Simultaneous Equations

Simultaneous Equations

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Simultaneous equations are two (or more) equations in the same variables, solved together to find values satisfying all of them at once. The standard form is a pair of linear equations: ax + by = c and dx + ey = f, where a, b, c, d, e, f are known constants and x, y are the unknowns.

• Three solution cases: a unique solution when a/d ≠ b/e; no solution (inconsistent) when a/d = b/e ≠ c/f; infinitely many solutions (dependent) when a/d = b/e = c/f.
• Cramer’s rule gives x = (ce − bf)/(ae − bd) and y = (af − cd)/(ae − bd) directly, provided ae − bd ≠ 0.
• NECO SSCE tests 1–3 questions: direct solving by substitution or elimination, graphical reading of intersection, and word problems on ages, costs, or rates.

---

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

What a System of Two Linear Equations Represents

Geometrically, each equation ax + by = c is a straight line. Solving two equations simultaneously means finding the point of intersection of those two lines. If the lines cross at exactly one point, the system is consistent and independent; if they never meet, the system is inconsistent; if they lie on top of each other, the system is dependent.

The Three Solution Cases

Condition	Meaning	Solution
a/d ≠ b/e	Lines cross at one point	Unique (x, y)
a/d = b/e ≠ c/f	Parallel, distinct lines	None (inconsistent)
a/d = b/e = c/f	Same line	Infinitely many (dependent)

Substitution Method

Solve one equation for one variable (e.g., make x the subject: x = (c − by)/a) and substitute into the second equation. This produces a single-variable equation you can solve directly, then back-substitute to get the other variable.

Elimination Method

Multiply one or both equations by suitable numbers so the coefficients of one variable match. Then add the equations to eliminate that variable (signs opposite) or subtract them (signs the same). Solve for the remaining variable, then back-substitute.

Cramer’s Rule

Write the system in matrix form and compute determinants:

• D = ae − bd
• Dx = ce − bf
• Dy = af − cd Then x = Dx/D and y = Dy/D. This fails when D = 0 (no unique solution).

NECO SSCE Question Patterns

Paper 1 (Objective) typically offers 1–2 multiple-choice items on determining the number of solutions, identifying the determinant, or computing x and y. Paper 2 (Essay) includes a 5–7 mark problem asking for a full solution by substitution or elimination, plus a 7–10 mark word problem translating sentences like “the sum of two numbers is…” or “A is twice as old as B, in 5 years…” into a solvable system.

---

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

When Methods Are Equivalent (and When They Differ)

Substitution, elimination, and Cramer’s rule all yield the same answer on a well-posed system, but they differ in efficiency. Substitution shines when one variable already has a coefficient of 1 or −1. Elimination is faster when both variables have large, awkward coefficients. Cramer’s rule is best for compact two-by-two systems where determinants can be written out in one line; for three-by-three systems (NECO Extension/electives) the determinant expansion becomes expensive and elimination is preferred.

Graphical Method in Detail

Plot each line by finding two points (the intercepts are convenient: set x = 0 to get the y-intercept, y = 0 for the x-intercept). Draw on graph paper using a 2 cm × 2 cm scale, read the point of intersection to the nearest unit, and state (x, y). In NECO, graphical solutions often come with a 1 cm × 1 cm grid; misreading coordinates by half a square is a common loss of a mark.

Worked Example

Solve 2x + 3y = 12 and 4x − y = 5. By elimination: multiply the second equation by 3 → 12x − 3y = 15. Add to the first: 14x = 27, so x = 27/14. Substitute back: y = 4x − 5 = 108/14 − 70/14 = 38/14 = 19/7. Solution: (27/14, 19/7). Check: 2(27/14) + 3(19/7) = 54/14 + 114/14 = 168/14 = 12 ✓ and 4(27/14) − 19/7 = 108/14 − 38/14 = 70/14 = 5 ✓.

Common Traps

• Sign errors when subtracting equations — flipping a sign silently negates the answer.
• Dividing by zero in Cramer’s rule: if ae − bd = 0, stop and classify the system instead of forcing a calculation.
• Word-problem translation errors: “twice as old” means 2 × (age), not 2 + (age); “sum is” means +; “exceeds” means −. Always define variables clearly first.

Connections

Simultaneous equations underpin linear programming (constraints), electrical circuit analysis (Kirchhoff’s laws), and economics (supply-demand equilibrium). Strong handling here carries directly into WAEC, JAMB, and beyond.

Practice Prompts

1. Solve the system 3x + 5y = 11 and 2x − y = 4 using (a) substitution and (b) Cramer’s rule. Confirm the answers match.
2. A man is three times as old as his son. In 12 years, he will be twice as old. Set up two equations in their current ages and solve.

---

Content adapted based on your selected roadmap duration. Switch tiers using the selector above.