# SOLUTION: The system of linear equations has a unique solution. Find the solution using Gaussian elimination or Gauss-Jordan elimination. x + y + 6z = 3 x + y + 3z = 3 x + 2y+ 4z = 7

Algebra ->  Algebra  -> Matrices-and-determiminant -> SOLUTION: The system of linear equations has a unique solution. Find the solution using Gaussian elimination or Gauss-Jordan elimination. x + y + 6z = 3 x + y + 3z = 3 x + 2y+ 4z = 7       Log On

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 Question 48491This question is from textbook College Algebra : The system of linear equations has a unique solution. Find the solution using Gaussian elimination or Gauss-Jordan elimination. x + y + 6z = 3 x + y + 3z = 3 x + 2y+ 4z = 7 I know this has to do with combining rows and eliminating variables until only one is left, then back-substituting into the system to find the remaining variables, but I honestly don't know where to begin the elimination. Thank you so very much for all your help!This question is from textbook College Algebra Answer by AnlytcPhil(1276)   (Show Source): You can put this solution on YOUR website!```x + y + 6z = 3 x + y + 3z = 3 x + 2y + 4z = 7 I will assume you have had matrices. If not, post again and I will show you how to solve it without matrices: The augmented matrix is [1 1 6 | 3] [1 1 3 | 3] [1 2 4 | 7] We need to get 0's in the three lower left positions, that is, in the positions below the upper left to lower-right diagonal: To get a 0 where the 1 is in row 2 column 1 is, we multiply row 1 temporarily by -1 and add it to 1 times row 2. This is easy to do mentally if you write -1 to the left of row 1 and 1 left of row 2: -1[1 1 6 | 3] 1[1 1 3 | 3] [1 2 4 | 7] [1 1 6 | 3] [0 0 -3 | 0] [1 2 4 | 7] To get a 0 where the 1 is in row 3 column 1 is, we multiply row 1 temporarily by -1 and add it to 1 times row 3. This is easy to do mentally if you write -1 to the left of row 1 and 1 left of row 3: -1[1 1 6 | 3] [0 0 -3 | 0] 1[1 2 4 | 7] [1 1 6 | 3] [0 0 -3 | 0] [0 1 -2 | 4] To get a 0 where the 1 is in row 3 column 2 is, we merely need to swap rows 2 and 3 [1 1 6 | 3] [0 1 -2 | 4] [0 0 -3 | 0] Now we need to get 1's on the upper left to lower right diagonal. Two of the diagonal elements are already 1's. To get a 1 where the -3 is, we multiply row 3 by -1/3 [1 1 6 | 3] [0 1 -2 | 4] -1/3[0 0 -3 | 0] [1 1 6 | 3] [0 1 -2 | 4] [0 0 1 | 0] This is the system: 1x + 1y + 6z = 3 0x + 1y - 2z = 4 0x + 0y + 1z = 0 or x + y + 6z = 3 y - 2z = 4 z = 0 The bottom equation tells us that z = 0 Substitute z = 0 into the 2nd equation x + y + 6z = 3 y - 2z = 4 y - 2·0 = 4 y = 4 Substitute z = 0 and y = 4 into the 1st equation x + 4 + 6·0 = 3 x + 4 + 0 = 3 x = -1 So the solution is (x, y, z) = (-1, 4, 0) Edwin```