Chapter 2.0, Example 4 on the TI-83

The TI-83 does not do symbolic algebra. That is, we can not ask the TI-83 to do the problem of Chapter 2, Section 0, Example 4: Solve for y: x²y + z(^–y – 1) = 4(x + z) The text does illustrate the steps that we need to take to arrive at the solution

y =	4x + 5z

	x² – z

However, we can use the TI-83 to check our answer. In particular, we can choose some arbitrary values for x and z. Then, we can use those values to determine a value for y by evaluating the right side of the answer. Once we have a value for each of the variables x, y, and z, we can find the value of both the left and the right side of the original problem. If we have done the problem correctly, then, the value that we find for the left side must be the same as the value that we fnd for the right side. Note that we choose arbitrary, and not easy, even, nice values for x and z because we do not want to have trivial computations. The calculator is going to do the work and it does not care how easy it is to multiply or divide two values. The Figures below demonstrate one such check of the problem.

Figure 1 In Figure 1 we take the first three steps in the check of the problem. In this example we have chosen the arbitrary value of 2.053 to store into the variable X. We would do this via the keys . Next we have assigned .1589 to the variable Z via the keys . Finally, we use the answer of the problem,
y = 4x + 5z

x² – z

to generate a value for y, which we store in the variable Y via . The display indicates that the value assigned to Y was 2.22058796.
Figure 2 Now that we have values assigned to X, Y, and Z, we evaluate the two sides of the original equation. Thus, we evaluate x²y + z(^–y – 1) via , to get 8.8476.
And, we evaluate 4(x + z) via , which also produces 8.8476. This equality for the two sides gives us confidence that the problem was solved correctly.