Question 3604
Let's consider this problem geometrically. We'll call Joe's starting point J, Bob's starting point B and the point at which they meet we'll call M. The angle JMB is a right angle because Joe is walking north and Bob is walking east. We also know that the line from Joe's starting point to Bob's starting point is 5 miles. This line forms the hypotenuse of the right triangle JMB. We also know that the distance JM is one mile more than the distance MB. <br>
Now let's collect these facts:<br>
<ol>
<li>The triangle with verticies JMB is a right triangle</li>
<li>JB is the hypotenuse of the triangle and is 5 miles long</li>
<li>JM = MB + 1</li>
</ol>

For simplicity, let's call JM x and JB y. From the Pythagorean theorem, we know that {{{x^2 + y^2 = h^2}}}, this problem tells us that h = 5 and {{{x = y + 1}}}. This allow's us to restate the formula for a right triamgle as:<br>
{{{(y + 1)^2 + y^2 = 25}}}<br>
completing the square, we get {{{y^2 + 2y + 1 + y^2 = 25}}}<br>
adding like terms we get {{{2y^2 + 2y + 1 = 25}}}<br>
subtrating 25 from both sides of the equation from we get {{{2y^2 + 2y - 24 = 0}}}.
Notice that we can factor 2 our of the equation so {{{y^2 + y - 12 = 0}}}.
This equation factors to {{{(y + 4)(y - 3) = 0}}} so y = -4 or y = 3. A distance of -4 does not make sense, so we set y to 3 and this tells us that {{{x = y + 1 = 3 + 1 = 4}}}. Remember y is the distance of the line MB which is the distance Bob walked and x is the distance of the line JM, which is the distance Joe walked.