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November 24, 2014

November 24, 2014

Posted by **Me** on Friday, April 9, 2010 at 3:15pm.

The sum of three numbers is 15. The sum of the first, the second, and twice the third is 20. The sum of the first, twice the second, and three times the third is 27.

This is what I did:

1st # = x

2nd # = y

3rs # = z

so

x+y+z=15

x+y+2z=20

x+2y+3z=27

then solved by elimination of z, and finally got (16,10,-1)

where is my mistake?

Where is my mistake?

- Math -
**Damon**, Friday, April 9, 2010 at 3:26pm2x+2y+2z = 30

x + y+2z = 20

---------------

x + y = 10

3x+3y+3z = 45

x +2y+3z = 27

-----------------

2x + y = 18

gives

2x+2y = 20

2x +y = 18

-----------

y = 2 (not 10)

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