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 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 step-wise fashion ? Thanks! wulfgarpro.
April 24th, 2010, 03:35 AM   #4
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Re: Simplify

Quote:
 Originally Posted by wulfgarpro Hm, Thanks for your prompt reply - although, i'm still confused. Could you show me in a step-wise fashion ? Thanks! wulfgarpro.
Given: (a^2 - 16)z = a - 4

1.) isolate z. Divide both sides by (a^2 - 16) and you get:

z = (a-4)/(a^2 - 16)

2.) expand (a^2 - 16). It's a difference of two squares (x^2 - y^2) = (x+y)(x-y). (a^2 - 16) becomes (a + 4)(a - 4). You get:

z = (a-4)/ [(a+4)(a-4)]

3.) You have a common factor in the numerator and denominator of the right side. (a-4). 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 = a-4 (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 = -10-14(1/(a+4)) -7y = -10-14/(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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