# 7.2.2 Algebraic Formulae, PT3 Practice

Question 6:

Solution:
$\begin{array}{l}2ab=3a-\frac{{b}^{2}}{2}\\ 3a-2ab=\frac{{b}^{2}}{2}\\ a\left(3-2b\right)=\frac{{b}^{2}}{2}\\ a=\frac{{b}^{2}}{2\left(3-2b\right)}\end{array}$

Question 7:

Solution:
$\begin{array}{l}a=\frac{\sqrt{b+1}}{a}\\ \sqrt{b+1}={a}^{2}\\ {\left[{\left(b+1\right)}^{\frac{1}{2}}\right]}^{2}={\left({a}^{2}\right)}^{2}\\ b+1={a}^{4}\\ b={a}^{4}-1\end{array}$

Question 8:

Solution:

Question 9:
Azmin is h years old. His father is twice his brother’s age. If Azmin is 3 years older than his brother, write a formula for the sum (S) of their age.

Solution:
Azmin’s age = h
Azmin’s brother’s age = h – 3
Azmin’s father’s age = (h – 3) × 2 = 2h – 6

Therefore, the sum (S) of their age
S = h + (h – 3) + (2h – 6)
S = h + h – 3 + 2h – 6
S = 4h – 9

Question 10:
Mei Ling is 12 years older than Ali. In the next four years, Raju will be two times older than Ali. If h represents Ali’s age, write the algebraic expressions that represent the total of their ages, in terms of h, in four years time.

Solution:
Ali’s age in the next four years = h + 4
Mei Ling’s age = (h + 4) + 12 = h + 16
Raju’s age = (h + 4) × 2 = 2h + 8

Therefore, the total (S) of their age
S = (h + 4) + (h + 16) + (2h + 8)
S = 4h + 28

# 7.2.1 Algebraic Formulae, PT3 Practice

7.2.1 Algebraic Formulae, PT3 Practice

Question 1:
Given m2+ 7 = k, express m in terms of k.

Solution:
$\begin{array}{l}{m}^{2}+7=k\\ \text{}{m}^{2}=k-7\\ \text{}m=\sqrt{k-7}\end{array}$

Question 2:
Given 5km = mn – 2k, express k in terms of m and n.

Solution:
$\begin{array}{l}5km=mn-2k\\ 5km+2k=mn\\ k\left(5m+2\right)=mn\\ \text{}k=\frac{mn}{5m+2}\end{array}$

Question 3:
Given $2k-\frac{m}{2n}=m$ , express m in terms of k and n.

Solution:
$\begin{array}{l}2k-\frac{m}{2n}=m\\ -m-\frac{m}{2n}=-2k\\ \text{}m+\frac{m}{2n}=2k\\ \frac{2nm+m}{2n}=2k\\ \frac{m\left(2n+1\right)}{2n}=2k\\ \text{}m=\frac{4kn}{2n+1}\end{array}$

Question 4:
Given $\frac{4m}{\sqrt{T}-k}=\frac{3h}{m}$  , express T in terms of h and m.

Solution:
$\begin{array}{l}\frac{4m}{\sqrt{T}-k}=\frac{3h}{m}\\ \text{}4{m}^{2}=3h\sqrt{T}-3hk\\ \text{}3h\sqrt{T}=4{m}^{2}+3hk\\ \text{}\sqrt{T}=\frac{4{m}^{2}+3hk}{3h}\\ \text{}T={\left(\frac{4{m}^{2}+3hk}{3h}\right)}^{2}←\overline{)\text{Square both sides}}\end{array}$

Question 5:
Given $\sqrt{\frac{8s-3h}{4}}=2$ , express s in terms of h.

Solution:
$\begin{array}{l}\sqrt{\frac{8s-3h}{4}}=2\\ \text{}\frac{8s-3h}{4}=4←\overline{)\text{Square both sides}}\\ \text{}8s-3h=16\\ \text{}8s=16+3h\\ \text{}s=\frac{16+3h}{8}\\ \text{}s=2+\frac{3h}{8}\end{array}$

# 7.1 Algebraic Formulae

7.1 Algebraic Formulae

7.1.1 Variables and Constant
1. A variable is a quantity without a fixed value.
2. A constant is a quantity with a fixed value.

7.1.2 Concept of Formulae
1. An algebraic formula is an equation which shows the relationship between several variables and/ or constant.
2. The subject of an algebraic formula is a linear variable expressed in terms of the other variables.

Example 1:

Solution:
$\begin{array}{l}y=\frac{2-3x}{x}\\ xy=2-3x\\ xy+3x=2\\ x\left(y+3\right)=2\\ x=\frac{2}{y+3}\end{array}$

Example 2:

Solution:
$\begin{array}{l}s=\frac{{p}^{2}-{q}^{2}}{2r}\\ 2rs={p}^{2}-{q}^{2}\\ {p}^{2}=2rs+{q}^{2}\\ p=\sqrt{2rs+{q}^{2}}\end{array}$