How to Solve: Simultaneous Equations (Linear & Quadratic)

How to Solve: Simultaneous Equations (Linear & Quadratic)

For GCSE & A-Level Maths – Ace Your Exam!

Introduction

"Mastering simultaneous equations unlocks 8–12 marks on your GCSE or A-Level exam—enough to boost your grade by a full level. Whether it’s calculating profit in business, designing bridges in engineering, or even cracking codes, this skill is your ticket to top marks and real-world problem-solving."

What You Need To Know First

Before diving in, ensure you’re confident with: 1. Solving linear equations (e.g., 2x + 3y = 6). 2. Solving quadratic equations (e.g., x² + 5x + 6 = 0). 3. Substitution method (replacing one variable with an expression).

If any of these feel shaky, pause and review them first—this guide builds on them!

Key Vocabulary

Formulas To Know

Step-by-Step Method

Problem Type: One linear equation + one quadratic equation (e.g., y = 2x + 1 and y = x² + 3x – 2).

Step 1: Identify the Linear Equation

• Look for the equation with no squared terms (e.g., y = 3x + 2 or 2x + y = 5).
• Goal: Solve the linear equation for y (or x if easier).

Step 2: Substitute into the Quadratic

• Take the expression for y (or x) from Step 1 and plug it into the quadratic equation.
• Example: If y = 2x + 1 and the quadratic is y = x² + 3, substitute 2x + 1 for y: 2x + 1 = x² + 3

Step 3: Rearrange into Standard Quadratic Form

• Move all terms to one side to set the equation to zero: x² + 3 – 2x – 1 = 0 → x² – 2x + 2 = 0
• Check: The equation should look like ax² + bx + c = 0.

Step 4: Solve the Quadratic

• Option 1: Factorise (if possible). x² – 5x + 6 = 0 → (x – 2)(x – 3) = 0 → x = 2 or x = 3.
• Option 2: Use the quadratic formula (if factorising is tricky). x = [-b ± √(b² – 4ac)] / (2a)
• Option 3: Complete the square (less common, but examiners love it!).

Step 5: Find the Matching y Values

• Take each x solution and plug it back into the linear equation to find y.
• Example: If x = 2 and the linear equation is y = 2x + 1, then y = 2(2) + 1 = 5.
• Write the solutions as coordinate pairs: (2, 5) and (3, 7).

Step 6: Check Your Solutions

• Plug both pairs into the quadratic equation to ensure they work.
• Example: For (2, 5), check y = x² + 3 → 5 = (2)² + 3 → 5 = 7? No! (This means you made a mistake—go back!)

Worked Examples

Example 1 – Basic (No Tricks)

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

Step 1: Linear equation is already solved for y: y = x + 2. Step 2: Substitute into quadratic: x + 2 = x² + 3x – 4 Step 3: Rearrange: x² + 3x – 4 – x – 2 = 0 → x² + 2x – 6 = 0 Step 4: Solve using quadratic formula (a=1, b=2, c=-6): x = [-2 ± √(4 + 24)] / 2 → x = [-2 ± √28] / 2 → x = -1 ± √7 Step 5: Find y for each x: - For x = -1 + √7: y = (-1 + √7) + 2 = 1 + √7 - For x = -1 – √7: y = (-1 – √7) + 2 = 1 – √7 Step 6: Check one solution (e.g., x = -1 + √7, y = 1 + √7): (1 + √7) = (-1 + √7)² + 3(-1 + √7) – 4 (1 + √7) = (1 – 2√7 + 7) – 3 + 3√7 – 4 (1 + √7) = (8 – 2√7) – 7 + 3√7 (1 + √7) = 1 + √7 ✅

Solutions: (-1 + √7, 1 + √7) and (-1 – √7, 1 – √7)

What we did and why: - We used substitution because one equation was already solved for y. - The quadratic didn’t factorise, so we used the quadratic formula. - Always check your answers—it’s easy to make sign errors!

Example 2 – Medium (Hidden Linear Equation)

Equations: 1. 2x + y = 5 (linear) 2. y = x² – 4x + 3 (quadratic)

Step 1: Solve linear equation for y: y = 5 – 2x Step 2: Substitute into quadratic: 5 – 2x = x² – 4x + 3 Step 3: Rearrange: x² – 4x + 3 – 5 + 2x = 0 → x² – 2x – 2 = 0 Step 4: Solve using quadratic formula (a=1, b=-2, c=-2): x = [2 ± √(4 + 8)] / 2 → x = [2 ± √12] / 2 → x = 1 ± √3 Step 5: Find y for each x: - For x = 1 + √3: y = 5 – 2(1 + √3) = 3 – 2√3 - For x = 1 – √3: y = 5 – 2(1 – √3) = 3 + 2√3 Step 6: Check one solution (e.g., x = 1 + √3, y = 3 – 2√3): (3 – 2√3) = (1 + √3)² – 4(1 + √3) + 3 (3 – 2√3) = (1 + 2√3 + 3) – 4 – 4√3 + 3 (3 – 2√3) = (4 + 2√3) – 1 – 4√3 (3 – 2√3) = 3 – 2√3 ✅

Solutions: (1 + √3, 3 – 2√3) and (1 – √3, 3 + 2√3)

What we did and why: - The linear equation wasn’t already solved for y, so we rearranged it first. - We simplified √12 to 2√3 (always simplify surds!). - Checking one solution is enough—if it works, the other likely does too.

Example 3 – Exam-Style (Disguised Problem)

Question: A rectangle has length l and width w. Its perimeter is 24 cm, and its area is 32 cm². Find the possible dimensions of the rectangle.

Step 1: Translate the problem into equations. - Perimeter: 2l + 2w = 24 → l + w = 12 (linear) - Area: l × w = 32 (quadratic)

Step 2: Solve linear equation for l: l = 12 – w Step 3: Substitute into quadratic: (12 – w) × w = 32 → 12w – w² = 32 Step 4: Rearrange: w² – 12w + 32 = 0 Step 5: Solve by factorising: (w – 4)(w – 8) = 0 → w = 4 or w = 8 Step 6: Find l for each w: - If w = 4, l = 12 – 4 = 8 - If w = 8, l = 12 – 8 = 4 Step 7: Check: - Perimeter: 2(8) + 2(4) = 24 ✅ - Area: 8 × 4 = 32 ✅

Solutions: The rectangle is 8 cm × 4 cm (order doesn’t matter).

What we did and why: - We translated the word problem into equations first. - The quadratic factorised easily, so we didn’t need the formula. - Always state the final answer in context (here, as dimensions).

Common Mistakes

Exam Traps

1-Minute Recap (Night Before the Exam)

"Listen up—this is your 60-second survival guide to simultaneous equations with a quadratic. First, spot the linear equation—it’ll have no squares. Solve it for y (or x). Then substitute that expression into the quadratic. Rearrange to ax² + bx + c = 0, then solve the quadratic—factorise if you can, otherwise use the formula. Don’t forget the ±! Once you’ve got x, find y using the linear equation. Finally, check your answers by plugging them back in. If the discriminant is negative, write ‘no real solutions.’ That’s it—go smash those 12 marks!