Chapter 2 — Equations and Inequalities

Part 1: Linear Equations

1.1 Solving Linear Equations

The goal is to isolate the variable. Apply the same operation to both sides.

Example: Solve $4x - 5 = 5$
$4x = 10$
$x = 2.5$

Example with fractions: Solve $\displaystyle \frac{2x+3}{5} = x - 1$
Multiply both sides by 5: $2x + 3 = 5(x - 1)$
$2x + 3 = 5x - 5$
$8 = 3x$
$\displaystyle x = \frac{8}{3}$

1.2 Equations with the Variable on Both Sides

Collect variable terms on one side, constants on the other.

Example: Solve $3(x + 2) = 5x - 8$
$3x + 6 = 5x - 8$
$14 = 2x$
$x = 7$

Part 2: Simultaneous Equations (Systems of Equations)

Graphically, solving a system of equations means finding where the lines (or curves) intersect.

2.1 Elimination Method

Make the coefficients of one variable equal, then add or subtract.

Example: Solve ${ 7A + 4C = 46; 8A + 6C = 57 }$
Multiply first by 3: $21A + 12C = 138$
Multiply second by 2: $16A + 12C = 114$
Subtract: $5A = 24 \rightarrow A = 4.8$
Substitute: $7(4.8) + 4C = 46 \rightarrow 33.6 + 4C = 46 \rightarrow C = 3.1$

2.2 Substitution Method

Express one variable in terms of the other, then substitute.

Example: Solve ${ 5x = 2 - y; x = 13 + 4y }$
Substitute x = 13 + 4y into first equation:
$5(13 + 4y) = 2 - y$
$65 + 20y = 2 - y$
$21y = -63$
$y = -3, x = 13 + 4(-3) = 1$

2.3 Non-linear Simultaneous Equations

One equation is quadratic, one is linear. Use substitution.

Example (ISMA Prep Q19): Solve ${ 3x^{2} + y^{2} - xy = 5; y = 2x - 3 }$
Substitute y = 2x − 3:
$3x^{2} + (2x-3)^{2} - x(2x-3) = 5$
$3x^{2} + 4x^{2} - 12x + 9 - 2x^{2} + 3x = 5$
$5x^{2} - 9x + 9 = 5$
$5x^{2} - 9x + 4 = 0$
Using the quadratic formula or factorising: $(5x - 4)(x - 1) = 0$
x = 4/5 or x = 1
When x = 4/5: y = 2(4/5) − 3 = −7/5
When x = 1: y = 2(1) − 3 = −1

Part 3: Quadratic Equations

3.1 Solution by Factorisation

If ab = 0, then a = 0 or b = 0.

Example: Solve $x^{2} - 2x - 35 = 0$
$(x + 5)(x - 7) = 0$
x = −5 or x = 7

Example: Solve $2x^{2} - 3x = 2$
$2x^{2} - 3x - 2 = 0$
$(2x + 1)(x - 2) = 0$
x = −1/2 or x = 2

3.2 The Quadratic Formula

$For ax^{2} + bx + c = 0:$
$\displaystyle x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$

The discriminant Δ = b² − 4ac tells us:

• Δ > 0: two distinct real solutions
• Δ = 0: one repeated (equal) root
• Δ < 0: no real solutions

Example: Solve $3x^{2} - 5x + 1 = 0$
$a = 3, b = -5, c = 1$
$\Delta = 25 - 12 = 13$
$\displaystyle x = \frac{5 \pm \sqrt{13}}{6}$

3.3 Completing the Square

Write ax² + bx + c in the form a(x + p)² + q.

$\text{Steps for } x^{2} + bx + c:$
1. Take half the coefficient of x: b/2
$\displaystyle 2. \text{Square it:}\ \left(\frac{b}{2}\right)^{2}$
$\displaystyle 3. \text{Write:}\ (x + \frac{b}{2})^{2} - \left(\frac{b}{2}\right)^{2} + c$

Example (ISMA Prep Q20a): Express 7 + 12x − 3x² in the form a + b(x + c)²
$= -3(x^{2} - 4x) + 7$
$= -3(x^{2} - 4x + 4 - 4) + 7$
$= -3(x - 2)^{2} + 12 + 7$
$= 19 - 3(x - 2)^{2}$

$So a = 19, b = -3, c = -2.$
The maximum value is 19 (when x = 2), giving point A = (2, 19).

3.4 Forming Quadratic Equations from Word Problems

Example (IB 2004, Q4): Richard's age squared equals his mother's age. When he is twice his current age, she will be 7/2 times his age then.
$\text{Let Richard} = r, Mother = r^{2}$
When he is 2r: she is r² + r (same years pass)
$\displaystyle r^{2} + r = \frac{7}{2}(2r)$
$r^{2} + r = 7r$
$r^{2} - 6r = 0$
$r(r - 6) = 0$
$r = 6 (reject r = 0)$
Richard is 6, Mother is 36.

Part 4: Inequalities

4.1 Linear Inequalities

Solve like equations, but reverse the inequality sign when multiplying/dividing by a negative number.

Example: Solve $5x - 7 \leq 2$
$5x \leq 9$
$\displaystyle x \leq \frac{9}{5}$

Example: Solve $\displaystyle \frac{2x}{1 - x} \leq 3$
Careful! We cannot simply multiply by (1 − x) because its sign depends on x.

