how do i calculate it?
Qd = 900 - 10p
p = 15 + 0.05Qs
This is what have tried.
making Qs the subject of formular
p - 15 = 0.05Qs
Qs= (p-15)/0.05
then now to solve the equlibrium
Qd=Qs
900 - 10p = (p-15)/0.05
(900 - 10p)* 0.05 = p - 15
45 - 0.5p = p - 15
45 + 15 = p + 0.05p
60 = 1.05p
p = 60/1.05 = 57.1
insert p into Qd
Qd = 900 - 10(57.1)
= 900 - 571
Qd = 329
The equilibrium of price and quantity is at the point $(p^*,q^*)$ where the following equation is satisfied: $$Q_d=q^*=Q_s$$
Thus we want to find the unique solution of this system of equations: $$ \left\{ \begin{array}{c} 10p+Q_d=900 \\ p-0.05Q_s=15 \end{array} \right. $$
Firstly we set $Q_d=q^*=Q_s$, which gives us: $$ \left\{ \begin{array}{c} 10p+q^*=900 \\ p-0.05q^*=15 \end{array} \right. $$ $$\Leftrightarrow \left\{ \begin{array}{c} q^*=900-10p \\ p-0.05\cdot(900-10p)=15 \end{array} \right. $$ $$\Leftrightarrow \left\{ \begin{array}{c} q^*=900-10p \\ \frac32p=45+15=60 \end{array} \right. $$ $$\Leftrightarrow \left\{ \begin{array}{c} q^*=900-10p \\ p^*=p=40 \end{array} \right. $$ $$\Leftrightarrow \left\{ \begin{array}{c} q^*=900-10p^*=900-10\cdot 40=500 \\ p^*=40 \end{array} \right. $$
Thus the equilibrium point is $(p^*,q^*)=(40,500)$.
Let's control if this is the right answer: