GCSE Maths Algebra - How to Solve: Simultaneous Equations (Linear & Quadratic) – Complete Guide

How to Solve: Simultaneous Equations (Linear & Quadratic) – Complete Guide

For GCSE/A-Level (Physics, Chemistry, Biology) – Ace Your Exam!

Introduction

"Mastering simultaneous equations lets you solve real-world problems like finding the exact point where a drug’s concentration peaks in the bloodstream—or where two moving objects collide in physics. On your exam, this topic is worth 8-12 marks—enough to boost your grade by a full level. Let’s break it down so you never lose marks again."

WHAT YOU NEED TO KNOW FIRST

Before starting, you must understand: 1. Solving linear equations (e.g., 2x + 3y = 6). 2. Solving quadratic equations (e.g., y = x² + 3x – 4). 3. Substitution method (replacing one variable with an expression).

If you’re shaky on these, pause and review them first.

KEY TERMS & FORMULAS

Key Terms

• Simultaneous equations: Two (or more) equations with the same variables that must be true at the same time.
• Linear equation: An equation where the highest power of x or y is 1 (e.g., y = 2x + 1).
• Quadratic equation: An equation where the highest power of x or y is 2 (e.g., y = x² + 3x – 4).
• Substitution: Replacing one variable with an expression from another equation.
• Intersection points: The (x, y) solutions where both equations meet.

Formulas

1. Linear equation (standard form):
   ax + by = c
2. a, b, c = constants
3. x, y = variables

4. Given on exam sheet (but memorise for speed).

5. Quadratic equation (standard form):
   y = ax² + bx + c

6. a, b, c = constants
7. x, y = variables

8. MEMORISE THIS (not always given).

9. Substitution method:

10. Step 1: Solve the linear equation for one variable (e.g., y = 2x + 1).
11. Step 2: Substitute this into the quadratic equation.
12. Step 3: Solve the resulting quadratic for x.
13. Step 4: Substitute x back into the linear equation to find y.

STEP-BY-STEP METHOD

Problem type: One linear equation + one quadratic equation. Goal: Find all (x, y) pairs that satisfy both equations.

Steps (Always Follow This Order)

1. Rearrange the linear equation to express y (or x) in terms of the other variable.

2. Example: If 2x + y = 5, rearrange to y = 5 – 2x.

3. Substitute this expression into the quadratic equation.

4. Replace y in the quadratic with the expression from Step 1.

5. Expand and simplify the new equation to form a quadratic in one variable (usually x).

6. Combine like terms and write in standard form: ax² + bx + c = 0.

7. Solve the quadratic equation for x using:

8. Factorising (if possible),
9. Quadratic formula: x = [–b ± √(b² – 4ac)] / (2a),

10. Or completing the square (rare for this topic).

11. Find y for each x by substituting back into the linear equation (not the quadratic!).

12. This avoids extra solutions that don’t satisfy both equations.

13. Write the solutions as coordinate pairs (x, y).

14. Check: Plug both pairs back into both original equations to verify.

15. State the number of solutions:

16. 2 solutions → curves intersect at two points.
17. 1 solution → curves touch at one point (tangent).
18. 0 solutions → curves never meet.

WORKED EXAMPLES

Example 1 – Basic (No Tricks)

Equations: 1. y = x + 3 (linear) 2. y = x² – 4x + 3 (quadratic)

Step 1: Linear equation is already solved for y. y = x + 3

Step 2: Substitute y = x + 3 into the quadratic: x + 3 = x² – 4x + 3

Step 3: Rearrange to standard quadratic form: x² – 4x + 3 – x – 3 = 0 x² – 5x = 0

Step 4: Factorise: x(x – 5) = 0 Solutions: x = 0 or x = 5

Step 5: Find y for each x using y = x + 3: - If x = 0, y = 0 + 3 = 3 - If x = 5, y = 5 + 3 = 8

Step 6: Solutions: (0, 3) and (5, 8)

Step 7: Check in both equations: - For (0, 3): - Linear: 3 = 0 + 3 ✔️ - Quadratic: 3 = 0 – 0 + 3 ✔️ - For (5, 8): - Linear: 8 = 5 + 3 ✔️ - Quadratic: 8 = 25 – 20 + 3 ✔️

What we did and why: We substituted the linear equation into the quadratic to eliminate y, solved for x, then found y using the linear equation. This ensures both equations are satisfied.

Example 2 – Medium (Added Complication)

Equations: 1. 2x + y = 4 (linear) 2. y = x² – 2x – 3 (quadratic)

Step 1: Rearrange linear equation for y: y = 4 – 2x

Step 2: Substitute into quadratic: 4 – 2x = x² – 2x – 3

Step 3: Rearrange to standard form: x² – 2x – 3 – 4 + 2x = 0 x² – 7 = 0

Step 4: Solve for x: x² = 7 x = ±√7

Step 5: Find y for each x using y = 4 – 2x: - If x = √7, y = 4 – 2√7 - If x = –√7, y = 4 – 2(–√7) = 4 + 2√7

Step 6: Solutions: (√7, 4 – 2√7) and (–√7, 4 + 2√7)

Step 7: Check (optional for exams, but good practice): - For (√7, 4 – 2√7): - Linear: 2√7 + (4 – 2√7) = 4 ✔️ - Quadratic: 4 – 2√7 = 7 – 2√7 – 3 ✔️

What we did and why: The linear equation had a coefficient of 2 for x, so we rearranged carefully. The quadratic simplified to x² = 7, giving irrational solutions. Always keep exact values (√7) unless the question asks for decimals.

Example 3 – Exam-Style (Disguised, Time Pressure)

Question: A particle moves along a path described by y = x² – 5x + 6. At the same time, a second particle moves along the line 3x – y = 2. Find the coordinates where the two particles meet.

Step 1: Rearrange the linear equation for y: 3x – y = 2 → y = 3x – 2

Step 2: Substitute into the quadratic: 3x – 2 = x² – 5x + 6

Step 3: Rearrange to standard form: x² – 5x + 6 – 3x + 2 = 0 x² – 8x + 8 = 0

Step 4: Solve using the quadratic formula (can’t factorise easily): a = 1, b = –8, c = 8 x = [8 ± √(64 – 32)] / 2 x = [8 ± √32] / 2 √32 = 4√2, so: x = [8 ± 4√2] / 2 = 4 ± 2√2

Step 5: Find y for each x using y = 3x – 2: - If x = 4 + 2√2, y = 3(4 + 2√2) – 2 = 12 + 6√2 – 2 = 10 + 6√2 - If x = 4 – 2√2, y = 3(4 – 2√2) – 2 = 12 – 6√2 – 2 = 10 – 6√2

Step 6: Solutions: (4 + 2√2, 10 + 6√2) and (4 – 2√2, 10 – 6√2)

What we did and why: The question was disguised as a physics problem, but it’s just simultaneous equations. We used the quadratic formula because factorising wasn’t obvious. Always simplify √ terms (e.g., √32 = 4√2).

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before the Exam)

"Listen up—this is your 60-second crash course for simultaneous equations with a quadratic. Here’s what you must remember:

1. Rearrange the linear equation to get y = or x =.
2. Substitute this into the quadratic. Now you’ve got one equation with one variable.
3. Solve the quadratic—factorise if you can, otherwise use the quadratic formula. You’ll get two x values (usually).
4. Find y by plugging x back into the linear equation (not the quadratic!).
5. Write both solutions as (x, y) pairs. Check them if you have time.
6. Watch for traps: Non-integer answers? Use the quadratic formula. Disguised question? Write the equations first.

That’s it. You’ve got this. Now go smash that exam!"