Simultaneous Equations (Non-Linear)

Welcome — Non-Linear Simultaneous Equations!

When one equation is non-linear (quadratic, circle, etc.) and the other is linear, we use substitution to reduce the system to a single quadratic. The solutions represent the intersection points of the two curves.

Method: isolate y (or x) from the linear equation → substitute into the non-linear → solve the quadratic → find both coordinate pairs

Learning Objectives

Solve simultaneous equations where one is quadratic
Solve simultaneous equations involving circles
Use the discriminant to find when a line is tangent to a curve
Interpret solutions geometrically (0, 1 or 2 intersection points)
Write solutions as coordinate pairs in the correct order

Substitution Method

6 steps from linear + quadratic to solution pairs

Geometric & Discriminant

Tangent lines, 0/1/2 intersections

Circle Intersections

x² + y² = r² and (x−a)² + (y−b)² = r²

Worked Examples

6 fully worked mark-scheme examples

Common Mistakes

5 traps to avoid

Visualiser

Interactive line + parabola intersection explorer

Learn 1 — The Substitution Method

The substitution method is the standard approach for non-linear simultaneous equations. The key rule is: always rearrange the LINEAR equation, then substitute into the NON-LINEAR one.

The 6-Step Process

Step 1: Label the equations (1) and (2). Identify which is linear, which is non-linear.

Step 2: From the linear equation, express y (or x) in terms of the other variable.

Step 3: Substitute this expression into the non-linear equation.

Step 4: Expand and rearrange to get a standard quadratic ax² + bx + c = 0.

Step 5: Solve the quadratic (factorise or use the formula). Find x₁ and x₂.

Step 6: Substitute each x-value back into the linear equation to find y₁ and y₂. Write as pairs (x₁, y₁) and (x₂, y₂).

Worked Example: y = 2x + 1 and y = x² − 3

Step 1–2: Linear: y = 2x + 1 already expressed.

Step 3: Sub into y = x² − 3: 2x + 1 = x² − 3

Step 4: Rearrange: x² − 2x − 4 = 0

Step 5: Quadratic formula: x = [2 ± √(4 + 16)] / 2 = 1 ± √5

Step 6: x₁ = 1 + √5 ≈ 3.24 → y₁ = 2(1+√5)+1 = 3 + 2√5 ≈ 7.47
x₂ = 1 − √5 ≈ −1.24 → y₂ = 2(1−√5)+1 = 3 − 2√5 ≈ −1.47

Always substitute back into the LINEAR equation for y-values — it's simpler and less error-prone than using the quadratic.

Worked Example 2: y = 3x − 4 and y = x² + x − 6

Sub: 3x − 4 = x² + x − 6 → x² − 2x − 2 = 0

Formula: x = [2 ± √(4+8)] / 2 = 1 ± √3

y-values: y = 3(1+√3)−4 = −1+3√3 and y = 3(1−√3)−4 = −1−3√3

Solutions: (1+√3, −1+3√3) and (1−√3, −1−3√3)

Learn 2 — Geometric Interpretation and the Discriminant

How Many Intersections?

When a line y = mx + c meets parabola y = ax² + bx + d, substituting gives a quadratic. The discriminant tells us:
• Discriminant > 0: 2 real roots → line crosses the parabola at 2 points
• Discriminant = 0: 1 repeated root → line is a tangent to the parabola (touches at exactly 1 point)
• Discriminant < 0: no real roots → line does not intersect the parabola

Finding k for a Tangent Line

A common exam question: find the value(s) of k such that a line is tangent to a curve.

Example: Find k so that y = kx − 3 is tangent to y = x² + 1.

Substitute: kx − 3 = x² + 1 → x² − kx + 4 = 0

Discriminant = 0 for tangent: k² − 16 = 0 → k² = 16 → k = ±4

Check: When k = 4: x² − 4x + 4 = (x−2)² = 0 → x = 2, y = 5. Tangent point (2, 5). ✓

Another Pattern: Finding k for a Tangent

Example: Line y = k is tangent to y = −x² + 4x − 1. Find k.

