# In 1979, a river was 27 feet below a bridge. By 1989 it was only 18 feet below. Assuming the rate of rise is constant, write an expression for the distance from the bridge to the river. In what year will the river reach the height of the bridge?

Jan 18, 2018

The answer is $27 - \frac{9}{10} t$ and $2009$

#### Explanation:

We are starting, at $t = 0$ (1979), with the river at 27 feet below the bridge, so it makes sense that this is our starting point.

From there, we discover that the level changed by $27 - 18 = 9$ feet in the $1989 - 1979 = 10$ year period between 1979 and 1989.

That means that the rate of rise is $9$ feet every $10$ years, so we can express that as $\frac{9}{10} t$.

To find the year in which the river reaches the bridge, we know that the distance between them will be $0$ at that time.

$0 = 27 - \frac{9}{10} t$

Solving this gives $t = 30$ years, which when added to the start year of $1979$ leads to $2009$.