April 24th, 2010, 12:01 AM  #1 
Senior Member Joined: Mar 2010 From: Melbourne Posts: 178 Thanks: 0  Simplify
Hi, I'm trying to simplify: (a^2  16)z = a  4 to solve 'z'. Is this as far as I can go ? z = a  4 / a^2  16 Any help would be greatly appreciated! wulfgarpro. 
April 24th, 2010, 12:41 AM  #2 
Newbie Joined: Apr 2010 Posts: 7 Thanks: 0  Re: Simplify
Nope. a^2  16 is a difference of two squares. Expand that to (a + 4)(a  4). Now you can cancel out (a  4) in the numerator and denominator. So you have z = 1/(a+4) And that's the final answer.

April 24th, 2010, 12:57 AM  #3 
Senior Member Joined: Mar 2010 From: Melbourne Posts: 178 Thanks: 0  Re: Simplify
Hm, Thanks for your prompt reply  although, i'm still confused. Could you show me in a stepwise fashion ? Thanks! wulfgarpro. 
April 24th, 2010, 03:35 AM  #4  
Newbie Joined: Apr 2010 Posts: 7 Thanks: 0  Re: Simplify Quote:
1.) isolate z. Divide both sides by (a^2  16) and you get: z = (a4)/(a^2  16) 2.) expand (a^2  16). It's a difference of two squares (x^2  y^2) = (x+y)(xy). (a^2  16) becomes (a + 4)(a  4). You get: z = (a4)/ [(a+4)(a4)] 3.) You have a common factor in the numerator and denominator of the right side. (a4). Cancel it out. You get: z = 1 / (a+4) cheers.  
April 24th, 2010, 03:45 AM  #5 
Senior Member Joined: Mar 2010 From: Melbourne Posts: 178 Thanks: 0  Re: Simplify
Thanks CarlosJesena! wulfgarpro. 
April 24th, 2010, 04:03 AM  #6 
Senior Member Joined: Mar 2010 From: Melbourne Posts: 178 Thanks: 0  Re: Simplify
The problem is that I have a system of linear equations with a constant 'a'. If the constant 'a' is 4, the system has no solution; if 'a' is 4, the system has infinitely many solutions; if 'a' is not 4 or 4, the system has exactly one solution. I've given forms of solutions for when the system has no solution and infinitely many solutions; I'm now just trying to show a form of solution for exactly one solution given we have the constant 'a'. The new system (post Gaussian Elimination) is: x + 2y  3z = 4 (i) 7y + 14z = 10 (ii) (a^2  16)z = a4 (iii)  Given what has been shown here, (iii) is: z = 1/(a+4)  I'm now just having trouble solving for y and x ? So far with y I have: 7y+14(1/(a+4))=10 7y = 1014(1/(a+4)) 7y = 1014/(a+4) .. How do I get 10 to have a common denominator?  Thank you kindly! wulfgarpro. 
April 24th, 2010, 12:33 PM  #7 
Senior Member Joined: Mar 2010 From: Melbourne Posts: 178 Thanks: 0  Re: Simplify
Hm, Still quite stuck on this simplification heh  . wulfgarpro. 
April 24th, 2010, 01:37 PM  #8 
Senior Member Joined: Mar 2010 From: Melbourne Posts: 178 Thanks: 0  Re: Simplify
OK, I think I have solved for 'y' and 'x'; it would be great if someone could confirm my thoughts. 7y + 14(1/(a+4))= 10 7y = 10  14(1/(a+4)) y = 10/7  2/(a+4) AND x + 2(10/7  2/(a+4))  3(1/(a+4)) = 4 x = 4  2(10/7  2/(a+4)) + 3(1/(a+4)) x = 4  20/7  4/(a+4) + 3/(a+4) x = 4  20/7  7/(a+4) x = 28/7  20/7  7/(a+4) x = 8/7  7/(a+4) Does this look correct ? Thanks! wulfgarpro. 
April 24th, 2010, 02:08 PM  #9 
Senior Member Joined: Mar 2010 From: Melbourne Posts: 178 Thanks: 0  Re: Simplify
Ah, 'y' is correct, 'x' had a sign error: x + 2(10/7  2/(a+4))  3(1/(a+4)) = 4 x = 4  2(10/7  2/(a+4)) + 3(1/(a+4)) x = 4  20/7  4/(a+4) + 3/(a+4) x = 4  20/7 + 7/(a+4) x = 28/7  20/7 + 7/(a+4) x = 8/7 + 7/(a+4) Thanks! wulfgarpro. 
April 24th, 2010, 02:55 PM  #10 
Global Moderator Joined: Dec 2006 Posts: 20,927 Thanks: 2205 
Your values for x and y are both incorrect. For a ? ±4, you correctly obtained 7y = 10  14/(a + 4), but that gives y = 10/7 + 2/(a + 4). Since x + 2y  3z = 4, x = 4  2(10/7 + 2/(a + 4)) + 3/(a + 4) = 8/7  1/(a + 4). If a = 4, z can have any value and you get x = 8/7  z and y = 10/7 + 2z. 

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