Create a flow chart for the equation. Consider n as the input and 8 as the output.
b.
Work backwards from your flow chart to find the value of n that produces 8 as an output.
This method of solving equations is called backtracking. Backtracking involves "undoing" operations to work backwards from the output to the input.
Video Segment In this video segment, Professor Cossey draws a flow chart for the equation in Problem B2, then demonstrates the method of backtracking. You can choose to do Problem B3 before or after watching the video segment. You can first try to do the problem on your own, then use the video segment to reinforce what you've learned. Or you can watch the video segment before doing the problem to help you get started making flow charts or doing the method of backtracking.
How are flow charts similar to the function machines you created in Session 3, Part C?
You can find this segment on the session video, approximately 9 minutes and 9 seconds after the Annenberg Media logo.
Can you find an equation that cannot be solved by backtracking?
Think about whether a function machine's operation could always be "undone." You might also look for problems in Part C that would be difficult to solve by backtracking. Close Tip
Look at the toothpick pattern below.
One of the stages needs 112 toothpicks to form the pattern. Can you use backtracking to find out which stage it was?
A similar problem appears in Session 2, Part B. If you find a formula for the number of toothpicks at a given stage, you can backtrack using that formula. Close Tip
Some equations lend themselves to a process called covering up. Covering up takes a complex equation and changes it into a series of one-step equations. For example, let's say we wanted to solve the equation
A solution by covering up would begin by covering the most complicated expression in the equation (in this case, 21 / (x + 1) is the expression). Then the equation reads (covered) - 6 = 1, an equation that is solved quickly. Now we know that 21 / (x + 1) = 7.
To continue, cover up the most complicated expression in the new equation, which is x + 1. The equation reads 21 / (covered) = 7. Now we know that x + 1 = 3, so x must be 2.
On Monday, the produce manager stocked his store's display case with 80 heads of lettuce. By the end of the day some heads of lettuce had been sold. On Tuesday, the manager counted the number of heads of lettuce that were left and decided to add an equal number of heads of lettuce, thereby doubling the leftovers. By the end of the day he had sold the same number of heads of lettuce as on Monday.
On Wednesday, the manager decided to triple the number of heads of lettuce that had been left in the case. He sold the same number of heads of lettuce that day, too. At the end of the day, though, there were no heads of lettuce left.
How many were sold each day?
Describe the strategies you used to solve this problem.
Which of the methods in this part would be useful here? What would be a useful variable? Don't forget: The same number of heads of lettuce are sold each day. Close Tip
Problem B10 taken from the Math Forum Project: Algebra Problem of the Week, posted March 29, 1999. Available online at http://mathforum.com.
