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

*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:**$\text{Given}a=\frac{\sqrt{b+1}}{a}.\text{Express}b\text{intermsof}a\text{.}$

*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:**$\begin{array}{l}\text{Given}x=\frac{5w}{2w-y}.\\ (i)\text{Express}w\text{intermsof}x\text{and}y\text{.}\\ \text{(ii)Findthevalueofwwhen}x=10\text{and}y=15.\end{array}$

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

**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 = 2

*h*– 6

Therefore, the sum (

*S*) of their age

*S*=

*h + (h – 3) +*(2

*h*– 6)

*S*=

*h + h – 3 +*2

*h*– 6

*S***= 4**

*h*– 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 = 2

*h*+ 8

Therefore, the total (

*S*) of their age

*S*= (

*h*+ 4) + (

*h*+ 16) + (2

*h*+ 8)

*S***= 4**

*h*+ 28