College Algebra, Chapter 1, 1.3, Section 1.3, Problem 94

Two ships $A$ and $B$ depart a harbor at the same time, $A$ is traveling east, the other south. $A$ travels $3 mi/hr$ faster than $B$. After two hours, the ships are $30 mi$ apart. Find the speed of $B$.

Let $x$ be the speed of $B$, so $x + 3$ will be the speed of $A$.

So the distance of $B$ after 2 hours is $2x$, while $A$ is $2(x + 3)$.







By using Pythagorean Theorem,


$
\begin{equation}
\begin{aligned}

(2x)^2 + (2(x + 3))^2 =& (30)^2
&& \text{Model}
\\
\\
4x^2 + (2x + 6)^2 =& 900
&& \text{Distribute 2}
\\
\\
4x^2 + 4x^2 + 24x + 36 =& 900
&& \text{Expand}
\\
\\
8x^2 + 24x + 36 =& 900
&& \text{Combine like terms}
\\
\\
8x^2 + 24x - 864 =& 0
&& \text{Subtract 900}
\\
\\
x^2 + 3x - 108 =& 0
&& \text{Divide both sides by 8}
\\
\\
(x + 12)(x - 9) =& 0
&& \text{Factor}
\\
\\
x + 12 =& 0 \text{ and } x - 9 = 0
&& \text{ZPP}
\\
\\
x =& -12 \text{ and } x = 9
&& \text{Solve for } x
\\
\\
x =& 9 mi/hr
&& \text{Choose } x > 0

\end{aligned}
\end{equation}
$


The speed of $B$ is $9 mi/hr$.