Answer
$y=104\,500t+5\,383\,500$
Work Step by Step
Let $t$ be the number of years after the year $2000$ and $y$ be the population of Washington State. A linear equation that gives the population of Washington 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 $2015$ (or $15$ years after $2000$), the population is projected at $6951$ thousand, this can be represented by the ordered pair $(t_1,y_1)=(15,6\,951\,000)$. Since in $2025$ (or $25$ years after $2000$), the population is projected at $7996$ thousand, this can be represented by the ordered pair $(t_2,y_2)=(25,7\,996\,000)$.
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,
$$\begin{aligned}
m&=\frac{y_1-y_2}{t_1-t_2}
\\&=
\frac{7\,996\,000-6\,951\,000}{25-15}
\\&=
\frac{1\,045\,000}{10}
\\&=
104\,500
.\end{aligned}
$$
With $m=104\,500$, then the linear equation that gives the population $t$ years since $2000$ takes the form of
$$
y=104\,500t+b
.$$
Since the line passes through the point $(15,6\,951\,000)$, substitute $t=15$ and $y=6\,951\,000$ in the equation above to solve for $b$. That is,
$$\begin{aligned}
y&=104\,500t+b
\\
6\,951\,000&=104\,500(15)+b
\\
6\,951\,000&=1\,567\,500+b
\\
6\,951\,000-1\,567\,500&=b
\\
b&=5\,383\,500
.\end{aligned}
$$
With $b=5\,383\,500$, then the linear equation that gives the population $t$ years since $2000$ is
$$
y=104\,500t+5\,383\,500
.$$
