Simultaneous Equations

At A-Level you must solve simultaneous equations where one equation is linear and one is quadratic (or involves other non-linear terms). This extends the GCSE skill of solving two linear simultaneous equations. The Edexcel 9MA0 specification requires both algebraic and graphical understanding.

---

Linear Simultaneous Equations — Recap

Two linear equations with two unknowns can be solved by elimination or substitution.

Example: Solve 2x + 3y = 12 and x − y = 1.

By substitution:

• From the second equation: x = y + 1
• Substitute into the first: 2(y + 1) + 3y = 12
• 2y + 2 + 3y = 12
• 5y = 10, so y = 2
• x = 2 + 1 = 3
• Solution: x = 3, y = 2

By elimination:

• Multiply the second equation by 3: 3x − 3y = 3
• Add to the first: 2x + 3y + 3x − 3y = 12 + 3
• 5x = 15, so x = 3, then y = 2

---

One Linear, One Quadratic

This is the key A-Level technique. You always use substitution: rearrange the linear equation to express one variable in terms of the other, then substitute into the quadratic equation.

Example: Solve y = x + 1 and x² + y² = 13.

Step 1: Substitute y = x + 1 into x² + y² = 13:

• x² + (x + 1)² = 13
• x² + x² + 2x + 1 = 13
• 2x² + 2x − 12 = 0
• x² + x − 6 = 0
• (x + 3)(x − 2) = 0
• x = −3 or x = 2

Step 2: Find corresponding y values:

• When x = −3: y = −3 + 1 = −2
• When x = 2: y = 2 + 1 = 3

Solutions: (−3, −2) and (2, 3).

---

Graphical Interpretation

The solutions of simultaneous equations are the points of intersection of their graphs.

• A line and a circle can intersect at 0, 1 or 2 points.
• A line and a parabola can intersect at 0, 1 or 2 points.

The number of intersections corresponds to the number of solutions of the resulting quadratic:

Discriminant of the quadratic	Intersections	Geometric meaning
b² − 4ac > 0	2 points	Line cuts through the curve
b² − 4ac = 0	1 point	Line is tangent to the curve
b² − 4ac < 0	0 points	Line does not meet the curve

Example: Show that the line y = x + 5 does not intersect the circle x² + y² = 4.

Substitute: x² + (x + 5)² = 4

• x² + x² + 10x + 25 = 4
• 2x² + 10x + 21 = 0
• Discriminant: 100 − 168 = −68 < 0
• No real solutions, so no intersection.

Exam Tip: When a question asks you to "show" or "prove" that a line is tangent to a curve, set up the simultaneous equation, form a quadratic, and show the discriminant equals zero. This is a very common exam question.

---

Linear and Other Non-Linear Equations

The same substitution technique works when the non-linear equation involves other terms.

Example: Solve y = 2x − 1 and y = x² − 3x + 5.

Set the right-hand sides equal:

• 2x − 1 = x² − 3x + 5
• 0 = x² − 5x + 6
• 0 = (x − 2)(x − 3)
• x = 2 or x = 3

When x = 2: y = 3. When x = 3: y = 5.

Solutions: (2, 3) and (3, 5).

---

Using the Discriminant to Find Unknown Parameters

A common A-Level question gives you an unknown constant and asks for the value(s) that produce a tangent (one intersection) or no intersection.

Example: The line y = kx + 2 is tangent to the curve y = x² + 3. Find the value(s) of k.

Set equal: kx + 2 = x² + 3

• x² − kx + 1 = 0

For tangency, discriminant = 0:

• k² − 4 = 0
• k² = 4
• k = 2 or k = −2

---

Key Terms

Term	Definition
Simultaneous equations	Two or more equations that must be satisfied at the same time
Substitution	Replacing one variable with an expression in terms of the other
Elimination	Adding or subtracting equations to remove a variable
Tangent (to a curve)	A line that touches a curve at exactly one point

---

Summary

• Solve one-linear-one-quadratic systems by substitution: rearrange the linear equation and substitute into the non-linear one.
• The resulting quadratic determines the number of solutions — use the discriminant to classify.
• Graphically, solutions correspond to intersection points of the two curves.
• Use discriminant = 0 to find conditions for tangency — this is a high-frequency exam question.

