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January 17th, 2017, 09:57 AM   #1
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Linear equation I can't seem to solve!

Any help with this appreciated:

x+a/x-b+d =c

How do I solve for x when there is an x in both numerator and denominator which means you keep getting x's on both sides?

Many thanks,

Simon
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January 17th, 2017, 10:30 AM   #2
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Quote:
x+a/x-b+d =c
as posted ...

$x + \dfrac{a}{x} - b + d = c$

is this what you meant to post? if not, use some parentheses to clarify numerator & denominator for the fraction.
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January 17th, 2017, 10:31 AM   #3
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This is not a linear equation. It is a quadratic as you can see by multiplying both sides of the equation by x to clear the fraction.

$x + \dfrac{a}{x} - b + d = c \implies x^2 + (d - b - c)x + a = 0 \implies$

$x = \dfrac{b + c - d \pm \sqrt{(d - b - c)^2 - 4a}}{2}.$
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January 17th, 2017, 10:36 AM   #4
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apologies

Sorry -haven't learnt how to use maths notation code yet-should be:

(x+a)/x-b+d = c

apologies!
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January 17th, 2017, 10:37 AM   #5
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apologies again!

(x+a)/(x-b+d) =c

Thanks
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January 17th, 2017, 10:56 AM   #6
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Quote:
Originally Posted by Simonsky View Post
(x+a)/(x-b+d) =c
$x+a = (x-b+d)c$

$x+a = xc-bc+dc$

$x - xc = dc-bc-a$

$x(1-c) = dc-bc-a$

$x = \dfrac{dc-bc-a}{1-c}$
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January 17th, 2017, 11:07 AM   #7
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Thanks Skeeter

I can follow what you did but couldn't have done it myself. I think I would stumble at the step where you go: x-xc becomes x(1-c)

Could you (or anyone else with the time) explain how x-xc goes to x(1-c)

I would have gone: x-x times c = just c! I'm obviously missing something
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January 17th, 2017, 11:22 AM   #8
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$x - xc = x(1) - x(c)$

common factor for both terms is $x$, so pull out the common factor from both terms (just doing the distributive property backwards) ...

$x(1) - x(c) = x(1 - c)$
Thanks from greg1313 and topsquark
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February 10th, 2017, 10:56 AM   #9
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I'll make a video on 2nd degree equations and put it up on youtube.

Last edited by Math for Gwen; February 10th, 2017 at 10:59 AM.
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