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We are given the function:

\[
f(x) = 1000(x - 1980) + 10,000
\]

This models the per capita income, where $x$ is the year, and we are asked to find the year when the per capita income was $18,000.

We can set $f(x) = 18,000$ and solve for $x$:

\[
18,000 = 1000(x - 1980) + 10,000
\]

### Step 1: Subtract 10,000 from both sides:
\[
18,000 - 10,000 = 1000(x - 1980)
\]
\[
8,000 = 1000(x - 1980)
\]

### Step 2: Divide both sides by 1000:
\[
\frac{8,000}{1000} = x - 1980
\]
\[
8 = x - 1980
\]

### Step 3: Solve for $x$:
\[
x = 1980 + 8 = 1988
\]

Thus, the per capita income was $18,000 in the year **1988**.

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Would you like further details or have any questions? Here are five related questions to expand on this:

1. How would the result change if the per capita income target was $20,000 instead of $18,000?
2. What would the per capita income be in the year 2000 according to this model?
3. How would you adjust the equation if the per capita income grew at a faster rate, say $1500 per year?
4. Can you interpret what the constants 1000 and 10,000 represent in the model?
5. What would be the per capita income in 2010 according to this function?

**Tip:** When solving equations, make sure to isolate the variable step by step to avoid mistakes in calculations.

The per capita income (per person) from 1980 to 2010 can be modeled by f(x) = 1000(x - 1980) + 10,000, where x is the year. Determine the year when the per capita income was $18,000.

Learn how to solve a linear equation to determine the year when the per capita income was $18,000, modeled by the function f(x) = 1000(x - 1980) + 10,000. This problem is ideal for students in Grades 9-11, covering concepts of linear functions and equation solving.

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