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### a. Find a linear equation that gives price $ y $ as a function of the quantity demanded $ x $.
We need to find a linear equation, $ y = mx + b $, where $ y $ represents price and $ x $ represents the quantity demanded.

The table gives us two points for $ x $ (quantity demanded) and $ y $ (price):
- $ (210, 50) $
- $ (130, 90) $

To find the equation of the line, we first calculate the slope $ m $:
\[
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{90 - 50}{130 - 210} = \frac{40}{-80} = -0.5
\]

Now, using the point-slope form $ y - y_1 = m(x - x_1) $, we substitute one point, say $ (210, 50) $:
\[
y - 50 = -0.5(x - 210)
\]
Simplifying:
\[
y - 50 = -0.5x + 105
\]
\[
y = -0.5x + 155
\]
So, the equation for price as a function of quantity demanded is:
\[
y = -0.5x + 155
\]

### b. Find a linear equation that gives price $ y $ as a function of the quantity supplied $ x $.
Similarly, we use two points for $ x $ (quantity supplied) and $ y $ (price):
- $ (0, 50) $
- $ (160, 90) $

Again, calculating the slope $ m $:
\[
m = \frac{90 - 50}{160 - 0} = \frac{40}{160} = 0.25
\]

Using point-slope form again with $ (0, 50) $:
\[
y - 50 = 0.25(x - 0)
\]
\[
y = 0.25x + 50
\]

Thus, the equation for price as a function of quantity supplied is:
\[
y = 0.25x + 50
\]

### c. Find the equilibrium price and the equilibrium quantity.
The equilibrium occurs where quantity demanded equals quantity supplied. Set the two equations from parts (a) and (b) equal to each other:
\[
-0.5x + 155 = 0.25x + 50
\]

Solving for $ x $:
\[
155 - 50 = 0.25x + 0.5x
\]
\[
105 = 0.75x
\]
\[
x = \frac{105}{0.75} = 140
\]

Now, substitute $ x = 140 $ into either equation to find the price:
\[
y = -0.5(140) + 155 = -70 + 155 = 85
\]

So, the equilibrium price is $ 85 $, and the equilibrium quantity is $ 140 $ (thousands).

### d. Use the demand price function found in part (a) to find a revenue function.
Revenue is given by $ R(x) = x \cdot y(x) $, where $ y(x) $ is the demand price function:
\[
R(x) = x(-0.5x + 155)
\]
Expanding:
\[
R(x) = -0.5x^2 + 155x
\]

Thus, the revenue function is:
\[
R(x) = -0.5x^2 + 155x
\]

### e. The cost of these calculators is given by $ C(x) = 5x + 500 $. Use this to help you find the profit function.
Profit is given by $ P(x) = R(x) - C(x) $. Using the revenue function from part (d) and the cost function:
\[
P(x) = (-0.5x^2 + 155x) - (5x + 500)
\]
Simplifying:
\[
P(x) = -0.5x^2 + 155x - 5x - 500
\]
\[
P(x) = -0.5x^2 + 150x - 500
\]

Thus, the profit function is:
\[
P(x) = -0.5x^2 + 150x - 500
\]

---

Would you like more details on any specific part?  
Here are 5 related questions to extend this topic:
1. How does the equilibrium price change if the supply curve becomes steeper?
2. What happens to the revenue if the demand for calculators decreases?
3. How can we interpret the slope of the demand function economically?
4. What is the break-even quantity, where the profit function $ P(x) = 0 $?
5. How would the profit function change if fixed costs increase from $500 to $1000?

**Tip:** When dealing with supply and demand equations, finding the slope gives insights into how sensitive price is to changes in quantity.

The table gives the quantity of calculators demanded and the quantity supplied for selected prices. Tasks include finding linear equations for price as a function of quantity demanded and supplied, determining equilibrium price and quantity, and finding revenue and profit functions.

This problem covers how to find the linear price equation as a function of quantity demanded and supplied, calculate equilibrium price and quantity, and determine revenue and profit functions based on supply and demand curves. Detailed step-by-step solution provided.

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