Linear Systems and Inequalities
The exam tests solving two-equation systems, recognizing when a system has no solution or infinitely many, and solving and graphing linear, compound, and absolute-value inequalities.
Solving 2 × 2 systems
- Substitution: solve one equation for a variable, substitute into the other, solve, then back-substitute.
- Elimination: multiply one or both equations so a variable's coefficients are opposites, add to cancel, then solve.
- Graphing: the solution is the intersection point.
- Three outcomes: one solution (lines intersect, different slopes); no solution (parallel lines, same slope, different intercepts); infinitely many (the same line written two ways).
Inequalities
- Solve like equations, except multiplying or dividing both sides by a negative number reverses the inequality symbol.
- On a number line: open circle for < or >, closed circle for ≤ or ≥. Interval notation uses parentheses for excluded endpoints and brackets for included ones, e.g. (2, 5].
- Compound statements: "and" means the overlap; "or" means the union.
- For a system of inequalities, graph each boundary (solid for ≤ or ≥, dashed for < or >), shade each half-plane, and keep the overlap - the feasible region. Testing (0, 0) is a quick check when it is not on a boundary.
- Absolute value: |x| < 3 gives −3 < x < 3; |x| > 3 gives x < −3 or x > 3.
Worked example
Solve 3x + 2y = 16 and x − y = 2.
- From the second equation, x = y + 2.
- Substitute: 3(y + 2) + 2y = 16 → 3y + 6 + 2y = 16 → 5y = 10 → y = 2.
- Then x = 2 + 2 = 4.
- Check: 3(4) + 2(2) = 16 ✓ and 4 − 2 = 2 ✓.
Inequality: −2x + 5 ≥ 11 → −2x ≥ 6 → divide by −2 and flip → x ≤ −3. Check x = −4: 8 + 5 = 13 ≥ 11 ✓.
Traps that cost marks
- Forgetting to flip the inequality sign when dividing by a negative.
- Sign errors when subtracting one equation from another during elimination.
- Reporting "no solution" when the lines are actually the same line.
- Using the wrong bracket or circle type for an endpoint.
- Shading the wrong side without testing a point.
- Accepting a pair that satisfies only one of the original equations.