# Math Help - system of equations

1. ## system of equations

hello everyone!

I have some questions about solving a system of equations by the Gaussian and Gauss-Jordan elimination methods.

1) how do i determine which method should i use for each system of equations?

2) using any of the two methods, do i have to do it by hand or is there a quicker way to do it (for example using a calculator)? If there is,how can i use my calculator for this? my calculator is a TI-84 plus.

3) how do i solve a system of equations with only the x variable. The variables for one of my systems of equations are x1, x2,x3,x4? Do i treat it as a system with x,y and z?

thank you so much

2. Originally Posted by kithy
hello everyone!

I have some questions about solving a system of equations ....

...

2) ... or is there a quicker way to do it (for example using a calculator)? If there is,how can i use my calculator for this? my calculator is a TI-84 plus.

3) how do i solve a system of equations with only the x variable. The variables for one of my systems of equations are x1, x2,x3,x4? Do i treat it as a system with x,y and z?

thank you so much
I'm going to show you how you can use your calculator:

1. You have a system of simultaneous equations:

$\left|\begin{array}{l}x_1+2x_2 = 3 \\ 4x_1+5x_2 = 6\end{array}\right.$

2. Go to MATRIX --> EDIT --> choose a matrix name (in my example I've taken 2:[ B ]) --> ENTER

3. You'll get an input screen: First type in the numbers of rows and afterwards the number of columns. The cursor is placed automatically on the correct place in the matrix.

4. Type in the coefficients of the variables + ENTER: The cursor will move one place ahead. After the last value + ENTER you quit this menu.

5. Go to MATRIX --> MATH --> B: rref( (+ ENTER of course)

6. Goto MATRIX --> NAMES --> [ B ] (+ ENTER of course)

7. Don't forget the finishing bracket and (+ ENTER of course)

8. You'll get the result screen: In the first row there is only a 1 (indicating one $x_1$) and the corresponding value. So this row reads $x_1 = -1$
In the second row there is no $x_1$ but one $x_2$ and the corresponding value. So this row reads $x_2 = 2$