Substitute: −x² + 4x − 1 = k → −x² + 4x − 1 − k = 0 → x² − 4x + 1 + k = 0

Disc = 0: 16 − 4(1 + k) = 0 → 12 = 4k → k = 3

Interpretation: The maximum value of −x² + 4x − 1 is 3 (the horizontal tangent line).

Using Discriminant to Count Intersections

Line y = 2x + k and parabola y = x² − 3:
Substitute: 2x + k = x² − 3 → x² − 2x − (3 + k) = 0
Disc = 4 + 4(3 + k) = 4 + 12 + 4k = 16 + 4k
• Disc > 0 when k > −4: 2 intersections
• Disc = 0 when k = −4: tangent (1 intersection)
• Disc < 0 when k < −4: no intersection

For exam questions asking "find the range of k for which the line intersects the curve", set discriminant > 0 and solve the resulting inequality.

Learn 3 — Circle Intersections

Circle Centred at Origin: x² + y² = r²

x² + y² = r² — all points distance r from the origin

Method: Express y from the linear equation, then substitute into the circle equation. You get a quadratic in x.

Example: x² + y² = 25 and y = x + 1

Step 1: From linear: y = x + 1

Step 2: Sub into circle: x² + (x+1)² = 25

Step 3: Expand: x² + x² + 2x + 1 = 25 → 2x² + 2x − 24 = 0 → x² + x − 12 = 0

Step 4: Factorise: (x+4)(x−3) = 0 → x = −4 or x = 3

Step 5: y-values: y = −4+1 = −3 and y = 3+1 = 4

Solutions: (−4, −3) and (3, 4) ✓ Check: 16+9=25 ✓ and 9+16=25 ✓

Circle Not Centred at Origin: (x − a)² + (y − b)² = r²

Example: (x−1)² + (y−2)² = 10 and y = x + 1

Sub y = x+1: (x−1)² + (x+1−2)² = 10 → (x−1)² + (x−1)² = 10 → 2(x−1)² = 10 → (x−1)² = 5

Solve: x − 1 = ±√5 → x = 1 ± √5

y: y = (1+√5)+1 = 2+√5 and y = 2−√5. Solutions: (1+√5, 2+√5) and (1−√5, 2−√5)

Checking if a Line Misses a Circle

If after substitution the discriminant is negative, the line does not intersect the circle at all — it's fully outside the circle. Always state "no real solutions — the line does not intersect the circle" rather than giving complex answers.

Distance from centre to line < r → 2 intersections. Distance = r → tangent. Distance > r → no intersection. You can verify using the formula: distance = |ax₀ + by₀ + c| / √(a²+b²).

Example 1 — y = x + 3 and y = x² − 5 M1M1A1A1

M1: Sub y = x+3 into y = x²−5: x+3 = x²−5 → x²−x−8 = 0

M1: Formula: x = [1 ± √33] / 2

A1: x₁ = (1+√33)/2 ≈ 3.37, x₂ = (1−√33)/2 ≈ −2.37

A1: y₁ = x₁+3 ≈ 6.37, y₂ ≈ 0.63. Solutions: (3.37, 6.37) and (−2.37, 0.63).

Example 2 — y = 2x − 1 and y = x² + x − 3 M1M1A1

M1: Sub: 2x−1 = x²+x−3 → x²−x−2 = 0

M1: Factorise: (x−2)(x+1) = 0 → x = 2 or x = −1

A1: y = 2(2)−1 = 3 and y = 2(−1)−1 = −3. Solutions: (2, 3) and (−1, −3)

Example 3 — Find k: y = kx − 3 tangent to y = x² + 1 M1M1A1

M1: Sub: kx−3 = x²+1 → x²−kx+4 = 0

M1: For tangent, discriminant = 0: k²−16 = 0

A1: k = ±4

Example 4 — x² + y² = 10 and y = x + 2 M1M1A1

M1: Sub y = x+2: x²+(x+2)² = 10 → 2x²+4x+4 = 10 → 2x²+4x−6 = 0 → x²+2x−3 = 0

M1: Factorise: (x+3)(x−1) = 0 → x = −3 or x = 1

A1: y = −3+2 = −1 and y = 1+2 = 3. Solutions: (−3, −1) and (1, 3)

