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November 28, 2014

November 28, 2014

Posted by **rayne** on Monday, October 1, 2012 at 10:26pm.

2x+2y+3z=-1

3x-5y-2z=21

7x+3y+5z=10

- math -
**Reiny**, Monday, October 1, 2012 at 10:35pmWhat method have you learned?

- math -
**rayne**, Monday, October 1, 2012 at 10:37pmEITHER SUBSTITUTION OR ELIMINATION

- math -
**Reiny**, Monday, October 1, 2012 at 10:55pmOk, elimination then

work on the z's

1st times 2 ---> 4x + 4y + 6z = -2

2nd times 3 ---> 9x - 15y - 6z = 63

add them

13x - 11y = 61 (#4)

2nd times 5 ----> 15x - 25y - 10z = 105

3rd times 2 ----> 14x + 6y + 10z = 20

add them

29x - 19y = 125 (#5)

going to be messy .....

#4 times 19 ----> 247x - 209y = 1159

#5 times 11 ----> 319x - 209y = 1375

subtract them

72x = 216

x = 216/72 = 3 , Yeahhh!

back in #4

13(3) - 11y = 61

-11y = 61-39

-11y = 22

y = -2

back in #1

2(3)+2(-2) + 3z = -1

3z = -1 -2

z = -1

**x=3 , y=-2 , z=-1**

- math -
**rayne**, Monday, October 1, 2012 at 11:26pmCAN ANYONE JUST GET ME HEADED IN THE RIGHT

DIRECTION ON THIS PROBLEM

2x+2y+3z=-1

3x-5y-2z=21

7x+3y+5z=10

- math -
**Reiny**, Tuesday, October 2, 2012 at 12:01amAnd exactly what did you not like about the solution which is staring at you right above you ?????

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