Mathematics notes, chapter 3
Algebraic Formulae
Textbook Mathematics Form 2 (DLP), pages 42–53
What this chapter covers
- Write a formula based on a situation, using variables and the basic operations (3.1.1)
- Change the subject of a formula, including formulae with squares and square roots (3.1.2)
- Find the value of a variable when the values of the other variables are given (3.1.3)
- Solve problems involving algebraic formulae (3.1.4)
3.1Algebraic formulae
A variable is a quantity whose value is not known or can change (such as m, n, x). A constant is a quantity whose value is fixed (such as 3 or RM6). An algebraic formula is an equation that connects a few variables using addition, subtraction, multiplication or division.
3.1.1 Forming a formula
A formula is built from the relationship between variables. Read the situation, give each unknown quantity a letter, then join the pieces with the correct operations.
Table from Mathematics Form 2 (DLP) textbook, p. 45.
From the table, add across each race or down each dance:
- s, Chinese dancers: s = 2c + b + 3a
- d, Kuda Kepang dancers: d = 2b + b + 5b = 8b
- w, Indian and Malay dancers: w = (2a + 5b + c) + (a + 2b + 2c) = 3a + 7b + 3c
The letter on its own on one side (s, d or w) is the subject of the formula. It can be written on the left or the right. Any letters from a to z can be variables.
Example 1 (textbook p. 45): selling price with a discount
Suzi sells chocolate cake at RM3 a slice and cheese cake at twice that price. All cakes have 10% discount. Find a formula for the selling price z when she sells m slices of chocolate cake and n slices of cheese cake.Cheese cake: 2 × RM3 = RM6 a slice
Before discount: 3m + 6n
After the 10% discount the price is 100% − 10% = 90% = 0.9 of the total
z = 0.9(3m + 6n) = 2.7m + 5.4n
A discount of 10% means paying 90%, not 10%. Multiply the whole bracket by 0.9, not just one term.
A tailor charges RM30 to sew a baju kurung and RM18 to sew a skirt. (a) Write a formula for the total charge C for k baju kurung and s skirts. (b) Find C when k = 2 and s = 3.
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(a) C = 30k + 18s
(b) C = 30(2) + 18(3) = 60 + 54 = RM114
(b) C = 30(2) + 18(3) = 60 + 54 = RM114
3.1.2 Changing the subject of a formula
The perimeter of an isosceles triangle with base a and two equal sides b is P = a + 2b. Here P is the subject. We can rewrite it with a as the subject, a = P − 2b, or with b as the subject, b = P − a2.
Treat the formula like a balance: do the same to both sides. To get the new subject alone, undo the operations in the reverse order.
| The letter has been | To undo it, do this to both sides |
|---|---|
| added to: + q | subtract q |
| subtracted: − q | add q |
| multiplied by q | divide by q |
| divided by q | multiply by q |
| squared: p² | take the square root |
| square root: √p | square both sides |
The coefficient of the subject must be 1, and the subject is written on the left. If you reach −m = something, multiply both sides by −1.
Example 2 (textbook p. 46): make m the subject
(b) b = 2s − mm + b = 2s (add m to both sides)
m = 2s − b
(d) t = m − n−3
t × (−3) = m − n, so −3t = m − n
m = −3t + n = n − 3t
Example 3 (textbook p. 47): make p the subject
(a) q = √p. Square both sides: p = q²(c) w = √(p3). Square both sides: w² = p3, so p = 3w²
(d) t = 1p². Then tp² = 1, p² = 1t, so p = √(1t)
Taking a square root gives ±. If x² = a then x = ±√a. The textbook takes the positive root for lengths and similar quantities. Also divide every term: from y = 3x − 5, x = y + 53, not y + 53.
Make x the subject. (a) y = 3x − 5 (b) T = √(2x)
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(a) y + 5 = 3x, so x = y + 53
(b) T² = 2x, so x = T²2
(b) T² = 2x, so x = T²2
3.1.3 Finding the value of a variable
If all the variables except one are known, substitute the numbers into the formula. To find a letter that is not the subject, either substitute first and solve the equation, or change the subject first.
Example 4 (textbook pp. 47–48): given w = 7t − 5u
(a) t = 3 and u = −2: w = 7(3) − 5(−2) = 21 + 10 = 31(b) w = 15 and u = 4: 7t − 5(4) = 15, so 7t = 35 and t = 5
Put brackets round a negative value when you substitute: 5(−2), not 5−2.
Example 5 (textbook p. 48): given m = 14(p − q)², find q when m = 16 and p = 3
16 = 14(3 − q)², so (3 − q)² = 64Take the positive square root: 3 − q = 8
q = −5
Check by substituting back: 14(3 − (−5))² = 14 × 64 = 16.
Given P = 3x² − 4y, find (a) P when x = −2 and y = 5, (b) y when P = 7 and x = 3.
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(a) P = 3(−2)² − 4(5) = 12 − 20 = −8
(b) 7 = 3(3)² − 4y = 27 − 4y, so 4y = 20 and y = 5
(b) 7 = 3(3)² − 4y = 27 − 4y, so 4y = 20 and y = 5
3.1.4 Solving problems
Follow four steps: understand the problem, plan a strategy (choose letters, write an equation), carry out the plan, then write a conclusion and check.
Example 6 (textbook pp. 48–49): fried chicken
A fried chicken costs twice the price of a bun. Azman buys 2 buns and 1 chicken with RM5, and the balance is RM1. With RM12 he buys 2 buns and as many chickens as he can. How many chickens?Let a bun cost RMx and a chicken RM2x.
2x + 2x + 1 = 5, so 4x = 4 and x = 1. Bun RM1, chicken RM2.
Let y be the number of chickens: 2(1) + 2y = 12, so y = 5.
Azman can buy 5 pieces of fried chicken.
A notebook costs RMx and a pen costs RM(x − 1). Maya pays RM13 for 3 notebooks and 2 pens. Find the price of a notebook and of a pen.
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3x + 2(x − 1) = 13, so 5x − 2 = 13 and x = 3.
Notebook RM3, pen RM2. Check: 9 + 4 = 13.
Notebook RM3, pen RM2. Check: 9 + 4 = 13.
Quick summary
| Task | What to do |
|---|---|
| Form a formula | Name the quantities, join them with operations, use brackets for discounts |
| Change the subject | Undo operations in reverse order, same on both sides, coefficient 1 |
| Find a value | Substitute with brackets, then solve |
| Word problem | Understand, plan, carry out, conclude |