Please post the answer. I am so rusty on this but I have a guess.I can't wait.
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Please post the answer. I am so rusty on this but I have a guess.I can't wait.
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I've been Boo'd...
Thanks Barry!
My daughter started replying and then she got frustrated, she hasn't had algebra 2 since she was a freshman so she is rusty also.
She may come back and finish her reply later..
don't breed or buy while shelter dogs die....
I have been frosted!
Thanks Kfamr for the signature!
Well it must be hard because she is a math wiz...she drafted an answer but she isn't sure its rights...i'm tempted to send it behind her back![]()
don't breed or buy while shelter dogs die....
I have been frosted!
Thanks Kfamr for the signature!
Maybe this will help?
http://www.coolschool.ca/lor/PMA11/unit5/U05L06.htm
Kristen & the Dynamutts...
I tried solving it as a matrix but linear algebra was over 30(!) years ago and I stunk on ice at it then! Good luck.
I've been finally defrosted by cassiesmom!
"Not my circus, not my monkeys!"-Polish proverb
If I remember correctly it is something like this...
1) equation 1 (8x + 2y - Z = -25 ) solve for X (you will end up with a Y and a Z in the answer))
2) substitute the answer from step 1 into equation 2 (3x - 3y + 5z = 10) and solve for Z (you will end up with a Y in the answer but no X)
3) substitute the Z answer from the previous steps into equation 3 (-5x + 6y - 2z = 17) and solve for Y ( you will end up with an X in the answer but no Z)
4) substitute the answer from step 2 and 3 in to equation 1, solving for X. This time you will get a numerical answer. Then you can solve for the remaining variables based on the answers from step 2 and 3.
Good luck!
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