Teacher2Teacher 
Q&A #9409 
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From: Loyd <loydlin@aol.com> To: Teacher2Teacher Public Discussion Date: 2005110507:45:42 Subject: Re: Re: HELP!; additional help Jen, Here is a little more help. Subtract equation 2 from equation 1 and that yields> eq 1A) x3y+5z=0 Multiply equation 2 by 2 and that yeilds> 2w+2x+4y4z=6 Subtract equation 3 from last result> 2w+2x++2yz=6 The result is: eq 2A) 2y3z=0 Multiply eq 3 by 3 and eq 4 by 2 and that yields: 6w+6x+6y3z=18 6w2x2y+8z=24 Subtract the last two equations and the result is: eq 3A) 8x+8y11z=6 Now you have three equations; eq 1A) x3y+5z=0 eq 2A) 2y3z=0 eq 3A) 8x+8y11z=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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