Flashcards · 4MA1 · Topic 2

Equations, formulae and identities flashcards

47 cards on Topic 2, subtopics 2.1 to 2.8. Pick a subtopic or study them all.

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2.1 Use of symbols

Expression, equation and formula: Expression: terms joined by + or −, no equals sign (3a − 2b + 7). Equation: true only for certain values of the letter (2x + 5 = 11). Formula: a rule linking variables (P = 2(l + w)).
Coefficient: The number multiplying a letter. In −4x the coefficient is −4.
Simplify 4x³ × 2x⁻¹: Multiply the numbers and add the powers: 8x².
Simplify a⁻³ × a⁸, and give 9⁰ and 9⁻¹: a⁻³⁺⁸ = a⁵. 9⁰ = 1 and 9⁻¹ = 1/9.
Work out 16^(−3/4)Higher: 1. Negative: take the reciprocal, 1/16^(3/4). 2. Denominator 4: fourth root of 16 = 2. 3. Numerator 3: 2³ = 8. Answer: 1/8

2.2 Algebraic manipulation

Factorise fully 14x²y + 21xy: Take out the HCF of the numbers and the letters: 7xy(2x + 3).
Factorise x² − x − 12: Two numbers with product −12 and sum −1: −4 and +3. (x − 4)(x + 3)
Factorising x² + bx + c: sign patterns: c positive, b positive: both numbers positive. c positive, b negative: both negative. c negative: one of each; the larger has the sign of b.
Factorise 6x² + 7x − 3Higher: ac = −18; two numbers with product −18 and sum 7: 9 and −2. 6x² + 9x − 2x − 3 = 3x(2x + 3) − (2x + 3) = (3x − 1)(2x + 3)
Write x² + 6x − 2 in the form (x + p)² + qHigher: (x + 3)² − 9 − 2 = (x + 3)² − 11 General: x² + bx + c = (x + b/2)² − (b/2)² + c
Simplify (x² − 6x + 9)/(x² − 9)Higher: Factorise fully: (x − 3)(x − 3) / ((x − 3)(x + 3)). Cancel the common factor: (x − 3)/(x + 3)
Prove that (2n + 1)² − 1 is a multiple of 4 for every integer nHigher: (2n + 1)² − 1 = 4n² + 4n + 1 − 1 = 4n² + 4n = 4n(n + 1). n(n + 1) is an integer, so this is a multiple of 4.

2.3 Expressions and formulae

Subject of a formula: The letter on its own on one side of a formula: P in P = 2l + 2w.
Evaluate 2a² − b when a = −3 and b = −0.5: 2(−3)² − (−0.5) = 2 × 9 + 0.5 = 18.5 Put each value in brackets when you substitute.
A plumber charges £45 call-out plus £30 per hour. Write a formula for the cost £C for h hours.: C = 45 + 30h "Per" means multiply; the fixed charge is added.
Make v the subject of E = ½mv²: 2E = mv² v² = 2E/m v = √(2E/m)
Make y the subject of x = (y + 2)/(y − 5)Higher: x(y − 5) = y + 2 xy − 5x = y + 2 xy − y = 5x + 2 y(x − 1) = 5x + 2 y = (5x + 2)/(x − 1)

2.4 Linear equations

Method for solving a linear equation with fractions and brackets: 1. Multiply every term by the LCM of the denominators. 2. Expand brackets, watching the sign in front. 3. Collect the unknown on one side and numbers on the other. 4. Divide by the coefficient, then check.
Solve 6 − (x − 4) = 2x + 1: 6 − x + 4 = 2x + 1 (the minus changes both signs in the bracket) 10 − x = 2x + 1 9 = 3x, so x = 3
Solve (x + 2)/4 = (x − 1)/3: Multiply both sides by 12: 3(x + 2) = 4(x − 1) 3x + 6 = 4x − 4 x = 10
Angles on a straight line are 2y°, (3y + 20)° and (y − 8)°. Find y.: 2y + 3y + 20 + y − 8 = 180 6y + 12 = 180, so y = 28. The angles are 56°, 104° and 20°.