---

A-Level Deep Dive: Simultaneous Equations

Spec mapping

Edexcel 9MA0-01 specification section 2 — Algebra and functions, sub-strand 2.7 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). This dovetails with section 2.10 on graphical interpretation and connects forward to section 3 (Coordinate geometry — intersections of lines and circles) and section 5 (Trigonometry — solving systems involving $\sin x$sinx and $\cos x$cosx via identities).

The Edexcel formula booklet does not list a dedicated formula for simultaneous equations — you are expected to manipulate by inspection. However, the discriminant $b^2 - 4ac$b2−4ac from the quadratic-formula listing is your principal analytical tool.

---

Worked example with full mark scheme

Question (8 marks): The line $l$l has equation $y = 2x - k$y=2x−k, where $k$k is a positive constant. The curve $C$C has equation $x^2 + 4y^2 = 20$x2+4y2=20.

(a) Show that the x-coordinates of any points of intersection of $l$l and $C$C satisfy the equation $17x^2 - 16kx + (4k^2 - 20) = 0$17x2−16kx+(4k2−20)=0. (3)

(b) Given that $l$l is a tangent to $C$C, find the exact value of $k$k. (3)

(c) For this value of $k$k, find the coordinates of the point of tangency. (2)

Solution with mark scheme:

(a) Substitute $y = 2x - k$y=2x−k into $x^2 + 4y^2 = 20$x2+4y2=20:

$x^2 + 4(2x - k)^2 = 20$x2+4(2x−k)2=20

M1 — substituting the linear equation into the quadratic. Examiners award this for any correct substitution, even if subsequent algebra is botched.

Expand $(2x - k)^2 = 4x^2 - 4kx + k^2$(2x−k)2=4x2−4kx+k2, so $4(2x - k)^2 = 16x^2 - 16kx + 4k^2$4(2x−k)2=16x2−16kx+4k2.

$x^2 + 16x^2 - 16kx + 4k^2 = 20$x2+16x2−16kx+4k2=20

M1 — correct expansion of the squared bracket. A common error here is to write $(2x - k)^2 = 4x^2 + k^2$(2x−k)2=4x2+k2, missing the cross term. That loses both M1 marks.

$17x^2 - 16kx + 4k^2 - 20 = 0$17x2−16kx+4k2−20=0

A1 — correct collected form, in the precise format requested. Note "show that" questions require the answer to appear exactly as stated; a sign slip (e.g. $+20$+20) loses A1 even if the method is right.

(b) For tangency, the quadratic has a repeated root, so the discriminant is zero:

$(-16k)^2 - 4(17)(4k^2 - 20) = 0$(−16k)2−4(17)(4k2−20)=0

M1 — applying $b^2 - 4ac = 0$b2−4ac=0. Some candidates write $b^2 = 4ac$b2=4ac; equally valid.

$256k^2 - 68(4k^2 - 20) = 0$256k2−68(4k2−20)=0 $256k^2 - 272k^2 + 1360 = 0$256k2−272k2+1360=0 $-16k^2 + 1360 = 0$−16k2+1360=0 $k^2 = 85$k2=85

M1 — correct algebraic simplification leading to a value for $k^2$k2.

Since $k > 0$k>0, $k = \sqrt{85}$k=85​.

A1 — exact value, with the constraint $k > 0$k>0 acknowledged. If a candidate writes $k = \pm\sqrt{85}$k=±85​ and fails to reject the negative root, they lose A1. The phrase "exact value" forbids decimals — $k \approx 9.22$k≈9.22 would also lose A1.

(c) When $k = \sqrt{85}$k=85​, the quadratic becomes $17x^2 - 16\sqrt{85}\,x + (4 \cdot 85 - 20) = 0$17x2−1685​x+(4⋅85−20)=0, i.e. $17x^2 - 16\sqrt{85}\,x + 320 = 0$17x2−1685​x+320=0.

Repeated root: $x = \dfrac{16\sqrt{85}}{2 \cdot 17} = \dfrac{8\sqrt{85}}{17}$x=2⋅171685​​=17885​​.

M1 — using $x = -b/(2a)$x=−b/(2a) (the formula for a repeated root) or equivalent.

