Teacher2Teacher Q&A #9409

Teachers' Lounge Discussion: Solving algebraic equations

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From: Loyd

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) 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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