\( \begin{array}{ll}p=\frac{x}{450}+\frac{1}{2} & \text { Supply equation } \\ p=\frac{6,300}{x} & \text { Demand equation }\end{array} \)

The given equations represent the supply and demand equations in the market. Supply equation: \( p = \frac{x}{450} + \frac{1}{2} \) Demand equation: \( p = \frac{6,300}{x} \) To find the equilibrium point, we need to set the supply equation equal to the demand equation and solve for \( x \). Setting the supply equation equal to the demand equation: \( \frac{x}{450} + \frac{1}{2} = \frac{6,300}{x} \) Now, we can solve this equation to find the equilibrium point. Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{x}{450}+\frac{1}{2}=\frac{6300}{x}\) - step1: Find the domain: \(\frac{x}{450}+\frac{1}{2}=\frac{6300}{x},x\neq 0\) - step2: Multiply both sides of the equation by LCD: \(\left(\frac{x}{450}+\frac{1}{2}\right)\times 450x=\frac{6300}{x}\times 450x\) - step3: Simplify the equation: \(x^{2}+225x=2835000\) - step4: Move the expression to the left side: \(x^{2}+225x-2835000=0\) - step5: Factor the expression: \(\left(x-1575\right)\left(x+1800\right)=0\) - step6: Separate into possible cases: \(\begin{align}&x-1575=0\\&x+1800=0\end{align}\) - step7: Solve the equation: \(\begin{align}&x=1575\\&x=-1800\end{align}\) - step8: Check if the solution is in the defined range: \(\begin{align}&x=1575\\&x=-1800\end{align},x\neq 0\) - step9: Find the intersection: \(\begin{align}&x=1575\\&x=-1800\end{align}\) - step10: Rewrite: \(x_{1}=-1800,x_{2}=1575\) The equilibrium point is at \( x = 1575 \) or \( x = -1800 \). Since the equilibrium point cannot be negative, the equilibrium point is at \( x = 1575 \).