Example 5 — x² + y² = 20 and y = 2x M1A1

M1: Sub y = 2x: x²+4x² = 20 → 5x² = 20 → x² = 4 → x = ±2

A1: x = 2 → y = 4; x = −2 → y = −4. Solutions: (2, 4) and (−2, −4)

Example 6 — Line y = 3x − 2 meets curve y = x² + x − 4 M1M1A1

M1: 3x−2 = x²+x−4 → x²−2x−2 = 0

M1: Formula: x = [2 ± √12] / 2 = 1 ± √3

A1: x ≈ 2.73 → y ≈ 6.20; x ≈ −0.73 → y ≈ −4.20. Solutions: (1+√3, 1+3√3) and (1−√3, 1−3√3)

Common Mistakes — Non-Linear Simultaneous Equations

Mistake 1 — Substituting into the linear equation instead of the non-linear

❌ y = x+3 and y = x²−2. Student expresses x from y=x+3 as x=y−3, then subs into y=x+3 again — getting a trivial identity!

✅ Always rearrange the linear equation, then sub into the non-linear. Here: sub y=x+3 into y=x²−2, giving x+3=x²−2 → a quadratic in x.

Mistake 2 — Only finding x-values, forgetting to find y-values

❌ x = 2 or x = −1 (student stops here)

✅ Always substitute back to find both y-values: y = 2(2)−1 = 3 and y = 2(−1)−1 = −3. Write as coordinate pairs: (2, 3) and (−1, −3).

Mistake 3 — Pairing x and y values incorrectly

❌ x = 2 gives y₁, x = −1 gives y₂, but student writes (2, y₂) and (−1, y₁)

✅ Each x must be paired with its own y, found by substituting that specific x into the linear equation. x₁ = 2 → y₁ = 3 → pair (2, 3). x₂ = −1 → y₂ = −3 → pair (−1, −3). Never mix the pairs.

Mistake 4 — Expanding brackets incorrectly (especially (ax + b)²)

❌ (2x+1)² = 4x² + 1 (missing the cross term)

✅ (2x+1)² = 4x² + 4x + 1. Always use FOIL: (2x+1)(2x+1) = 4x² + 2x + 2x + 1 = 4x² + 4x + 1.

Mistake 5 — Not stating "no real solutions" when discriminant < 0

❌ After getting a negative discriminant, student writes "x = [2 ± √−8] / 2" as if this is valid.

✅ When discriminant < 0, the quadratic has no real roots. State: "Discriminant < 0, so the line and curve do not intersect — no real solutions." This is worth marks in an exam.

Key Formulas

Given in Exam

Quadratic formula: x = [−b ± √(b²−4ac)] / 2a
Discriminant: Δ = b² − 4ac

The quadratic formula and discriminant are on the Cambridge IGCSE formula sheet.

Must Memorise

Circle: x² + y² = r²
General circle: (x−a)² + (y−b)² = r²
Δ > 0 → 2 intersection points
Δ = 0 → tangent (1 point)
Δ < 0 → no real intersection
Always substitute into the LINEAR equation for y-values
Write solutions as coordinate pairs

6-Step Summary

1. Label equations. Identify linear vs non-linear.

2. Rearrange the linear equation: y = ... (or x = ...)

3. Substitute into the non-linear equation.

4. Expand and rearrange to ax² + bx + c = 0.

5. Solve the quadratic (factorise or formula). Find x-values.

6. Sub each x back into the linear equation. Write solution pairs.

Line + Parabola Intersection Visualiser

Fixed parabola: y = x² − 4. Adjust the line y = mx + c. Watch intersection points move.

m: 2 c: 1

Adjust sliders to see intersections.

Exercise 1 — Line + Parabola (Factorisable)

Find the x-coordinates of the intersection points. Enter the larger x-value.

1. y = x + 2 and y = x². Larger x-value?

