Module 23 of 30Lesson 1 of 44 · A-E0 Algebra Essentials at Extended Level: Forming and Substituting
Algebra strand · Unit A-E0 · Syllabus E2.1

Algebra Essentials at Extended Level: Forming and Substituting

🟠 Extended level
Negatives, fractions, powers — and formulas you build yourself

1 Learn it

Quick recap. A letter stands for a number that can change. \(3a\) means \(3 \times a\), \(ab\) means \(a \times b\), and \(a^{2}\) means \(a \times a\). To substitute, you swap each letter for its number and then follow the order of operations: brackets, powers, × and ÷, then + and −.

🔑 When you substitute a negative number or a fraction, put it in brackets first. Then the signs look after themselves: if \(x = -3\), then \(x^{2} = (-3)^{2} = 9\), but \(-x^{2} = -(-3)^{2} = -9\).

What Extended adds. On Papers 2 and 4 the numbers are harder and the formulas are longer:

  • Negatives and powers together: \((-2)^{3} = -8\), but \((-2)^{4} = 16\). An odd power keeps the minus; an even power removes it.
  • Fractions: if \(x = \dfrac{2}{3}\), then \(x^{2} = \dfrac{4}{9}\) — square the top and the bottom.
  • Science-style formulas such as \(v^{2} = u^{2} + 2as\), \(s = ut + \dfrac{1}{2}at^{2}\) and \(E = \dfrac{1}{2}mv^{2}\). Work out the right-hand side first, then finish off (for example, take a square root to get \(v\)).
  • Forming formulas from words or from a diagram. Choose a letter for each quantity that changes, write the rule in words, then turn the words into algebra. Simplify at the end.
  • Interpreting answers. A negative height, a time of −2 seconds or a length of 0 tells you something about the situation. Say what it means.

Expression or formula? An expression has no equals sign (\(6w + 6\)). A formula links a subject to an expression (\(P = 6w + 6\)). If a question says “write a formula for \(P\)”, your answer must start “\(P =\)”.

2 Watch it work
EXAMPLE 1

Find the value of \(3a^{2} – 2ab\) when \(a = -2\) and \(b = 5\).

\(3(-2)^{2} – 2(-2)(5) = 3(4) – (-20) = 12 + 20 = 32\)

Brackets around −2 make the square positive, and “minus a negative” becomes plus.

EXAMPLE 2

Use \(v^{2} = u^{2} + 2as\) to find \(v\) when \(u = 12\), \(a = -2.5\) and \(s = 20\). Give your answer to 3 significant figures.

\(v^{2} = 12^{2} + 2(-2.5)(20) = 144 – 100 = 44\)

\(v = \sqrt{44} = 6.633\ldots = 6.63\)

The formula gives \(v^{2}\), not \(v\). The last step is always the square root.

EXAMPLE 3

Use \(\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}\) to find \(f\) when \(u = 12\) and \(v = -4\).

\(\dfrac{1}{f} = \dfrac{1}{12} + \dfrac{1}{-4} = \dfrac{1}{12} – \dfrac{3}{12} = -\dfrac{2}{12} = -\dfrac{1}{6}\)

so \(f = -6\)

You found \(\dfrac{1}{f}\) first. Flip it to get \(f\). Do not add the bottoms (12 + (−4) = 8 is wrong).

EXAMPLE 4

A rectangle has width \(w\) cm. Its length is 3 cm more than twice its width. (a) Write a formula for its perimeter \(P\) cm. (b) Find \(w\) when \(P = 51\).

(a) Length \(= 2w + 3\). Perimeter = 2 × (length + width):

\(P = 2(2w + 3 + w) = 2(3w + 3) = 6w + 6\)

(b) \(6w + 6 = 51 \Rightarrow 6w = 45 \Rightarrow w = 7.5\) cm.

EXAMPLE 5

The time, \(T\) seconds, for one swing of a pendulum of length \(l\) metres is \(T = 2\pi\sqrt{\dfrac{l}{9.8}}\). Find \(T\) when \(l = 0.8\).

