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Unformatted text preview: 3d=5c d=5c/3 4. Substitute into scalar equation and solve for c a+b+c+d=1 (2c)+(c)+c+(5c/3)=1 6c+3c+3c+5c=3 17c=3 c=3/17 5. Solve for b b=c b=3/17 6. Solve for a a=2c a=2(3/17) a=6/17 7. Solve for d d=5c/3 d=5(3/17)/3 d=15/51 d=5/17 thus: a=6/17 b=3/17 c=3/17 d=5/17...
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This note was uploaded on 02/20/2012 for the course 790 373 taught by Professor Boros during the Fall '09 term at Rutgers.
- Fall '09