2.5 Proportion

Constant of proportionalityHigher: The number k. It is the same for every pair of values.
Write formulae: y ∝ x², y ∝ √x, y inversely proportional to x², y inversely proportional to √xHigher: y = kx² y = k√x y = k/x² y = k/√x
P is inversely proportional to √Q. P = 6 when Q = 25. Find P when Q = 36.Higher: P = k/√Q, so 6 = k/5 and k = 30. P = 30/√36 = 30/6 = 5
y ∝ x³. x is increased by 10%. Find the percentage increase in y.Higher: Multiplier for x: 1.1. For y: 1.1³ = 1.331, so y increases by 33.1%.
Graph of y ∝ xHigher: A straight line through the origin, with gradient k.
Is y = 3x + 2 an example of direct proportion?Higher: No. It is a linear relationship, but the line does not pass through the origin. In proportion, k multiplies and there is no added constant.

2.6 Simultaneous linear equations

Simultaneous equations: Two equations that must both be true for the same values of the unknowns.
Elimination: when do you add and when do you subtract the equations?: Make one letter have the same size of coefficient in both. Same signs: subtract. Different signs: add.
Solve 3x + 4y = 10 and 5x − 6y = 4: Multiply by 3 and 2: 9x + 12y = 30 and 10x − 12y = 8. Add: 19x = 38, so x = 2. Then 6 + 4y = 10, so y = 1.
Solve x = 2y + 1 and 3x − 4y = 7: Substitute, with brackets: 3(2y + 1) − 4y = 7 2y + 3 = 7, so y = 2 and x = 5.
What does the solution of two simultaneous linear equations show on a graph?Higher: The coordinates of the point where the two straight lines intersect.
Why do 2x + y = 3 and 4x + 2y = 10 have no solution?Higher: They are y = −2x + 3 and y = −2x + 5: the same gradient with different intercepts, so the lines are parallel and never meet.

2.7 Quadratic equations

The rule behind solving a quadratic by factorising: If A × B = 0, then A = 0 or B = 0. It only works with zero on one side, so rearrange to = 0 first.
Solve x² + 11x + 28 = 0: (x + 4)(x + 7) = 0 x = −4 or x = −7
Solve x² = 7x: Do not divide by x (you lose x = 0). Rearrange: x² − 7x = 0, so x(x − 7) = 0. x = 0 or x = 7
The quadratic formulaHigher: For ax² + bx + c = 0: x = (−b ± √(b² − 4ac)) / (2a) It is on the Higher formulae sheet. Work out b² − 4ac first.
What does b² − 4ac tell you about ax² + bx + c = 0?Higher: b² − 4ac > 0: two different solutions. b² − 4ac = 0: two equal solutions. b² − 4ac < 0: no real solutions.
Solve x² − 4x − 7 = 0 by completing the square, giving exact answersHigher: (x − 2)² − 4 − 7 = 0, so (x − 2)² = 11. x − 2 = ±√11, so x = 2 ± √11.
Solve y = 3x and x² + y² = 40Higher: Substitute the linear equation into the quadratic: x² + 9x² = 40, so x² = 4 and x = ±2. Pairs from y = 3x: (2, 6) and (−2, −6).

2.8 Inequalities

Strict and inclusive inequalities: Strict: > or <; the boundary value is not included (open circle). Inclusive: ≥ or ≤; the boundary value is included (closed circle).
Solve 9 − 2x ≤ 1: −2x ≤ −8. Dividing by −2 reverses the sign: x ≥ 4.
Solve 1 < 7 − 2x ≤ 9: Subtract 7 from all three parts: −6 < −2x ≤ 2. Divide by −2 and reverse both signs: 3 > x ≥ −1. Write in ascending order: −1 ≤ x < 3.
Boundary line: The line you get by replacing the inequality sign with =.
How do you decide which side of a boundary line to shade?: Substitute a test point not on the line, such as (0, 0). If it makes the inequality true, the region is on its side. The required region is where all the conditions overlap.
Solve x² − 7x + 10 ≤ 0Higher: (x − 2)(x − 5) ≤ 0; critical values 2 and 5. "≤ 0" means between the roots: 2 ≤ x ≤ 5.
Solve x² > 16Higher: Outside the roots: x < −4 or x > 4. (x² < 16 would be −4 < x < 4.)