User Name Remember Me? Password

 Algebra Pre-Algebra and Basic Algebra Math Forum

 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
Newbie

Joined: Apr 2010

Posts: 7
Thanks: 0

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. Tags simplify Search tags for this page

### 3(3z-4)-( 20 z -5) show me how do i solve

Click on a term to search for related topics.
 Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Kinroh Algebra 4 March 26th, 2014 06:09 AM Ronaldo Calculus 1 December 27th, 2012 07:56 AM Valar30 Elementary Math 3 April 21st, 2011 11:27 PM xdeathcorex Calculus 2 August 31st, 2010 01:37 PM Ronaldo Complex Analysis 1 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.      