Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 1 - Linear Functions - 1.5 Finding Equations of Lines - 1.5 Exercises - Page 75: 20

Answer

$y=\frac{234\,400}{3}t+\frac{5\,729\,000}{3}$

Work Step by Step

Let $t$ be the number of years after the year $2000$ and $y$ be the population of Nevada State. A linear equation that gives the population of Nevada State $t$ years since $2000$ takes the form of $$ y=mt+b, $$ where $m$ is the slope and $b$ is the $y$-intercept. Since in $2010$ (or $10$ years after $2000$), the population is projected at $2691$ thousand, this can be represented by the ordered pair $(t_1,y_1)=(10,2\,691\,000)$. Since in $2025$ (or $25$ years after $2000$), the population is projected at $3863$ thousand, this can be represented by the ordered pair $(t_2,y_2)=(25,3\,863\,000)$. The formula for finding the slope, $m$, of the line passing through two points, $(t_1,y_1)$ and $(t_2,y_2)$ is given by $m=\frac{y_1-y_2}{t_1-t_2}$. That is, $$\begin{aligned} m&=\frac{y_1-y_2}{t_1-t_2} \\&= \frac{3\,863\,000-2\,691\,000}{25-10} \\&= \frac{1\,172\,000}{15} \\&= \frac{234\,400}{3} .\end{aligned} $$ With $m=\frac{234\,400}{3}$, then the linear equation that gives the population $t$ years since $2000$ takes the form of $$ y=\frac{234\,400}{3}t+b .$$ Since the line passes through the point $(10,2\,691\,000)$, substitute $t=10$ and $y=2\,691\,000$ in the equation above to solve for $b$. That is, $$\begin{aligned} y&=\frac{234\,400}{3}t+b \\ 2\,691\,000&=\frac{234\,400}{3}(10)+b \\ 2\,691\,000&=\frac{2\,344\,000}{3}+b \\ \frac{8\,073\,000}{3}-\frac{2\,344\,000}{3}&=b \\ b&=\frac{5\,729\,000}{3} .\end{aligned} $$ With $b=\frac{5\,729\,000}{3}$, then the linear equation that gives the population $t$ years since $2000$ is $$ y=\frac{234\,400}{3}t+\frac{5\,729\,000}{3} .$$
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