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This lesson covers advanced simultaneous equations as required by the AQA A-Level Mathematics specification (7357). At A-Level, you must be able to solve systems involving one linear and one quadratic equation (or other non-linear equation). You also need to understand the geometric interpretation of solutions and use the discriminant to determine the number of intersection points.
A linear-quadratic system consists of one linear equation and one quadratic equation (which may represent a parabola, circle, or other conic). The solution involves substituting the linear equation into the quadratic.
Example: Solve simultaneously: y = 2x + 1 and x² + y² = 10.
Substitute y = 2x + 1 into x² + y² = 10:
x² + (2x + 1)² = 10
x² + 4x² + 4x + 1 = 10
5x² + 4x − 9 = 0
(5x + 9)(x − 1) = 0
x = −9/5 or x = 1
When x = 1: y = 2(1) + 1 = 3. When x = −9/5: y = 2(−9/5) + 1 = −13/5.
Solutions: (1, 3) and (−9/5, −13/5).
Geometric interpretation: The line y = 2x + 1 intersects the circle x² + y² = 10 at two points.
Always substitute from the linear equation into the non-linear equation. If one substitution leads to a simpler quadratic, prefer that approach.
Example: Solve y = x + 1 and y = x² − 3x + 5.
x + 1 = x² − 3x + 5
0 = x² − 4x + 4
0 = (x − 2)²
x = 2 (repeated root)
When x = 2: y = 3.
Only one solution: (2, 3). The line is tangent to the parabola.
Sometimes both equations are quadratic (e.g., two circles). In such cases, subtract one from the other to obtain a linear equation, then proceed as above.
Example: Solve x² + y² = 25 and x² + y² − 6x − 2y = 3.
Subtracting the first from the second:
−6x − 2y = 3 − 25
−6x − 2y = −22
6x + 2y = 22
3x + y = 11
y = 11 − 3x
Substitute into x² + y² = 25:
x² + (11 − 3x)² = 25
x² + 121 − 66x + 9x² = 25
10x² − 66x + 96 = 0
5x² − 33x + 48 = 0
(5x − 16)(x − 3) = 0 — but let's check with the formula:
x = (33 ± √(1089 − 960))/10 = (33 ± √129)/10
Actually, let us re-check: 5(3)² − 33(3) + 48 = 45 − 99 + 48 = −6 ≠ 0, so we should use the quadratic formula. With discriminant = 1089 − 960 = 129, the solutions are irrational. In this case x = (33 ± √129)/10, and we substitute back for y.
Exam Tip: When you have two circles, subtracting their equations always eliminates the x² and y² terms, giving a linear equation called the radical axis. This technique simplifies the problem significantly.
The solutions of a linear-quadratic system correspond to the intersection points of the line and the curve. The discriminant of the resulting quadratic tells us the nature of the intersection:
| Discriminant | Number of Intersections | Geometric Meaning |
|---|---|---|
| b² − 4ac > 0 | Two distinct points | Line crosses the curve twice |
| b² − 4ac = 0 | One point (repeated root) | Line is tangent to the curve |
| b² − 4ac < 0 | No real points | Line does not meet the curve |
Example: For what values of k does the line y = kx + 1 meet the parabola y = x²?
kx + 1 = x²
x² − kx − 1 = 0
Discriminant: k² + 4.
Since k² ≥ 0 for all real k, the discriminant k² + 4 ≥ 4 > 0 for all values of k.
Therefore the line meets the parabola at two distinct points for all values of k.
Example: Find the value of m for which y = mx + 3 is tangent to the circle x² + y² = 9.
Substitute: x² + (mx + 3)² = 9
x²(1 + m²) + 6mx + 9 − 9 = 0
x²(1 + m²) + 6mx = 0
x[(1 + m²)x + 6m] = 0
One solution is x = 0 (where y = 3). For tangency to the circle at (0, 3), the line touches at exactly one point, which gives x = 0 as a repeated root. This happens when the other factor also gives x = 0, i.e. 6m = 0, so m = 0.
Alternatively, for a general tangent to x² + y² = 9, the perpendicular distance from the centre (0, 0) to the line y = mx + 3 (i.e., mx − y + 3 = 0) must equal the radius 3:
|3|/√(m² + 1) = 3
1/√(m² + 1) = 1
m² + 1 = 1
m = 0
The line y = 3 is tangent to the circle x² + y² = 9 at the point (0, 3).
Many A-Level problems require you to form your own simultaneous equations from a description.
Example: The sum of two numbers is 12 and the sum of their squares is 80. Find the two numbers.
Let the numbers be x and y.
x + y = 12 ... (1)
x² + y² = 80 ... (2)
From (1): y = 12 − x. Substitute into (2):
x² + (12 − x)² = 80
x² + 144 − 24x + x² = 80
2x² − 24x + 64 = 0
x² − 12x + 32 = 0
(x − 4)(x − 8) = 0
x = 4 or x = 8
The two numbers are 4 and 8 (or equivalently 8 and 4).
Example: For what values of c does the system y = 2x + c and y = x² have exactly one solution?
2x + c = x²
x² − 2x − c = 0
For exactly one solution, the discriminant = 0:
4 + 4c = 0
c = −1
When c = −1, the line y = 2x − 1 is tangent to y = x² at the point (1, 1).
