Complex 3 equations in 2 variables, unique solution

The main page for solving systems of linear equations on the TI-85 and TI-86.
The previous example page covers a Complex 3 equation 2 variable situation, where 2 lines are parallel.
The next example page covers a Complex 3 equation 2 variable situation, where two lines are identical and there is a solution.

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 3x + 4y = 43
5x - 2y = 11
11x - 13y = -36 As was the case in the previous example, 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: 3x + 4y = 43
5x - 2y = 11

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, 3, 4, and 43, 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, 5, -2, and 11, 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=5 and x₂=7, translates, in the restricted version of the equations as x=5 and y=7. 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 (5,7) solves both of the first two equations. If we were to graph the first two equations, they would cross at the point (5,7). What about the third equation? If we were to graph the third equation,

11x - 13y = -36 would it cross the other two at the same point? One way to answer this is to try the values (5,7) in the third equation. If it works, then (5,7) will solve all three equations and it would be the solution to the problem. In fact, 11(5)-13(7) is -36. Therefore, (5,7) does satisfy the third equation, and it is the unique solution to the system of three equations 3x + 4y = 43
5x - 2y = 11
11x - 13y = -36 .

Another way to see this is to re-use the "SIMULT" operation to see where the third equation intersects with the second equation. 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 3.

Figure 6 Now we need to enter the coefficients and the constant for the third equation, namely, 11, -13, and -36. However, the 5 and the -3 are already in place. Therefore, we can use the down arrow key to move to the b₁ entry. At that point we can enter the value 17. 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 (5,7) in common. Since this is the point where the first and the second equations cross, we can see that the three equations do have a single point in common.

At this point we have shown in two ways that the three original equations do have a single point in common, that they do 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. We know that these equations each intersect the second at (5,7). Therefore, it must be the case that the first and third intersect at (5,7). Let us see what the calculator produces.

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 first 1 in the value 11. 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 do have a single point of intersection at (5,7). This is exactly the result that we expect.

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, 3, 4, and 43, 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, 5, -2, and 11, 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=5 and x₂=7, translates, in the restricted version of the equations as x=5 and y=7. This is the point of intersection for the first two equations.

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 3.
Figure 6	Now we need to enter the coefficients and the constant for the third equation, namely, 11, -13, and -36. However, the 5 and the -3 are already in place. Therefore, we can use the down arrow key to move to the b₁ entry. At that point we can enter the value 17. 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 (5,7) in common. Since this is the point where the first and the second equations cross, we can see that the three equations do have a single point in common.

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 first 1 in the value 11. 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 do have a single point of intersection at (5,7). This is exactly the result that we expect.