2. y = x + 6 and y = x². Larger x-value?

3. y = 2x + 3 and y = x² − 1. Larger x-value?

4. y = x + 6 and y = x² − 4x. Larger x-value?

5. y = 3x + 4 and y = x² + 2x. Larger x-value?

6. y = 2x + 8 and y = x². Larger x-value?

7. y = x + 12 and y = x² − 2x. Larger x-value?

8. y = 2 and y = x² − 7. Larger x-value? (2 d.p.)

Exercise 2 — Line + Parabola (Quadratic Formula Needed)

Larger x-coordinate of intersection (2 d.p.).

1. y = x + 3 and y = x² − 5. Larger x (2 d.p.)?

2. y = 2x − 1 and y = x² + 1. Larger x (2 d.p.)?

3. y = x + 1 and y = x² − 2x + 1. Larger x (2 d.p.)?

4. y = 3x − 2 and y = x² + x − 4. Larger x (2 d.p.)?

5. y = x + 5 and y = x² − 2. Larger x (2 d.p.)?

6. y = 2x + 3 and y = x² − x. Larger x (2 d.p.)?

7. y = 4x − 1 and y = x² + 2. Larger x (2 d.p.)?

8. y = x − 1 and y = x² − 3x − 1. Larger x (2 d.p.)?

Exercise 3 — Discriminant and Tangent Lines

Find the value(s) of k. Enter the positive value of k.

1. y = kx + 1 tangent to y = x². Find k (positive).

2. y = kx − 5 tangent to y = x² + 3. Find k (positive).

3. y = kx + 4 tangent to y = x² + 2x. Find k (positive).

4. y = k tangent to y = −x² + 6x − 5. Find k.

5. y = k tangent to y = −x² + 4x + 1. Find k.

6. y = 2x + k tangent to y = x². Find k.

7. Line y = x + k meets y = x² − x + 2. For k such that disc = 0, find k.

8. Minimum value of x² − 2x + k is 0. Find k.

Exercise 4 — Circle Intersections (Exam Focus)

Find the larger x-coordinate of the intersection points.

1. x² + y² = 25 and y = x − 1. Larger x (2 d.p.)?

2. x² + y² = 10 and y = x + 2. Larger x?

3. x² + y² = 20 and y = 2x. Larger x?

4. x² + y² = 50 and y = x. Larger x (2 d.p.)?

5. x² + y² = 13 and y = x + 1. Larger x?

6. x² + y² = 5 and y = 2x. Larger x (2 d.p.)?

7. x² + y² = 17 and y = x + 1. Larger x?

8. x² + y² = 8 and y = x + 2. Larger x (2 d.p.)?

Exercise 5 — Cross-Topic Mixed Questions

Mixed — use whichever method is needed.

1. y = x + 4 and x² + y² = 26. Sum of x-coordinates of intersections?

2. y = 2x + 1 and y = x² − 2x + 3. Sum of x-coordinates?

3. y = kx + 2 tangent to y = x² + 4x + 5. Find k (positive).

4. x² + y² = 45 and y = 2x. Larger x value?

5. y = −x + 6 and y = x² − 2x. Sum of y-coordinates of both solutions?

6. Line y = x + k meets parabola y = x² − 4 at exactly 1 point. Find k.

7. y = 3 and x² + y² = 25. Larger x (2 d.p.)?

8. x² + y² = 34 and y = x + 2. Larger x?

Practice — 25 Questions

🔵 Non-calculator 🟢 Calculator allowed

🔵 1. y=x+2 and y=x². Larger x?

🔵 2. y=3x+4 and y=x²+2x. Smaller x?

🔵 3. y=2 and y=x²−7. Larger x (2 d.p.)?

🔵 4. x²+y²=25 and y=2x. Larger x (2 d.p.)?

🟢 5. y=x+3 and y=x²−5. Larger x (2 d.p.)?

🔵 6. x²+y²=10 and y=x+2. Sum of x-coords?

🟢 7. y=kx−3 tangent to y=x²+1. Positive k?

🔵 8. y=x−1 and y=x²−5. Sum of x-coords?

