 Let us look now at intelligent guessing and testing. This much maligned trial and error strategy is often loved by students yet loathed by many teachers. However, this is a legitimate strategy that merits our attention. In fact, this is not about taking wild guesses but about estimating an answer and then verifying our guess to refine our approximation until we arrive at the answer. Often this strategy is not efficient, but sometimes it is a good start. Indeed, when you might lack appropriate mathematical tools to solve a problem, intelligent guessing and testing is godsend. Let us look now at the following problem. Bernie has 40 yards of 6 foot high cyclone fence and wants to build two separate chicken coops and closing a total area of 58 square yards. How much fence does he need for each pin if Bernie wants the coops to have square bases? Let X be the side of the base of one square pin and let Y be the side of the base of the other square pin. Since we have 40 yards of cyclone fence, the sum of the perimeters of both pins must be 40. Likewise, the sum of the areas of both pins must be 58 square yards. We can solve this system of equations but students might encounter some problems. They may not be familiar with solving systems of equations, especially one that has a linear equation and a quadratic equation. What to do? Let's use some intelligent guessing and testing. This equation, when you divide it by 4, becomes X plus Y equals 10. Let us try then two numbers that add up to 10. Let's start with 5 and 5. 7 and 5 will be here 25 plus 25, that's not 58. So let us try 6 and 4. Will that work? Well, this is 36 and 16. Still doesn't work. Why don't we try 7 and 3? 7 squared is 49, 3 squared is 9, that's it, 58. So these are the dimensions we were looking for. We want a pin with side 3 yards and another one with side 7 yards. So the first one will have a perimeter of 12 yards and the second one will have a perimeter of 28 yards for a total of 40 yards. The area of the first one will be 9 square yards and the area of the second one will be 49 square yards. We can also use some algebra to confirm that this solution is unique. If we solve the first equation for X, then we can substitute that and see that this becomes X squared minus 10X plus 21 equals 0, so X minus 3 times X minus 7 is 0, so X equals 3 or X equals 7. When X is 3, Y is 7, when X is 7, Y is 3 and they are both covered in this case. Thank you.