Answer
$w=5h+100$
Work Step by Step
Let $w$ be the weight in pounds and $h$ be the height in inches above $5$ feet. A linear equation that gives the optimal weight of women based on the number of inches above $5$ feet takes the form of
$$
w=mh+b,
$$
where $m$ is the slope and $b$ is the $y$-intercept.
Since the optimal weight at $5$ feet (i.e. $0$ inches more than $5$ feet) is $100$ pounds, this can be represented by the ordered pair $(h_1,w_1)=(0,100)$. Since the optimal weight at $5.5$ feet (i.e. $6$ inches more than $5$ feet) is $130$ pounds, this can be represented by the ordered pair $(h_2,w_2)=(6,130)$.
The formula for finding the slope, $m$, of the line passing through two points, $(h_1,w_1)$ and $(h_2,w_2)$ is given by $m=\frac{w_1-w_2}{h_1-h_2}$. That is,
$$\begin{aligned}
m&=\frac{w_1-w_2}{h_1-h_2}
\\&=
\frac{100-130}{0-6}
\\&=
\frac{-30}{-6}
\\&=
5
.\end{aligned}
$$With $m=5$, then the linear equation that gives the optimal weight of women based on the number of inches above $5$ feet takes the form of
$$
w=5h+b
.$$Since the line passes through the point $(0,100)$, substitute $h=0$ and $w=100$ in the equation above to solve for $b$. That is,
$$\begin{aligned}
w&=5h+b
\\
100&=5(0)+b
\\
100&=b
.\end{aligned}
$$With $b=100$, then the linear equation that gives the optimal weight of women based on the number of inches above $5$ feet is
$$
w=5h+100
.$$
