2.1 Introduction to algebra: letters for numbers, substitution
p = 3 and q = −3.
Find 3p² − 2q.
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- 33
Show the answer
33. 3 × (3)² = 27. 2 × −3 = −6. 27 − (−6) = 33.
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Topic 2 of the Cambridge IGCSE Mathematics (0580) syllabus: from substitution to differentiation. Here is what each part asks, a practice question for each, and the mistakes that cost the most marks.
Each part with one practice question, made by the same engine your child plays. Parts marked Extended are for the Extended paper only.
p = 3 and q = −3.
Find 3p² − 2q.
33. 3 × (3)² = 27. 2 × −3 = −6. 27 − (−6) = 33.
Expand and simplify
(x − 6)(x − 5)
x² − 11x + 30. x × x = x². The x terms: −6x and −5x make −11x. The numbers: −6 × −5 = 30.
Simplify (x/5) ÷ (x²/2).
2/(5x). Dividing is multiplying by the flipped SECOND fraction: (x/5) × (2/x²) = 2x/(5x²) = 2/(5x).
Simplify x⁵ × x⁴
x⁹. Multiplying: add the powers. 5 + 4 = 9.
Solve the simultaneous equations:
y = x² − 1
y = 7x − 13
(3, 8) and (4, 15). Set them equal: x² − 1 = 7x − 13, so x² − 7x + 12 = 0 = (x − 3)(x − 4). x = 3 or 4; put each back in for y. A line can cross a curve twice.
How many integers satisfy
−5 ≤ x < 1 ?
6. From −5 up to 0: that is 6 integers. < leaves the end number out, ≤ keeps it.
4, 9, 16, 25, 36, …
Which is the nth term of this sequence?
n² + 2n + 1. The differences are 5, 7, 9, 11: they go up by 2 each time, so the rule has n² in it. Take n² away from each term and what is left is linear: n² + 2n + 1.
y is directly proportional to x.
When x = 7, y = 42.
Find y when x = 6.
36. y = kx. From the first pair, k = 42 ÷ 7 = 6. So y = 6 × 6 = 36.
The questions for this part come with a picture or are played as a game, so they are best tried in Huh Lab.
The curve y = a × bˣ passes through (0, 5) and (1, 10).
Find b.
2. At x = 0, y = a × b⁰ = a, so a = 5. At x = 1, y = a × b = 10, so b = 10 ÷ 5 = 2.
y = (x + 1)² − 2
Where is its turning point?
(−1, −2). The bracket is zero, and y is smallest, when x = −1. Then y = −2. The sign inside the bracket is the OPPOSITE of the x-coordinate.
y = x² + 8x + 4
Find the x-coordinate of the stationary point.
x = −4. dy/dx = 2x + 8. At a stationary point the gradient is 0: 2x = −8, so x = −4.
f(x) = 1 / (x + 6)
Which value of x must be excluded from the domain?
x = −6. You cannot divide by zero. The bottom is zero when x = −6.
Huh Lab checks every wrong answer against known mistakes like these, and goes back to the step behind the one your child keeps making.
Use a number line: below zero the numbers keep counting, in the other direction.
< and > leave the end number OUT. ≤ and ≥ keep it in. Check each end separately.
Whatever is done to x must be UNDONE on the other side: + becomes −, × becomes ÷.
Take a point in the middle of the shaded region: it must make every inequality true.
Check the gaps between terms. If the gaps themselves change, the rule is not "add the same each time".
Find k first from y = kx² (or k√x). Doubling x does not double y here.
After removing the number, there is still the multiplying to undo: divide as well.
A power of 2/3 means "cube root, then square". It is not "times two thirds".
Short rounds on any topic, a new question every time, the exact mistake found, and a report to you after every session.
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