Then $y = 2x - k = \dfrac{16\sqrt{85}}{17} - \sqrt{85} = \dfrac{16\sqrt{85} - 17\sqrt{85}}{17} = -\dfrac{\sqrt{85}}{17}$y=2x−k=171685​​−85​=171685​−1785​​=−1785​​.

A1 — both coordinates correct: $\left(\dfrac{8\sqrt{85}}{17},\, -\dfrac{\sqrt{85}}{17}\right)$(17885​​,−1785​​).

Total: 8 marks (M4 A3 — 7 marks of method/accuracy split as shown).

---

Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): The curve $C$C has equation $y = x^2 - 4x + 7$y=x2−4x+7 and the line $l$l has equation $y = 2x + k$y=2x+k, where $k$k is a constant.

(a) Show that the x-coordinates of any points of intersection of $l$l and $C$C satisfy $x^2 - 6x + (7 - k) = 0$x2−6x+(7−k)=0. (2)

(b) Given that $l$l and $C$C intersect at two distinct points, find the range of values of $k$k. (4)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1b) — equating $x^2 - 4x + 7 = 2x + k$x2−4x+7=2x+k.
• A1 (AO1.1b) — rearranging to printed form $x^2 - 6x + (7 - k) = 0$x2−6x+(7−k)=0.

(b)

• M1 (AO2.1) — recognising that "two distinct intersections" means $b^2 - 4ac > 0$b2−4ac>0. This is an AO2 mark because it requires reasoning about the link between geometry (two intersections) and algebra (positive discriminant). Many candidates default to $\geq 0$≥0 and lose this mark.
• M1 (AO1.1b) — substituting correctly: $36 - 4(7 - k) > 0$36−4(7−k)>0.
• A1 (AO1.1b) — simplifying to $36 - 28 + 4k > 0$36−28+4k>0, i.e. $4k > -8$4k>−8, $k > -2$k>−2.
• A1 (AO2.2a) — stating the range correctly with strict inequality, justifying with reference to "distinct" requiring $> 0$>0 not $\geq 0$≥0. Candidates writing $k \geq -2$k≥−2 lose this.

Total: 6 marks split AO1 = 4, AO2 = 2. Note the heavy AO2 weighting — Edexcel deliberately shifts marks toward reasoning when discriminant conditions are involved.

---

Synoptic links

Connects to:

1. Section 3 — Coordinate geometry (lines and circles): the standard "show that the line is a tangent to the circle" question is exactly the technique above. You will see this in Year 2 with circle equations $(x-a)^2 + (y-b)^2 = r^2$(x−a)2+(y−b)2=r2 where simultaneous solving with a linear equation produces a quadratic in $x$x whose discriminant must be zero for tangency.

2. Section 7 — Differentiation, gradient of curves: an alternative tangent test is to find the curve's gradient at a candidate point and equate it to the line's gradient. Pairing both methods (algebraic discriminant vs. calculus) is a common Year 2 synoptic question worth 8–10 marks.

3. Section 10 — Vectors: the intersection of two lines in 3D requires solving three simultaneous equations in two parameters $\lambda$λ and $\mu$μ. The system is over-determined and consistency must be checked — exactly the same logic as detecting "no solution" via a discriminant in 2D.

4. Sections 5 and 6 — Trigonometry: equations like $\sin x = \frac{1}{2}x$sinx=21​x or $2\cos^2 x = 1 - \sin x$2cos2x=1−sinx are simultaneous in nature; the second can be reduced to a quadratic in $\sin x$sinx using $\cos^2 x = 1 - \sin^2 x$cos2x=1−sin2x.

---

Mark-scheme literacy

Edexcel 9MA0 weights its assessment objectives across three tiers: AO1 (knowledge and procedural fluency), AO2 (reasoning, interpretation and communication), AO3 (problem-solving in unstructured contexts). For simultaneous-equation questions specifically:

Mark type	Typical weighting	What earns it
AO1	50–60%	Correct substitution, accurate algebraic manipulation, correct application of $b^2 - 4ac$b2−4ac formula
AO2	25–35%	Translating "tangent" → "discriminant = 0", recognising "distinct" → "strictly positive discriminant", justifying acceptance/rejection of roots
AO3	10–20%	Modelling problems where the simultaneous system arises from a context (e.g. projectile reaching a target curve)