Sec 3 E-Math Equations & Inequalities notes

An equation is a balance scale. Solving means finding every value of the letter that makes the balance true. You may add, subtract, multiply or divide both sides by anything (except zero) — as long as both sides get the same treatment. When letters appear on both sides, collect them on the side where the coefficient is larger, so you subtract the smaller coefficient and avoid negative letters.

1. 1Expand brackets

   3(2x − 1) − 4(x + 2) = 5 becomes 6x − 3 − 4x − 8 = 5. The −4 flips both signs.

2. 2Collect like terms on each side

   2x − 11 = 5.

3. 3Move the constant

   2x = 16.

4. 4Divide by the coefficient

   x = 8.

Checking is cheap insurance: substitute your answer into the original equation and confirm both sides match. An answer that fails the check has a sign slip somewhere — find it before moving on.

When the unknown sits in a denominator, multiply every term by the lowest common denominator (LCD) to clear the fractions. Before you start, note any value that makes a denominator zero — that value can never be a solution, no matter how neatly the algebra works out.

1. 1Find the LCD

   For (2x + 1)/3 − (x − 2)/4 = 1 the LCD of 3 and 4 is 12.

2. 2Multiply every term

   4(2x + 1) − 3(x − 2) = 12 — the denominators disappear.

3. 3Expand carefully

   8x + 4 − 3x + 6 = 12, so 5x + 10 = 12.

4. 4Solve the linear equation

   5x = 2, so x = 0.4.

Cross-multiplication works when each side is a single fraction: if a/b = c/d then ad = cb. With more than one fraction per side, use the LCD of all denominators instead — it is the same idea done in one step.

A pair of linear equations in x and y pins down one point (x, y) that satisfies both. Three methods are on the syllabus: elimination (add or subtract multiples of the equations to wipe out one letter), substitution (rearrange one equation and feed it into the other), and graphical (draw both lines; the solution is their intersection). Elimination is fastest when coefficients line up; substitution shines when one equation already has a letter isolated.

1. 1Line up the coefficients

   2x + 3y = 16 and 5x − y = 6: multiply the second by 3 → 15x − 3y = 18.

2. 2Add or subtract

   Adding kills y: 17x = 34, so x = 2.

3. 3Back-substitute

   5(2) − y = 6 gives y = 4.

4. 4Check in the unused equation

   2(2) + 3(4) = 4 + 12 = 16 ✓.

Graphically, parallel lines (same gradient) mean no solution; the same line drawn twice means infinitely many. Word problems often hide simultaneous equations: name the unknowns, translate each sentence into an equation, then solve and interpret — you cannot sell 3.5 tickets, so round or reject as the context demands.

A quadratic equation ax² + bx + c = 0 (a ≠ 0) has up to two real roots. Your first instinct should be factorisation: if it factors nicely, that is the quickest route. If it does not, use the quadratic formula x = (−b ± √(b² − 4ac)) / 2a, valid for every quadratic. Completing the square rewrites x² + px + q as (x + p/2)² + q − p²/4, which exposes the minimum point and gives an alternative solving route when the question demands it.

1. 1Factorisation: x² − 5x + 6 = (x − 2)(x − 3) = 0 gives x = 2 or x = 3 — a product is zero only when a factor is zero.
2. 2Formula: for 3x² + 5x − 1 = 0, x = (−5 ± √37)/6 ≈ 0.180 or −1.85 (3 s.f.).
3. 3Completing the square: x² − 6x + 2 = (x − 3)² − 7, so the minimum value is −7 at x = 3.

The discriminant b² − 4ac counts the real roots before you solve: positive → two distinct roots, zero → exactly one (repeated) root, negative → none. Graphically, the roots of ax² + bx + c = 0 are the x-intercepts of y = ax² + bx + c, so a question saying "hence solve graphically" wants those intercepts read off the sketch.

Inequalities behave like equations under addition and subtraction, and under multiplication or division by a positive number. Multiply or divide by a negative number and the sign reverses: from > to <, from ≤ to ≥. The solution is usually a range, drawn on a number line with filled dots for ≤ / ≥ and open dots for < / >.

1. 1Solve 5 − 2x ≥ 9

   Subtract 5: −2x ≥ 4. Divide by −2 and flip: x ≤ −2.

2. 2Alternative (no flip)

   Move the x term first: −4 ≥ 2x, i.e. x ≤ −2 — same answer, no reversal needed.

3. 3Compound inequality

   −3 ≤ 2x − 1 < 7: add 1 to all three parts → −2 ≤ 2x < 8, then divide by 2 → −1 ≤ x < 4.

4. 4Draw it

   Filled dot at −1, open dot at 4, line shaded between.

In context problems, inequalities answer "how much at most / at least" questions: translate the wording (at most → ≤, more than → >, no less than → ≥), solve for the unknown, and state the range with units. A distance answer like 5 < d ≤ 10 km should be read as "more than 5 km but at most 10 km".