Question 6:
$\text{Given }2ab=3a-\frac{b^2}{2}.\text{ Express }a\text{ in terms of }b\text{.}$

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:
$\text{Given }a=\frac{\sqrt{b+1}}{a}.\text{ Express }b\text{ in terms of }a\text{.}$

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:
$\begin{array}{l}\text{Given }x=\frac{5w}{2w-y}.\\ (i)\text{ Express }w\text{ in terms of }x\text{ and }y\text{.}\\ \text{(ii) Find the value of w when }x=10\text{ and }y=15.\end{array}$

Solution:
$\begin{array}{l}\text{(i)}\\ x=\frac{5w}{2w-y}\\ 2wx-xy=5w\\ 2wx-5w=xy\\ w\left(2x-5\right)=xy\\ w=\frac{xy}{2x-5}\\ \\ \text{(ii)}\\ w=\frac{\left(10\right)\left(15\right)}{2\left(10\right)-5}\\ =\frac{150}{15}\\ =10\end{array}$

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

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

  

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

 

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

$\begin{array}{l}\sqrt{\frac{8s-3h}{4}}=2\\ \frac{8s-3h}{4}=4\leftarrow \boxed{\text{Square both sides}}\\ 8s-3h=16\\ 8s=16+3h\\ s=\frac{16+3h}{8}\\ s=2+\frac{3h}{8}\end{array}$

2. A constant is a quantity with a fixed value.