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March 3, 2015

March 3, 2015

Posted by **Terri** on Tuesday, February 24, 2009 at 10:34pm.

- algebra attn Reiny -
**Reiny**, Tuesday, February 24, 2009 at 10:40pmif there is a solution, it works

the method I used is "elimination"

you have to look for the variable with the "easiest" number combination

It was luck that in the second set of equations we had 5x in both, that made it easy to simply subtract them.

Had the coefficients been different you would have to multiply by suitable numbers to get them the same.

- algebra attn Reiny -
**Terri**, Tuesday, February 24, 2009 at 10:44pmthanks for the clarification!!

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