πŸ“Š Equations & Inequalities

Mastering equation solving and inequality techniques is crucial for AMC success. This includes absolute value, systems, and advanced inequality methods.

🎯 Key Ideas

Absolute Value: $|x| = \begin{cases} x & \text{if } x \geq 0 \ -x & \text{if } x < 0 \end{cases}$

Systems: Multiple equations with multiple variables. Common methods: substitution, elimination, graphing.

Inequalities: AM-GM inequality and other techniques for optimization problems.

πŸ“ Absolute Value Equations

Basic Strategy

  1. Isolate the absolute value expression
  2. Split into two cases: positive and negative
  3. Solve each case separately
  4. Check solutions in original equation

Example: $|2x - 3| = 7$

  1. Already isolated: $|2x - 3| = 7$
  2. Split: $2x - 3 = 7$ or $2x - 3 = -7$
  3. Solve: $2x = 10$ or $2x = -4$
  4. Solutions: $x = 5$ or $x = -2$
  5. Check: $|2(5) - 3| = |7| = 7$ βœ“ and $|2(-2) - 3| = |-7| = 7$ βœ“

Pitfall: Extraneous Solutions

Always check solutions! Some may not work in the original equation.

πŸ”„ Systems of Equations

Substitution Method

  1. Solve one equation for one variable
  2. Substitute into other equation(s)
  3. Solve for remaining variable(s)
  4. Back-substitute to find all variables

Elimination Method

  1. Multiply equations to get same coefficients
  2. Add/subtract to eliminate one variable
  3. Solve for remaining variable(s)
  4. Back-substitute

Example: Solve $\begin{cases} x + y = 5 \ 2x - y = 1 \end{cases}$

By Elimination:

  1. Add equations: $(x + y) + (2x - y) = 5 + 1$
  2. Simplify: $3x = 6$, so $x = 2$
  3. Substitute: $2 + y = 5$, so $y = 3$
  4. Solution: $(2, 3)$

βš–οΈ AM-GM Inequality

Statement: For positive real numbers $a_1, a_2, \ldots, a_n$: $$\frac{a_1 + a_2 + \cdots + a_n}{n} \geq \sqrt[n]{a_1 a_2 \cdots a_n}$$

Equality: Holds when all numbers are equal.

Applications

  • Optimization: Find maximum/minimum values
  • Proofs: Establish bounds and relationships
  • Problem Solving: Recognize when AM-GM applies

Example: Find minimum of $x + \frac{1}{x}$ for $x > 0$

Solution:

  1. Apply AM-GM to $x$ and $\frac{1}{x}$: $$\frac{x + \frac{1}{x}}{2} \geq \sqrt{x \cdot \frac{1}{x}} = \sqrt{1} = 1$$
  2. Multiply by 2: $x + \frac{1}{x} \geq 2$
  3. Equality when $x = \frac{1}{x}$, so $x = 1$
  4. Minimum value is 2, achieved when $x = 1$

🎯 AMC-Style Worked Example

Problem: Find all real solutions to $|x^2 - 4| = 3x$.

Solution:

  1. Case 1: $x^2 - 4 \geq 0$ (i.e., $x \leq -2$ or $x \geq 2$)

    • Then $|x^2 - 4| = x^2 - 4$
    • Equation: $x^2 - 4 = 3x$
    • Rearrange: $x^2 - 3x - 4 = 0$
    • Factor: $(x-4)(x+1) = 0$
    • Solutions: $x = 4$ or $x = -1$
    • Check constraints: $x = 4$ works (since $4 \geq 2$), but $x = -1$ doesn’t (since $-1$ is not $\leq -2$ or $\geq 2$)
  2. Case 2: $x^2 - 4 < 0$ (i.e., $-2 < x < 2$)

    • Then $|x^2 - 4| = -(x^2 - 4) = 4 - x^2$
    • Equation: $4 - x^2 = 3x$
    • Rearrange: $x^2 + 3x - 4 = 0$
    • Factor: $(x+4)(x-1) = 0$
    • Solutions: $x = -4$ or $x = 1$
    • Check constraints: $x = 1$ works (since $-2 < 1 < 2$), but $x = -4$ doesn’t (since $-4$ is not in $(-2, 2)$)
  3. Final Answer: $x = 4$ and $x = 1$

πŸ” Common Traps & Fixes

Trap: Forgetting to check case constraints

Fix: Always verify that solutions satisfy the original case conditions.

Trap: Extraneous solutions from squaring

Fix: Check all solutions in the original equation.

Trap: Wrong AM-GM application

Fix: Ensure all numbers are positive before applying AM-GM.

Trap: Missing solutions in systems

Fix: Check all possible combinations and don’t assume unique solutions.

πŸ“‹ Quick Reference

Absolute Value Properties

  • $|a| \geq 0$ (always non-negative)
  • $|a| = |-a|$ (symmetric)
  • $|ab| = |a||b|$ (multiplicative)
  • $|a + b| \leq |a| + |b|$ (triangle inequality)

AM-GM Special Cases

  • Two variables: $\frac{a + b}{2} \geq \sqrt{ab}$
  • Three variables: $\frac{a + b + c}{3} \geq \sqrt[3]{abc}$
  • Equality when all variables are equal

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