Complex 3 equations in 2 variables, non-intersecting

The main page for solving systems of linear equations on the TI-85 and TI-86.
The previous example page covers a Simple 4 equation 4 variable situation, with no unique solution.
The next example page covers a Complex 3 equation 2 variable situation, where 2 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 is
-2x + y = 0
4x - 9y = -14
1x + 3y = 28
Having looked at the previous examples of using the "SIMULT" function on the calculator, we should recognize that this problem is different from all of the earlier problems. In those problems we always had exactly the same number of equations as we had variables. For example, in the previous page we had 4 equations in 4 variables. Here we have just two variables, but we have 3 equations.

The "SIMULT" function always expects the same number of equations as we have variables. In a real sense, we have an extra equation here. If we only had two equations then we could use "SIMULT" to find their point of intersection. Let us move ahead in that direction. At first we will restrict our analysis to the first two equations:

-2x + y = 0
4x - 9y = -14

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 restricted version of the problem, we have 2 variables and 2 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. In that figure we have also entered the desired values, -2, 1, and 0, via the and keys.
Figure 3
We leave Figure 2 by pressing the key. That will cause the display to change to Figure 3. In that figure we have also entered the desired values, 4, -9, and -14, via the and keys. After pressing those keys the screen should appear as in Figure 3.
Figure 4
At this point we are ready to ask the calculator to solve the problem. We press the key, and the calculator responds with the solution as shown in Figure 4. The solution, x1=1 and x2=2, translates, in the restricted version of the equations as x=1 and y=2. This is the point of intersection for the first two equations.

It is nice to have a solution to the first two equations. Having that intersection means that the point (1,2) solves both of the first two equations. If we were to graph the first two equations, they would cross at the point (1,2). What about the third equation? If we were to graph the third equation,

1x + 3y = 28
would it cross the other two at the same point? One way to answer this is to try the values (1,2) in the third equation. If it works, then (1,2) will solve all three equations and it would be the solution to the problem. However, 1(1)+3(2) is 8 not 28. Therefore, (1,2) does not satisfy the third equation, and there is no unique solution to the system of three equations
-2x + y = 0
4x - 9y = -14
1x + 3y = 28
.

Another way to see this is to re-use the "SIMULT" operation to see where the third equation intersects with the second equation. If the calculator were to determine that the third intersects the second at (1,2) then that would be a common solution to all three equations and, therefore, it would be a solution. The keystrokes and screens needed to do this are given below.

Figure 5
We had the solution to the first two equations displayed in Figure 4. Now we want to return to the data entry screen. To do this we select the "COEFS" command from the menu by pressing the key. The result is shown in Figure 5. The calculator has returned to the screen where we enter the coefficients and constant for the first equation. The blinking cursor is covering the minus sign in the -2. Note that we can tell that this is the first equation by looking at the subscripts of the a's and the b. The first subscript of each a is 1 and the subscript of the b is 1. This indicates that we are looking at the first equation.
Figure 6
Now we need to enter the coefficients and the constant for the third equation, namely, 1, 3, and 28. We do this via the keys and . The result is shown in Figure 6.
Figure 7
Figure 6 shows that we have entered the values of the third equation. The coefficients for the second equation are already in the machine. Therefore, we need only select the "SOLVE" command from the menu, via the key, to have the calculator find the intersection of the third and the second equations. Figure 7 shows that those two equations have the point (10,6) in common. Since this is not the point where the first and the second equations cross, we can see that the three equations do not have a single point in common.

At this point we have shown in two ways that the three original equations do not have a single point in common, that they do not have a unique solution. For completeness, the presentation below verifies this yet again by looking at the intersection of the first and the third equations.

Figure 8
We leave Figure 7 by selecting the "COEFS" menu item via the key. This returns the calculator to the data entry screen shown in Figure 8. The blinking cursor is covering the value 1. We recognize these values as the coefficients of the third equation. We want these values. Therefore, we will leave them here and move to the next screen to re-enter the coefficients and constant for the original first equation. We do this by selecting the "NEXT" menu item via the key.
Figure 9
Figure 9 shows the values for the second equation. We know that the previous screen held the coefficents and constants for the third equation. We want to replace the values shown in Figure 9 with the values for the first equation.
Figure 10
We re-enter the first equation values via the and keys. Figure 10 shows those values. Agan, the calculator is ready to solve the system of two equations, this time for the third and the first equation. We press to select the "SOLVE" menu item.
Figure 11
Figure 11 shows the result of the computation. The third and first equations cross at (4,8). Again, this demonstrates that the three equations do not have a single common point of intersection.

The main page for solving systems of linear equations on the TI-85 and TI-86.
The previous example page covers a Simple 4 equation 4 variable situation, with no unique solution.
The next example page covers a Complex 3 equation 2 variable situation, where 2 lines are parallel.

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