SOLUTION: "Solve the system of linear equations, using row echelon form." Eq 1. 2x + 2z = 2 Eq 2. 5x + 3y = 4 Eq 3. 3y - 4z = 4 I am getting different answers everytime, while the bo

Algebra ->  Matrices-and-determiminant -> SOLUTION: "Solve the system of linear equations, using row echelon form." Eq 1. 2x + 2z = 2 Eq 2. 5x + 3y = 4 Eq 3. 3y - 4z = 4 I am getting different answers everytime, while the bo      Log On


   

Question 719021: "Solve the system of linear equations, using row echelon form."
Eq 1. 2x + 2z = 2
Eq 2. 5x + 3y = 4
Eq 3. 3y - 4z = 4
I am getting different answers everytime, while the book says the answers are (-4,8,5).
Help me please.
Found 2 solutions by solver91311, Edwin McCravy:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Show me the matrix you started with.

John

My calculator said it, I believe it, that settles it
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Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Eq 1. 2x + 2z = 2
Eq 2. 5x + 3y = 4
Eq 3. 3y - 4z = 4 

Line them up:

2x      + 2z = 2
5x + 3y      = 4
     3y - 4z = 4

2x + 0y + 2z = 2
5x + 3y + 0z = 4
0x + 3y - 4z = 4




Multiply the first row thru by 1%2F2,
abbreviated:  1%2F2R1->R1



Multiply the first row through by -5, add to
the second row, then restore the first row,
abbreviated   -5R1+R2->R2

 

Multiply the second row through by -1, add to
the third row, then restore the second row,
abbreviated   -R2+R3->R3 



Convert back to equations:

1x + 0y + 1z =  1
0x + 3y - 5z = -1
0x + 0y + 1z =  5

x + z =  1
3y - 5z = -1
z = 5

Substitute z = 5 into 

3y - 5z   = -1
3y - 5(5) = -1
  3y - 25 = -1
       3y = 24
        y = 8

Substitute z = 5 into

x + z = 1
x + 5 = 1
    x = -4

Edwin