\(T = 2\pi\sqrt{\dfrac{0.8}{9.8}} = 2\pi \times 0.2857\ldots = 1.795\ldots = 1.80\) s (3 s.f.)

On a calculator, put the whole fraction inside the square root. Then multiply by \(2\pi\).

3 Your turn — try first, then reveal
🚫 No calculator[2 marks]

Q1. Find the value of \(5x – 2y^{2}\) when \(x = -3\) and \(y = -4\).

👀 Show answer
✍️ How your working should look in the exam
5(−3) − 2(−4)2 = −15 − 2(16)   ✓ M1
= −15 − 32 = −47   ✓ A1
Why this works
Square first: \((-4)^{2} = 16\). Then \(2 \times 16 = 32\) is subtracted. The answer is −47, not −15 + 32 = 17.

🚫 No calculator[2 marks]

Q2. Find the value of \(\dfrac{a^{3} – b}{2a}\) when \(a = -2\) and \(b = 4\).

👀 Show answer
✍️ How your working should look in the exam
(−2)3 − 42(−2) = −8 − 4−4   ✓ M1
= −12−4 = 3   ✓ A1
Why this works
An odd power keeps the minus: \((-2)^{3} = -8\). Work out the top and the bottom separately, then divide. Negative ÷ negative is positive.

🚫 No calculator[3 marks]

Q3. Find the value of \(x^{2} + xy\) when \(x = \dfrac{2}{3}\) and \(y = -\dfrac{3}{4}\). Give your answer as a fraction.

👀 Show answer
✍️ How your working should look in the exam
x2 = 49    xy = 23 × (−34) = −12   ✓ M1
49 − 12 = 818 − 918   ✓ M1
= −118   ✓ A1
Why this works
Do each term on its own, then use a common denominator (18) to subtract. A small negative answer is fine — \(xy\) is just a little bigger than \(x^{2}\).

🔢 Calculator[2 marks]

Q4. Use \(s = ut + \dfrac{1}{2}at^{2}\) to find \(s\) when \(u = 4.2\), \(t = 3.5\) and \(a = -1.8\).

👀 Show answer
✍️ How your working should look in the exam
s = 4.2 × 3.5 + 12 × (−1.8) × 3.52   ✓ M1
= 14.7 − 11.025 = 3.675   ✓ A1
Why this works
Only \(t\) is squared, not \(\dfrac{1}{2}at\). Square 3.5 to get 12.25, then multiply by \(\dfrac{1}{2} \times (-1.8) = -0.9\) to get −11.025.

🔢 Calculator[3 marks]

Q5. A truck on the A1 near Mahalapye is travelling at 25 m/s. The driver brakes and the truck slows down with \(a = -4\) m/s². Use \(v^{2} = u^{2} + 2as\) to find the distance \(s\) metres the truck travels before it stops. Give your answer to 3 significant figures.

👀 Show answer
✍️ How your working should look in the exam
stops → v = 0   ✓ B1
0 = 252 + 2(−4)s → 0 = 625 − 8s   ✓ M1
s = 78.125 = 78.1 m   ✓ A1
Why this works
The hidden value is in the words: “stops” means the final speed \(v\) is 0. Then substitute and solve \(8s = 625\).

🚫 No calculator[3 marks]

Q6. An isosceles triangle has two equal sides of length \((2x + 1)\) cm and a base of length \((x – 3)\) cm.
(a) Write a formula for the perimeter \(P\) cm, in its simplest form.
(b) Find \(x\) when \(P = 37\).

👀 Show answer
✍️ How your working should look in the exam
(a) P = 2(2x + 1) + (x − 3)   ✓ M1
P = 5x − 1   ✓ A1
(b) 5x − 1 = 37 → 5x = 38 → x = 7.6   ✓ B1
Why this works
There are TWO equal sides, so \(2x + 1\) is counted twice. A formula needs “\(P =\)” at the start. Check: the base is \(7.6 – 3 = 4.6\) cm, which is positive, so the answer makes sense.

