Chapter 1 Section 0 Example 3 on the TI-83

Note that the TI-83 and the TI-83 plus have slightly different keys. This page uses the keys associated with the TI-83. The differences are that the TI-83 key is replaced by the TI-83 Plus key, and the TI-83 key is replaced by the TI-83 Plus key.

Example 3 in the textbook gives the problem:

If x =

c – ab

a – b

find the value of the expression a(x+b)

The book example produces an answer by using the appropriate algebraic manipulations. The result is another expression, namely

a ( c – b² )

a – b

Although we can not do the symbolic manipulations on the TI-83 calculator, we can use that machine to check our work. To do this we will choose some arbitrary values for a, b, and c. Then, we will determine the value of x from the expression (c-ab)/(a-b). Knowing the value of x we will determine the value of a(x+b). Then we can determine the value of the algebraic answer to our problem.

Figure 1	We start by assigning some arbitrary values to A, B, and C. We use "wierd" values so that we are less likely to produce accidentally identical answers even if we had an error. Using values such as 1, 0, and 2 would not give a good test of our work. In any case, the calculator is going to do the arithmetic! We do not care if it has to work a little harder to use wierd values.
Figure 2	In Figure 2 we find the value of the expression *(C-AB) / (A-B) Remember that we need to explicitly include the multiplication of A and B. Then, we assign the value, 2.647593177, to the variable X. The keystrokes needed to do this could be as short as . If we start a line with the "store" command then the calculator automatically supplies the Ans variable. And, we conclude Figure 2 by finding the value of A(X+B)**, namely, 29.5811932.
Figure 3	The answer should be A(C-B²) / (A-B) Figure 3 concludes with a test of that expression, and it does evaluate to the correct 29.5811932 We have used the calculator to show that the original problem and our answer produce the same answer when we evaluate both expressions with the same "wierd" values for the variables. This does not prove that we have the correct answer, but it does give us much more confidence that we have not made some error in our algebraic manipulations. Had we made an algebraic error, then it is most likely that the original problem and our "incorrect" answer would have produced different answers.