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August 30, 2014

August 30, 2014

Posted by **Jacob** on Friday, March 12, 2010 at 12:31am.

- math -
**MathMate**, Friday, March 12, 2010 at 9:43amLet's set up the equations:

C=A+B+10 ....(1)

B=2A .....(2)

A+B+C=490 ....(3)

Substitute (2) in (1) to get:

C=A+2A+10=3A+10 .....(1A)

Substitute (1A) & (2) in (3) to get

A+2A+3A+10 = 490 ......(3A)

Solve for A and back-substitute in (2) and (1) to get B and C.

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