🚫 No calculator[2 marks]

Q7. At a cattle post near Serowe, it costs P35 a month to feed one cow and P12 a month to feed one goat. The farmer has \(c\) cows and \(g\) goats. Write a formula for \(T\), the total cost in pula of feeding all the animals for one year.

👀 Show answer
✍️ How your working should look in the exam
one month: 35c + 12g   ✓ B1
T = 12(35c + 12g) = 420c + 144g   ✓ B1
Why this works
Build the monthly cost first, then multiply by 12 months. The trap is to give the monthly formula and forget “one year”.

🔢 Calculator[2 marks]

Q8. The kinetic energy \(E\) joules of a car is \(E = \dfrac{1}{2}mv^{2}\). Find \(E\) when \(m = 1250\) kg and \(v = 13.6\) m/s.

👀 Show answer
✍️ How your working should look in the exam
E = 12 × 1250 × 13.62   ✓ M1
= 625 × 184.96 = 115 600 J   ✓ A1
Why this works
Only \(v\) is squared. Squaring \(mv\) together gives a far too big answer.

🚫 No calculator[3 marks]

Q9. The diagram shows a rectangle \(x\) cm wide and \((x + 4)\) cm long. A square of side 2 cm has been cut from one corner.
(a) Write a formula for the area \(A\) cm² of the shape that is left.
(b) Find \(A\) when \(x = 6\).

(x + 4) cmx cm2 cm2 cm

👀 Show answer
✍️ How your working should look in the exam
(a) A = x(x + 4) − 22   ✓ M1
A = x2 + 4x − 4   ✓ A1
(b) A = 36 + 24 − 4 = 56 cm2   ✓ B1
Why this works
Whole rectangle minus the missing square. Check (b) with real numbers: 6 × 10 = 60, take away 4, gives 56 ✔.

🔢 Calculator[2 marks]

Q10. A ball is thrown upwards from the edge of a cliff. Its height above the edge after \(t\) seconds is \(h = 30t – 5t^{2}\) metres. Find \(h\) when \(t = 6.5\) and explain what your answer means.

👀 Show answer
✍️ How your working should look in the exam
h = 30(6.5) − 5(6.5)2 = 195 − 211.25 = −16.25 m   ✓ B1
The ball is 16.25 m below the edge of the cliff   ✓ B1
Why this works
A negative height is not an error. Heights are measured from the cliff edge, so −16.25 means the ball has fallen past the edge and is 16.25 m below it.

4 Common mistakes — don’t lose easy marks
  • No brackets round negatives: typing −3² on a calculator gives −9. Type (−3)² to get 9.
  • Squaring too much: in \(\dfrac{1}{2}mv^{2}\) and \(2as\), only the letter with the power is squared.
  • Stopping at \(v^{2}\) or \(\dfrac{1}{f}\): the question asks for \(v\) or \(f\). Square-root or flip at the end.
  • Missing the hidden value: “starts from rest” means \(u = 0\); “stops” means \(v = 0\).
  • Writing an expression when a formula is asked for: start with the subject, e.g. “\(P = 5x – 1\)”, or you can lose the mark.
  • Forgetting the time units: “per month” in the question but “one year” in the answer — multiply by 12.
5 How the exam asks this
  • Paper 2 (no calculator): “Find the value of…” with negatives or fractions, usually 2 marks. Show the substituted line — it earns the M1 even if the arithmetic slips.
  • Paper 4 (calculator): science formulas such as \(v^{2} = u^{2} + 2as\) inside a longer question, often followed by “rearrange the formula” — the next skill in the Algebra strand.
  • Command words: “Find the value of”, “Write down a formula for”, “Write an expression for”, “Show that”. “Write down” means little or no working is expected.
  • Forming expressions from a diagram’s perimeter or area often starts a 4–6 mark algebra question that ends with solving an equation.
✅ Ready? Take the Check Yourself quiz below to complete this unit.

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