🎯 Systems Linear & Nonlinear — Line-Circle, Parabola-Line

Essential for geometric problems and advanced equation solving in AMC contests.

🎯 Recognition Cues

• “Find all solutions” — Systems with multiple equations
• “Intersection points” — Geometric problems involving curves
• Line and circle — $x^2 + y^2 = r^2$ with $ax + by = c$
• Parabola and line — $y = ax^2 + bx + c$ with $y = mx + b$

📚 Template Solutions

Linear Systems (2×2)

Nonlinear Systems

🎯 Worked Examples

Example 1: Line-Circle System

Problem: Find all real solutions to $\begin{cases} x^2 + y^2 = 25 \ x + y = 7 \end{cases}$

Solution:

1. From second equation: $y = 7 - x$
2. Substitute into first: $x^2 + (7-x)^2 = 25$
3. Expand: $x^2 + 49 - 14x + x^2 = 25$
4. Simplify: $2x^2 - 14x + 24 = 0$ → $x^2 - 7x + 12 = 0$
5. Factor: $(x-3)(x-4) = 0$ → $x = 3$ or $x = 4$
6. Find $y$: When $x = 3$: $y = 4$; when $x = 4$: $y = 3$
7. Check: $(3,4)$: $3^2 + 4^2 = 25$ and $3 + 4 = 7$ ✓; $(4,3)$: $4^2 + 3^2 = 25$ and $4 + 3 = 7$ ✓
8. Answer: $(3, 4)$ and $(4, 3)$

Example 2: Parabola-Line System

Problem: Find all real solutions to $\begin{cases} y = x^2 - 2x + 1 \ y = 2x - 3 \end{cases}$

Solution:

1. Substitute: $x^2 - 2x + 1 = 2x - 3$
2. Rearrange: $x^2 - 4x + 4 = 0$
3. Factor: $(x-2)^2 = 0$ → $x = 2$
4. Find $y$: $y = 2(2) - 3 = 1$
5. Check: $(2,1)$: $1 = 2^2 - 2(2) + 1 = 1$ and $1 = 2(2) - 3 = 1$ ✓
6. Answer: $(2, 1)$

Example 3: Two Circles

Problem: Find all real solutions to $\begin{cases} x^2 + y^2 = 25 \ x^2 + y^2 = 9 \end{cases}$

Solution:

1. Subtract equations: $(x^2 + y^2) - (x^2 + y^2) = 25 - 9$ → $0 = 16$
2. This is impossible: No real solutions
3. Answer: No real solutions

⚠️ Common Pitfalls

Pitfall: Forgetting to check all solutions

• Fix: Always substitute solutions back into original equations
• Example: $(3, 4)$ and $(4, 3)$ both satisfy $x^2 + y^2 = 25$ and $x + y = 7$

Pitfall: Incorrect substitution

• Fix: Be careful with variable replacement
• Example: In $y = 2x + 1$ and $3x + 2y = 7$, substitute $y = 2x + 1$ into second equation

Pitfall: Missing solutions

• Fix: Consider all possible cases
• Example: $x^2 = 4$ has solutions $x = \pm 2$, not just $x = 2$

🎯 AMC-Style Worked Example

Problem: Find all real solutions to the system: $$\begin{cases} x^2 + y^2 = 13 \ xy = 6 \end{cases}$$

Solution:

1. Strategy: Use substitution since both equations are symmetric
2. From second equation: $y = \frac{6}{x}$ (assuming $x \neq 0$)
3. Substitute: $x^2 + \left(\frac{6}{x}\right)^2 = 13$
4. Simplify: $x^2 + \frac{36}{x^2} = 13$
5. Multiply by $x^2$: $x^4 + 36 = 13x^2$
6. Rearrange: $x^4 - 13x^2 + 36 = 0$
7. Substitute $u = x^2$: $u^2 - 13u + 36 = 0$
8. Factor: $(u-4)(u-9) = 0$ → $u = 4$ or $u = 9$
9. Solve: $x^2 = 4$ gives $x = \pm 2$; $x^2 = 9$ gives $x = \pm 3$
10. Find $y$: When $x = 2$, $y = 3$; when $x = -2$, $y = -3$; when $x = 3$, $y = 2$; when $x = -3$, $y = -2$
11. Check all solutions:
    • $(2, 3)$: $2^2 + 3^2 = 13$ and $2 \cdot 3 = 6$ ✓
    • $(-2, -3)$: $(-2)^2 + (-3)^2 = 13$ and $(-2)(-3) = 6$ ✓
    • $(3, 2)$: $3^2 + 2^2 = 13$ and $3 \cdot 2 = 6$ ✓
    • $(-3, -2)$: $(-3)^2 + (-2)^2 = 13$ and $(-3)(-2) = 6$ ✓
12. Answer: $(2, 3)$, $(-2, -3)$, $(3, 2)$, $(-3, -2)$

Key insight: Symmetric systems often have symmetric solutions.

🔗 Related Patterns

• Geometric Problems — Systems often represent geometric intersections
• Substitution — Essential technique for nonlinear systems
• Symmetry — Look for symmetric solutions
• Verification — Always check solutions in original equations

📝 Practice Checklist

• Master linear system solving
• Practice nonlinear system techniques
• Learn geometric applications
• Practice substitution methods
• Understand symmetry patterns
• Work on speed and accuracy

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