Method: Rearrange to have 0 on one side:
$\displaystyle \frac{2x}{1-x} - 3 \leq 0$
$\displaystyle \frac{2x - 3(1-x)}{1-x} \leq 0$
$\displaystyle \frac{2x - 3 + 3x}{1-x} \leq 0$
$\displaystyle \frac{5x - 3}{1-x} \leq 0$

Critical values: x = 3/5 and x = 1
Test intervals:

• $\displaystyle x < \frac{3}{5}: \frac{-}{+} = negative \leq 0 \checkmark $
• $\displaystyle \frac{3}{5} < x < 1: \frac{+}{+} = positive ✗$
• $\displaystyle x > 1: \frac{+}{-} = negative \leq 0 \checkmark $

Answer: x ≤ 3/5 or x > 1

4.2 Quadratic Inequalities

Example: Solve $x^{2} - 5x + 6 > 0$
Factorise: $(x - 2)(x - 3) > 0$
The parabola opens upwards.
Solution: x < 2 or x > 3

Key method: Factorise, find roots, sketch the parabola, read off where it satisfies the inequality.

Part 5: Problem Solving with Equations

5.1 Speed, Distance, Time

distance = speed × time

Example (IB 2010): A fast train leaves Manchester for London (330 km) at noon. A slow train at half the speed leaves London for Manchester at the same time. They meet at 2pm.
Let fast train speed = v km/h. Slow train = v/2.
In 2 hours: fast covers 2v, slow covers 2(v/2) = v
Together: $2v + v = 330$
$3v = 330, v = 110 km/h$
Fast: $110 km/h, Slow: 55 km/h$

5.2 Mixture and Cost Problems

Example: A basketball team plays 6 games. After 5 games, mean score = 21 points. After 6 games, mean score = 23. Points in 6th game?
Total after 5 games: $5 \times 21 = 105$
Total after 6 games: $6 \times 23 = 138$
$6th game: 138 - 105 = 33 points$

5.3 Ratio Problems

Example (ISMA Prep Q2): Danil, Gabriel and Hadley share money in ratio 3:5:9. The difference between Gabriel and Hadley is 196 euros. Find Danil's share.
Difference = 9 − 5 = 4 parts = 196 euros
$1 part = 49 euros$
Danil = 3 parts = 147 euros

Exam Traps / Ловушки на экзамене

❌ Trap 1: Forgetting to flip the inequality sign
$-3x > 12 \rightarrow x > -4 — WRONG!$
✅ −3x > 12 → x < −4 — REVERSE the sign when dividing by a negative.
Ловушка: при делении/умножении неравенства на отрицательное число знак ПЕРЕВОРАЧИВАЕТСЯ. Это самая частая ошибка!

❌ Trap 2: Dividing both sides by a variable
$x^{2} = 5x \rightarrow x = 5 — WRONG! Lost x = 0!$
✅ x² − 5x = 0 → x(x − 5) = 0 → x = 0 or x = 5. Move to one side and factor.
Ловушка: НИКОГДА не дели на переменную — потеряешь решение. Перенеси на одну сторону и разложи.

❌ Trap 3: Sign errors in the quadratic formula
$For x^{2} - 5x + 3 = 0: b = 5 \rightarrow x = (-5 \pm √...)/2 — WRONG!$
✅ Here b = −5, so −b = +5. x = (5 ± √(25−12))/2. Always write −b carefully.
Ловушка: в формуле x = (−b ± √Δ)/2a аккуратно подставляй знак b. Если b = −5, то −b = +5.

❌ Trap 4: Substituting back into the wrong equation
Found x from one equation, substituting back into the SAME equation to find y.
✅ Substitute into the OTHER (simpler) equation — catches errors and is usually easier.
Ловушка: найдя x, подставляй в ДРУГОЕ (более простое) уравнение. Подстановка в то же — скрывает ошибки.

❌ Trap 5: Quadratic inequality — writing one interval instead of two
$(x-2)(x-3) > 0 \rightarrow x > 3 — WRONG! Forgot x < 2.$
✅ Parabola opens upwards. Positive regions: x < 2 OR x > 3. Sketch the parabola!
Ловушка: квадратичное неравенство обычно даёт ДВА интервала. Нарисуй параболу: > 0 → снаружи корней, < 0 → между корнями.

❌ Trap 6: "At least" vs "more than"
"At least 5" means x > 5 — WRONG!
✅ "At least 5" = x ≥ 5. "More than 5" = x > 5. "At most 5" = x ≤ 5. "Fewer than 5" = x < 5.
Ловушка: «не менее 5» (at least) → x ≥ 5. «Более 5» (more than) → x > 5. Одна единица — один балл.

❌ Trap 7: Non-linear simultaneous — forgetting to expand correctly
Substituting y = 2x − 3 into y²: writing (2x − 3)² = 4x² − 9 — WRONG!
✅ (2x − 3)² = 4x² − 12x + 9 — don't forget the middle term.
Ловушка: при подстановке в нелинейные системы раскрывай скобки ПОЛНОСТЬЮ. (2x−3)² = 4x² − 12x + 9.