Question 1085270
<font color="black" face="times" size="4">Let 
x = Brian's swimming speed in still water (no current)
t1 = time it takes to swim against the current
t2 = time it takes to swim with the current


Make a table to get
<table border=1 cellpadding=3><tr><td></td><td>Distance</td><td>Rate</td><td>Time</td></tr><tr><td>Against Current</td><td>2</td><td>x-4</td><td>t1</td></tr><tr><td>With Current</td><td>2</td><td>x+4</td><td>t2</td></tr></table>
Against the current, we can say


d = r*t
2 = (x-4)*t1
t1 = 2/(x-4)


with the current, we can say


d = r*t
2 = (x+4)*t2
t2 = 2/(x+4)


The two time values (t1 and t2) add to 3 hours, so


t1 + t2 = 3
2/(x-4) + 2/(x+4) = 3


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Let's solve for x


2/(x-4) + 2/(x+4) = 3


[2/(x-4) + 2/(x+4)](x-4)(x+4) = 3(x-4)(x+4) ... <font color=blue>multiply both sides by (x-4)(x+4)</font>


2(x+4)+2(x-4) = 3(x-4)(x+4)


2x+8 + 2x-8 = 3(x^2-16)


4x = 3x^2-48


4x-4x = 3x^2-48-4x


0 = 3x^2-4x-48


3x^2-4x-48 = 0


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Use a calculator or the quadratic formula to solve 3x^2-4x-48 = 0 to get the solution set {x = -3.388508353532, x = 4.721841686865} which are approximate values.


Toss out the negative x value because negative speeds don't make sense.


The only practical solution is approximately <font color=red>4.721841686865</font> which is the final answer. The units are in km per hour.


So this means that Brian's speed in still water (with no current) is approximately 4.721841686865 km/hr.</font>