# Ben's boat will take 1½ hours longer to go 12 miles up a river than to return. What is the rate of his boat if the rate of the current is 2 miles an hour?

Rate of the boat $v = 6 \text{ }$miles/hour

#### Explanation:

From the given data:
Let $d = 12$ miles
Let $t =$time going down with the river current
Let $v =$the rate of Ben's boat
Let ${v}_{c} = 2$ miles/hour-rate of the current

Let $v - {v}_{c} =$the rate going up the river
Let $v + {v}_{c} =$the rate going down the river

$\left(v + {v}_{c}\right) \left(t\right) = 12 \text{ }$ the equation for distance going down
$\left(v - {v}_{c}\right) \left(t + \frac{3}{2}\right) = 12 \text{ }$the equation for distance going up

Let us do the simultaneous solution of these equations

From $\left(v + {v}_{c}\right) \left(t\right) = 12 \text{ }$ the equation for distance going down

$\left(v + {v}_{c}\right) \left(t\right) = 12 \text{ }$

$t = \textcolor{red}{\frac{12}{v + {v}_{c}} \text{ }}$substitute this in the equation for distance going up

$\left(v - {v}_{c}\right) \left(t + \frac{3}{2}\right) = 12 \text{ }$equation for distance going up
$\left(v - {v}_{c}\right) \left(\textcolor{red}{\frac{12}{v + {v}_{c}}} + \frac{3}{2}\right) = 12 \text{ }$
Now change ${v}_{c} = 2$

$\left(v - 2\right) \left(\textcolor{red}{\frac{12}{v + 2}} + \frac{3}{2}\right) = 12 \text{ }$

We can now solve for v:
Multiply both sides by (v+2)

$\left(v - 2\right) \left(\frac{12}{v + 2} + \frac{3}{2}\right) \left(v + 2\right) = 12 \left(v + 2\right)$

$\frac{12 \left(v - 2\right) \left(v + 2\right)}{v + 2} + \frac{3}{2} \cdot \left(v - 2\right) \left(v + 2\right) = 12 v + 24$

Simplify

$\frac{12 \left(v - 2\right) \cancel{v + 2}}{\cancel{v + 2}} + \frac{3}{2} \cdot \left(v - 2\right) \left(v + 2\right) = 12 v + 24$

$12 \left(v - 2\right) + \frac{3}{2} \cdot \left(v - 2\right) \left(v + 2\right) = 12 v + 24$
Expand
$12 v - 24 + \frac{3}{2} \left({v}^{2} - 4\right) = 12 v + 24$

Simplify

$\frac{3}{2} \left({v}^{2} - 4\right) = 48$

${v}^{2} - 4 = 32$

${v}^{2} = 32 + 4$

${v}^{2} = 36$

$v = 6 \text{ }$miles/hour