Question 10887
Ok, let the distance traveled by the west-going boat = x miles.  Then the distance traveled by the north-going boat is (x+2) miles.

You can apply the Pythagorean theorem to this problem.

In the right triangle you get when you diagram the paths of these two boats, the hypotenuse is their distance from each other and that is 10 miles. So, c = 10 miles.  Now you can set up the equation per Pythagorus: {{c^2 = a^2 + b^2}}}

{{{10^2 = x^2 + (x+2)^2}}}

{{{100 = x^2 + (x^2 + 4x + 4)}}} Simplify. Subtact 100 from both sides and collect like-terms.

{{{2x^2 + 4x - 96 = 0}}} Factor out a 2 to simplify this.

{{{x^2 + 2x - 48 = 0}}} Factor this quadratic equation.

{{{(x - 6)(x + 8) = 0}}} 

{{{x - 6 = 0}}}, then {{{x = 6}}} or
{{{x + 8 = 0}}}, then {{{x = -8}}}  Discard this solution as the distance can't be negative.

So, x = 6 miles and x+2 = 8 miles.

The north-going boat traveled 8 miles while the west-going boat traveled 6 miles. 

You can see that 6, 8, and 10 are "Pythagorean numbers becuase {{{6^2 + 8^2 = 10^2}}}