Simple 4 equation 4 variable with missing and other variables

The main page for solving systems of linear equations on the TI-85 and TI-86.
The previous example page covers a Simple 3 equation 3 variable situation, with missing variables.
The next example page covers a Simple 2 equation 2 variable situation, where the lines are parallel.


WARNING: The TI-85 and TI-86 are almost identical in their use of the SIMULT function. The major difference is the labels that are on certain keys. On the TI-85, SIMULT is the 2nd function on the key, whereas on the TI-86 SIMULT is the 2nd function on the key. When a difference is important it will be presented in the text below. The exception to this is the "3" key. On the TI-85 it appears as , while on the TI-86 it is . To save some space, and to ignore this difference, the numeric keys (the gray ones) have been changed here to only show the key face, as in . In addition, the key will be shown as , again to save space.


The problem we will use on this page has both non-standard equations and missing variables. These topics have been covered in earlier pages. Therefore, we will move through the problem fairly quickly. The original statement of the problem is
8r + 37 = 3n + 5s
6m + 7s = 16
2n - 3r = 60 - 9m
2r - 6s = 38 -8m

>Our first step is to identify the variables. In this case, the variables are r, n, s and m. Second we need to decide on the order of the varaibles, and the traditional m, then n, then r, then s seems reasonable. Third, we can rewrite the equations, with the variables in order, and not leaving any out. We will add terms with the coefficient set to 0 to make up for any missing terms. The rewrite of the problem produces

0m - 3n + 8r - 5s = -37
6m + 0n + 0r +7s = 16
9m + 2n - 3r + 0s = 60
8m + 0n + 2r - 6s = 38

By adding the extra terms (having 0 as the coefficient) we have not changed any equation but we have changed the form of the equation so that it fits the same pattern that we have been using in the earlier pages. We are ready to use the calculator.
Figure 1
The keystrokes to start this process are the same on the two calculators, although the keys have a different name. For the TI-85 we start with and , but for the TI-86 we start with and . On either calculator this selects the "SIMULT" function. The calculator responds with a request for the value of "Number" as shown in Figure 1. The SIMULT function expects to have exactly the same number of equations as we have variables. For our problem, we have 4 variables and 4 equations. Therefore we respond with the key to complete Figure 1.
Figure 2
We leave Figure 1 by pressing the key. That will cause the display to change to Figure 2. Notice in Figure 2 that the calculator is requesting values for each of the coefficients and constants that we have in the general standard form for our first equation. The first subscript on each of the "a's" and the subscript on the "b" indicates that we are looking at values for the first equation.

Remember that we need to put the values in according to the standard form. Therefore we want the values 0, -3, 8, -5, and -37. The key sequence accomplishes this and leaves the display as in Figure 2.

Figure 3
We move from Figure 2 to Figure 3 by first pressing the key. Here we need to enter the coefficients and constants for the second equation, in our standard form, namely 6, 0, 0, 7, and 16. We use the keys to complete the image of Figure 3.

Now we can move to the next screen by pressing the key.

Figure 4
Here we need to enter the coefficients and constants for the third equation, in our standard form, namely 9, 2, -3, 0, and 60. We use the keys to complete the image of Figure 4.

Now we can move to the next screen by pressing the key.

Figure 5
For our final data entry screen for this problem, we need to enter the coefficients and constants for the fourth equation, in our standard form, namely 8, 0, 2, -6, and 38. We use the keys to complete the image of Figure 5.

We are ready to get the solution.

Figure 6
We request a solution by pressing the key. The calculator determines the correct answer and displays it as Figure 5. Note that x2 was calculated as -2.99999999999. Expecting the answers to be "nice" and realizing that calculators are not exact, we will assume that x2 is really -3.

Once again we need to return to our standard form and recognize that x1 is m, x2 is n, x3 is r, and x4 is s. Therefore, we have a unique solution to all three equations when m=5, n=-3, r=-7, and s=-2.

The main page for solving systems of linear equations on the TI-85 and TI-86.
The previous example page covers a Simple 3 equation 3 variable situation, with missing variables.
The next example page covers a Simple 2 equation 2 variable situation, where the lines are parallel.

©Roger M. Palay
Saline, MI 48176
October, 1998