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Solve the system using the Row-Echelon method. (Round your answer to decimal place.) 2X1 - 11X2 + 7X3 + 9X4 = -16 -6X1 + 7X2 - 12X3 + 11X4 = 37 2X1 - X2 + X3 - 8X4 = -2 10X1 - 9X2 - 7X3 - 3X4 = 69 X1 = Number X2 = Number X3 = Number X4 = Number

          Solve the system using the Row-Echelon method. (Round your answer to
decimal place.)
2X1 - 11X2 + 7X3 + 9X4 = -16
-6X1 + 7X2 - 12X3 + 11X4 = 37
2X1 - X2 + X3 - 8X4 = -2
10X1 - 9X2 - 7X3 - 3X4 = 69
X1 = Number
X2 = Number
X3 = Number
X4 = Number
        
Show more…
Solve the system using the Row-Echelon method. (Round your answer to
decimal place.)
2X1 - 11X2 + 7X3 + 9X4 = -16
-6X1 + 7X2 - 12X3 + 11X4 = 37
2X1 - X2 + X3 - 8X4 = -2
10X1 - 9X2 - 7X3 - 3X4 = 69
X1 = Number
X2 = Number
X3 = Number
X4 = Number

Added by Kimberly G.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Solve the system using the Row-Echelon method.(Round your answer to decimal place.) 2X1-11X2+7X3+9X4=-16 -6X1+7X2-12X3+11X4=37 2X1-X2+X3-8X4=-2 10X1-9X2-7X3-3X4=69 X1= Number X2 = Number X3 Number X4 - Number
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Transcript

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00:02 So in this problem, we want to solve the system of linear equations.
00:06 X1 minus 3x2 plus 4x3 is equal to negative 4.
00:12 3x1 minus 7x2 plus 7x3 is negative 8, and negative 4x1 plus 6x2 minus x3 is 7.
00:21 So a trick to solving systems of linear equations is to try and eliminate terms.
00:27 So we can eliminate the term x1 for a bit by multiplying this by three and then subtracting it from this.
00:38 See, if we multiply this by three, then we'll have three x1s, and then when we subtract, we won't have any more x1 terms anymore.
00:48 So once we have subtracted the second line from three times the first line, then the left -hand side will.
00:58 We've already eliminated this because 3x1 minus 3x1 is 0.
01:03 So we're left with negative 7x2 plus 7x3 minus minus 3 times negative 3x2 plus 4x3.
01:17 And we know from the right hand side that that will equal to negative 8 minus 3 times negative 4.
01:25 So simplifying that we get a new equation we can use...
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