solve the system using gaussian elimination. i need help filling in the blank

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Jul 26th, 2015

x + 2y + 3z = 11,.......................1

3x + 8y + 5z = 27.........................2

- x + y + z = 2...............................3

now

Multiply eq 1 by 3 and subtracting 2 from 1, we get

3x + 6y + 9z = 33

3x + 8y + 5z = 27
-     -       -        - 
-------------------------

- 2y + 4z = 6

- y + 2z = 3

y - 2z  = - 3 ................................4

Now multiply eq 3 by 3 and adding with eq 2, we get

3x + 8y + 5z = 27

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

  11y + 8z = 29 .........................5

Now multiplying eq 4 by 4 and adding with eq 5, we get

4y - 8z = - 12

11y + 8z = 29
------------------

15y = 17

y = 15/17

For z, put y = 15/17 in eq 4, we get

15/17 - 2z = - 3

15/17 + 3 = 2z

2z = (15 + 51) /17

2z = 66/17

z = 33/17

Now for x, put y= 15/17 and z = 33/17 in eq 3, we get

- x + 15/17 + 33/17 = 2

x = 15/17 + 33/17 - 2

L.C.M is 17, then

x = (15 + 33 - 34)/17

x = 14/17

Hence the solution is 

{x, y, z} = {14/17, 15/17, 33/17}

Jul 26th, 2015

thanks for all the work. what would i fill the blank in with?

Jul 26th, 2015

thanks for all the work. what would i fill the blank in with?

Jul 26th, 2015

I think there's a mistake in it. Let me check and write a modified solution. Okay?

Jul 26th, 2015

okay thanks i just need helping filling in the blank

Jul 26th, 2015

x + 2y + 3z = 11,.......................1

3x + 8y + 5z = 27.........................2

- x + y + 2z = 2...............................3

now

Multiply eq 1 by 3 and subtracting 2 from 1, we get

3x + 6y + 9z = 33

3x + 8y + 5z = 27
-     -       -        - 
-------------------------

- 2y + 4z = 6

- y + 2z = 3

y - 2z  = - 3 ................................4

Now multiply eq 3 by 3 and adding with eq 2, we get

3x + 8y + 5z = 27

- 3x + 3y + 6z = 6
-----------------------

  11y + 11z = 33

y + z = 33/11

y + z = 3 ............................................5

Now multiplying eq 5 by 2 and adding with eq 4, we get

y - 2z = - 3

2y + 2z = 6
------------------

3y = 3

y = 1

For z, put y = 1 in eq 5, we get

1 + z = 3

z = 3 - 1

z = 2

Now for x, put y= 1 and z = 2 in eq 3, we get

- x + 1 + 2(2) = 2

- x + 1 + 4 = 2

- x + 5 = 2

- x = 2 - 5

- x = - 3

x = 3

Hence the Solution is 

{x, y, z} = {3, 1, 2}

Jul 26th, 2015

if there are three blank

then

for x blank put 3 because x = 3

for y blank put 1 because y = 1

for z blank put 2 because z = 2

Jul 26th, 2015

Okay?

Jul 26th, 2015

i dont understand. I'm just asking what part E would be

Jul 26th, 2015

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Jul 26th, 2015
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Jul 26th, 2015
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