Precalculus, Chapter 8, 8.1, Section 8.1, Problem 67

rsWe began by stating our equations as follows:
3x - 2y = -27
x + 3y = 13
(NB: Note that I wrote the values of the x variable below each other, values with the y variable below each other and the constants below each other. Also note that the sign before each number is also part of the variable.)
Now we will put the equations in matrix form:

[[3,-2 : -27],[1,3 : 13]]
Now swap Row 1 with Row 2: We want the numbers in the main diagonal to have the number '1':
[[1,3 : 13],[3 ,-2 : -27]]
In Gaussian elimination we want all values below the main diagonal to equal zero. Hence, we shall subtract 3 to Row 3:
[[1, 3 : 13],[0 ,-5 : -30]]
Now we want the main diagonal to contain the number '1', hence we divide row 2 by -5:
[[1,3 : 13],[0, 1: 6]]
Now that we have the number 1 across the main diagonal and zero's below the main diagonal we can perform back substitution by substituting the value of y=6 into the first equation:
x + 3y = 13
y = 6
Therefore x = -5
Lets double check our answers
3*-5 + -2*6 = -15 - 12 = -27
-5 + 3 *6 = -5 + 18 = 13