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Q&A #9409

Teachers' Lounge Discussion: Solving algebraic equations

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From: Loyd <loydlin@aol.com>
To: Teacher2Teacher Public Discussion
Date: 2005110508:45:42
Subject: Re: Re: HELP!; additional help

Jen, Here is a little more help.

Subtract equation 2 from equation 1 and that yields--->  eq 1A) 
x-3y+5z=0

Multiply equation 2 by 2 and that yeilds---> 2w+2x+4y-4z=6
Subtract equation 3 from last result-------> 2w+2x++2y-z=6  The result
is:
eq 2A)  2y-3z=0

Multiply eq 3 by 3 and eq 4 by 2 and that yields:

6w+6x+6y-3z=18
6w-2x-2y+8z=24
Subtract the last two equations and the result is:

eq 3A)  8x+8y-11z=-6

Now you have three equations;

eq 1A)  x-3y+5z=0
eq 2A)    2y-3z=0
eq 3A) 8x+8y-11z=-6

Solve the last three equations and you will find the answers are:

w,x,y,z = 2,-1,3,2 respectively. 

You can multiply eq 1A) by 8 and then subtract eq 3A from that to
start.  

Good luck, hope this helps.

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