College Algebra, Chapter 1, 1.3, Section 1.3, Problem 92

Christian drove from Tortula to Cactus, a distance of $250 mi$. He increased his speed by $10 mi/hr$ for the $360 mi$ trip from Cactus to Dry Junction. If the total trip took $11 h$, what was his speed from Tortula to Cactus?

Recall that the formula for speed is $\displaystyle V = \frac{d}{t}$, so $\displaystyle t = \frac{d}{v}$. Let $t_1$ be the time consumed from Tortula to Cactus and $t_2$ be the time from Cactus to Dry Junction. Thus,


$
\begin{equation}
\begin{aligned}

t_T =& t_1 + t_2
&& \text{Model}
\\
\\
t_T =& \frac{d_1}{V_1} + \frac{d_2}{V_2}
&& \text{Substitute the formula for } t
\\
\\
11 =& \frac{250}{V_1} + \frac{360}{V_1 + 10}
&& \text{Substitute the given}
\\
\\
11 =& \frac{250(V_1 + 10) + 360 (V_1)}{V_1 (V_1 + 10)}
&& \text{Take the LCD}
\\
\\
11 =& \frac{250 V_1 + 2500 + 360V_1}{V_1 (V_1 + 10)}
&& \text{Simplify}
\\
\\
11V_1^2 + 110 V_1 =& 610V_1 + 2500
&& \text{Simplify the numerator and apply cross multiplication}
\\
\\
11V_1^2 - 500V_1 - 2500 =& 0
&& \text{Combine like terms and subtract } 2500
\\
\\
(V_1 - 50) \left(V_1 + \frac{50}{11} \right) =& 0
&& \text{Factor}
\\
\\
V_1 - 50 =& 0 \text{ and } V_1 + \frac{50}{11} = 0
&& \text{ZPP}
\\
\\
V_1 =& 50 \text{ and } V_1 = \frac{-50}{11}
&& \text{Solve for } V_1
\\
\\
V_1 =& 50 \frac{mi}{hr}
&& \text{Choose } V_1 > 0

\end{aligned}
\end{equation}
$


Thus, Christian's speed from Tortula to Cactus is $V_1 = 50 mi/hr$.