1. We will use Gauss-Jordan elimination to determine the solution set of this linear system.

2. STEP 1: multiply the first row by -2 and add it to the second row. Do the same for the third row.

3. STEP 2: Multiply row 2 by 1/5 to get a 1 in the pivot position

4. STEP 3: Add the second row to the first to eliminate the entry above the pivot. Multiply the second row by -4 and add it to the third to eliminate the entry below the pivot

5. STEP 4: Multiply the third row by -5/14 to get a 1 in the pivot position

6. STEP 5: Multiply the third row by 4/5 and add it to the second row to eliminate the entry above the pivot. Multiply the third row by -6/5 and add it to the first row to do the same We now have a matrix in Reduced Row Echelon Form.

7. Let us suppose that the columns corresponded to variables x 1, x 2, x 3, and x 4. Then we have one free variable, x 4 We set it to an arbitrary parameter, m 1 And we have the following equations for one representation of this system's solutions x 4 = m 1 x 1 = 8/7 x 2 = -3/7 x 3 = 3/14 - 1/2m 1

8. x 4 = m 1 x 1 = 8/7 x 2 = -3/7 x 3 = 3/14 - 1/2m 1 We may now express the system's solutions as a set of vectors. Solutions are thus linear combinations of the above two vectors and for any value of our free variable m 1. That is, when we have free variables, there are infinitely many solutions to the system.