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April 24th, 2010, 12:01 AM   #1
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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.
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April 24th, 2010, 12:41 AM   #2
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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.
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April 24th, 2010, 12:57 AM   #3
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Re: Simplify

Hm,

Thanks for your prompt reply - although, i'm still confused.

Could you show me in a step-wise fashion ?

Thanks!

wulfgarpro.
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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.
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April 24th, 2010, 03:45 AM   #5
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Re: Simplify

Thanks CarlosJesena!

wulfgarpro.
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April 24th, 2010, 04:03 AM   #6
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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.
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April 24th, 2010, 12:33 PM   #7
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Re: Simplify

Hm,

Still quite stuck on this simplification heh - .

wulfgarpro.
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April 24th, 2010, 01:37 PM   #8
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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.
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April 24th, 2010, 02:08 PM   #9
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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.
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April 24th, 2010, 02:55 PM   #10
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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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