My Math Forum Linear equation I can't seem to solve!

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 January 17th, 2017, 10:57 AM #1 Newbie   Joined: Jan 2017 From: Britain Posts: 11 Thanks: 0 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
January 17th, 2017, 11:30 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.

 January 17th, 2017, 11:31 AM #3 Senior Member   Joined: May 2016 From: USA Posts: 591 Thanks: 251 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. $x + \dfrac{a}{x} - b + d = c \implies x^2 + (d - b - c)x + a = 0 \implies$ $x = \dfrac{b + c - d \pm \sqrt{(d - b - c)^2 - 4a}}{2}.$
 January 17th, 2017, 11:36 AM #4 Newbie   Joined: Jan 2017 From: Britain Posts: 11 Thanks: 0 apologies Sorry -haven't learnt how to use maths notation code yet-should be: (x+a)/x-b+d = c apologies!
 January 17th, 2017, 11:37 AM #5 Newbie   Joined: Jan 2017 From: Britain Posts: 11 Thanks: 0 apologies again! (x+a)/(x-b+d) =c Thanks
January 17th, 2017, 11:56 AM   #6
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Quote:
 Originally Posted by Simonsky (x+a)/(x-b+d) =c
$x+a = (x-b+d)c$

$x+a = xc-bc+dc$

$x - xc = dc-bc-a$

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

$x = \dfrac{dc-bc-a}{1-c}$

 January 17th, 2017, 12:07 PM #7 Newbie   Joined: Jan 2017 From: Britain Posts: 11 Thanks: 0 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
 January 17th, 2017, 12:22 PM #8 Math Team   Joined: Jul 2011 From: Texas Posts: 2,507 Thanks: 1234 $x - xc = x(1) - x(c)$ 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
 February 10th, 2017, 11:56 AM #9 Newbie   Joined: Feb 2017 From: Ontario Posts: 4 Thanks: 1 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 11:59 AM.

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