🟢 9. y=2x+1 and y=x²−3. Larger x (2 d.p.)?

🔵 10. x²+y²=20 and y=2x. Larger x?

🔵 11. y=k tangent to y=−x²+8x−10. Find k.

🔵 12. y=x+6 and y=x². Larger x?

🟢 13. y=3x−2 and y=x²+x−4. Larger x (2 d.p.)?

🔵 14. y=2x+8 and y=x². Larger x?

🟢 15. x²+y²=50 and y=x. Larger x (2 d.p.)?

🔵 16. y=2x+3 and y=x²−1. Larger x?

🔵 17. y=2x−1 and y=x²+x−3. Larger x?

🟢 18. y=x+5 and y=x²−2. Larger x (2 d.p.)?

🔵 19. x²+y²=13 and y=x+1. Larger x?

🟢 20. y=x+1 and y=x²−2x+1. Larger x (2 d.p.)?

🔵 21. y=kx+1 tangent to y=x². Positive k?

🔵 22. y=x+12 and y=x²−2x. Larger x?

🟢 23. x²+y²=34 and y=x+2. Larger x?

🔵 24. y=2 and y=x²−2. Larger x?

🟢 25. y=4x−1 and y=x²+2. Larger x (2 d.p.)?

Challenge — 12 Multi-Step Questions

1. y=x+k and y=x²−3 have exactly 2 solutions. What is the minimum value of k (integer)?

2. y=2x−1 meets y=x²+x−3 at points A and B. Sum of both x-coords?

3. Circle x²+y²=r² passes through (3,4). Find r.

4. y=kx and y=x²+1. Find positive k for tangency.

5. y=x+a and x²+y²=18 have no solutions. Find minimum positive integer a.

6. y=mx+1 tangent to y=x²+3x+4. Find negative m.

7. Curve y=x²−4x+k and line y=x. For 2 intersections, what is the maximum integer k?

8. x²+y²=25 and y=x+1. Product of both x-coords?

9. y=x+4 and x²+y²=26. Sum of both y-coords?

10. y=2x+k and y=x². For tangency, find k (negative value).

11. y=3x−2 meets y=x²+x−4. Larger x value (2 d.p.)?

12. x²+y²=29 and y=2x+1. Larger x (2 d.p.)?

Exam Style — 5 Structured Questions

Q1 [4 marks] — y = 2x − 1 and y = x² + x − 3

(a) After substituting, the quadratic is x² − x − 2 = 0. Factorise: (x − ?)(x + 1) = 0. Enter ?.

(b) Larger x-value?

(c) Larger y-value? (use linear equation)

(d) Smaller y-value?

Q2 [3 marks] — x² + y² = 10 and y = x + 2

(a) After substituting: 2x² + 4x − 6 = 0 → x² + 2x − 3 = 0. Larger x?

(b) Smaller x?

(c) y-value when x = 1?

Q3 [4 marks] — Find k so that y = kx − 3 is tangent to y = x² + 1.

(a) After substituting: x² − kx + 4 = 0. For tangent, b² − 4ac = 0. Find k² (positive).

(b) Positive value of k?

(c) For k = 4, find the x-coordinate of the tangent point.

(d) For k = −4, find the x-coordinate of the tangent point.

Q4 [5 marks] — y = x + 4 and x² + y² = 26.

(a) After substituting: 2x² + 8x − 10 = 0 → x² + 4x − 5 = 0. Larger x?

(b) Smaller x?

(c) y-value when x = 1?

(d) y-value when x = −5?

(e) Check: does (1, 5) lie on the circle x² + y² = 26? (Enter 1 for yes, 0 for no)

Q5 [4 marks] — y = 2x + k and y = x² − 4.

(a) After substituting: x² − 2x − (4 + k) = 0. Discriminant expression = 4 + 4(4+k) = 4k + ?. Find ?.

(b) For tangency (disc = 0), find k.

(c) For 2 intersections, the condition is k > ? Enter the critical value.

(d) For k = 0, how many intersection points? (Enter 0, 1 or 2)