Exam Tip: Always substitute from the linear equation into the quadratic — never the other way round. Show all algebraic steps clearly. If the question asks you to "find the coordinates of the points of intersection", make sure you give both x and y values as coordinate pairs. If you get a repeated root, state that the line is tangent to the curve.
AQA 7357 specification, Paper 1 — Pure Mathematics, Section B (Algebra and Functions) covers simultaneous equations in two variables by elimination and by substitution, including one linear and one quadratic equation (refer to the official specification document for exact wording). Although Section B is the natural home of this topic, advanced simultaneous equations are tested across all three papers. They appear in Section H (Coordinate geometry) when finding intersections of a line and a circle, in Section J (Differentiation) via tangent and normal conditions where a tangent line meets the curve at a repeated root, and in Section K (Integration) when computing areas bounded by two curves. The AQA formula booklet provides the quadratic formula and the equation of a circle; it does not provide the discriminant condition for tangency — that must be reasoned from b2−4ac.
Question (8 marks):
The line ℓ has equation y=2x+k, where k is a constant. The circle C has equation x2+y2−4x−6y+9=0.
(a) Find the coordinates of the centre and the radius of C. (2)
(b) Find the value(s) of k for which ℓ is tangent to C. (6)
Solution with mark scheme:
(a) Step 1 — complete the square in x and y.
x2−4x=(x−2)2−4,y2−6y=(y−3)2−9
Substituting:
(x−2)2−4+(y−3)2−9+9=0⟹(x−2)2+(y−3)2=4
M1 — completing the square in both variables.
A1 — centre (2,3), radius r=2.
(b) Step 1 — substitute y=2x+k into the circle equation.
x2+(2x+k)2−4x−6(2x+k)+9=0
M1 — correct substitution of the linear into the circle equation.
Step 2 — expand and collect.
x2+4x2+4kx+k2−4x−12x−6k+9=0 5x2+(4k−16)x+(k2−6k+9)=0
M1 — correct expansion and collection of like terms.
A1 — correct quadratic in x: 5x2+(4k−16)x+(k−3)2=0, recognising k2−6k+9=(k−3)2.
Step 3 — apply the tangency condition.
A line is tangent to a circle precisely when the resulting quadratic has a repeated root, i.e. discriminant zero:
Δ=(4k−16)2−4⋅5⋅(k−3)2=0
M1 — invoking b2−4ac=0 as the tangency condition.
Step 4 — solve for k.
(4k−16)2=16(k−4)2=16(k2−8k+16) 20(k−3)2=20(k2−6k+9)
Setting equal:
16k2−128k+256=20k2−120k+180 0=4k2+8k−76 0=k2+2k−19
M1 — correct simplification to a quadratic in k.
By the quadratic formula:
k=2−2±4+76=2−2±80=−1±25
A1 — exact answers k=−1+25 and k=−1−25.
Total: 8 marks (M5 A3, split as shown).
Question (6 marks): The curves C1:y=x2−2x+3 and C2:y=−x2+6x−5 intersect at points P and Q.
(a) Find the coordinates of P and Q. (4)
(b) The midpoint of PQ lies on the line y=mx. Find m exactly. (2)
Mark scheme decomposition by AO:
(a)
(b)
Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1. This problem rewards reading the algebra: a discriminant of zero is information, not an error.
Connects to:
Quadratic discriminant (Section B): the tangency condition b2−4ac=0 for a line meeting a curve in a repeated root is the same discriminant condition that classifies the number of real roots of any quadratic. Two distinct intersections give Δ>0; tangency gives Δ=0; no real intersection gives Δ<0. The same trichotomy classifies all line-conic intersections.
Coordinate geometry of circles (Section H): the equation (x−a)2+(y−b)2=r2 comes from completing the square. Once a line is substituted in, the resulting quadratic in x encodes whether the line is a secant (two roots), tangent (repeated root) or non-intersecting (no real roots). The geometry and the algebra are saying the same thing in different languages.
Parametric curves (Section H, Year 2): when a curve is given parametrically as (x(t),y(t)), finding intersections with another curve becomes a system in two parameters. Eliminating the parameter recovers the Cartesian form, returning the problem to the standard simultaneous-equation framework.
Calculus tangent condition (Section J): a tangent line at (x0,y0) on y=f(x) has gradient f′(x0). Substituting this tangent line back into y=f(x) produces a polynomial with a repeated root at x=x0 — the algebraic and calculus definitions of "tangent" agree exactly. This equivalence is why tangency can always be detected by a discriminant.
Areas between curves (Section K): before integrating to find the area enclosed between two curves, you must locate their intersections — directly a simultaneous-equation problem. The bounds of integration are the solutions of f(x)=g(x).
Advanced simultaneous-equation questions on 7357 split AO marks as follows:
| AO | Typical share | Earned by |
|---|---|---|
| AO1 (knowledge / procedure) | 55–65% | Substituting linear into quadratic, expanding, collecting like terms, applying the quadratic formula |
| AO2 (reasoning / interpretation) | 25–35% | Identifying the tangency condition, interpreting a repeated root geometrically, justifying the choice of substitution direction |
| AO3 (problem-solving) | 5–15% | Multi-step modelling — e.g. finding the locus of a line tangent to a family of circles |
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