c)
At \(x=2\): \(f(2)=2+5=7\). Min at \(x=3\) is 5. For \(x\geq2\): range is \(\boldsymbol{f\geq5}\)
(d)
\(2(x-3)^2+5\leq29 \Rightarrow (x-3)^2\leq12\)
\(-2\sqrt3\leq x-3\leq2\sqrt3\)
\(\boldsymbol{3-2\sqrt3\leq x\leq3+2\sqrt3}\)
\(-2\sqrt3\leq x-3\leq2\sqrt3\)
\(\boldsymbol{3-2\sqrt3\leq x\leq3+2\sqrt3}\)
0606 Style
Quadratic Inequality & Graph
[8]
(a) Solve the inequality \(x^2 - x - 12 < 0\). [3]
(b) Hence solve \(e^{2t} - e^t - 12 < 0\). [3]
(c) On the same axes, sketch \(y = x^2 - x - 12\) and \(y = |x^2 - x - 12|\), marking all key points. [2]
(b) Hence solve \(e^{2t} - e^t - 12 < 0\). [3]
(c) On the same axes, sketch \(y = x^2 - x - 12\) and \(y = |x^2 - x - 12|\), marking all key points. [2]
FULL WORKED SOLUTION
(a) Factorise
\((x-4)(x+3)<0\)
Roots: \(x=4,\; x=-3\). U-shape below zero between roots.
\(\boldsymbol{-3 < x < 4}\)
Roots: \(x=4,\; x=-3\). U-shape below zero between roots.
\(\boldsymbol{-3 < x < 4}\)
(b) Substitution
Let \(u=e^t\). Then \(u^2-u-12<0 \Rightarrow (u-4)(u+3)<0\)
From (a): \(-3<u<4\). But \(u=e^t>0\) always, so \(0<e^t<4\)
\(t<\ln4\). Combined with \(e^t>0\) (always true):
\(\boldsymbol{t < \ln 4}\)
From (a): \(-3<u<4\). But \(u=e^t>0\) always, so \(0<e^t<4\)
\(t<\ln4\). Combined with \(e^t>0\) (always true):
\(\boldsymbol{t < \ln 4}\)
(c) Shape
Parent: U-shape, roots at \(-3\) and \(4\), vertex at \(x=0.5\), \(y=-12.25\)
Modulus: reflect the portion below x-axis upward. Vertex becomes local max at \((0.5, 12.25)\).
y-intercept for both: \(|{-12}|=12\).
Modulus: reflect the portion below x-axis upward. Vertex becomes local max at \((0.5, 12.25)\).
y-intercept for both: \(|{-12}|=12\).