
Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion 
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January 17th, 2017, 09:57 AM  #1 
Newbie Joined: Jan 2017 From: Britain Posts: 18 Thanks: 0  Linear equation I can't seem to solve!
Any help with this appreciated: x+a/xb+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, 10:30 AM  #2  
Math Team Joined: Jul 2011 From: Texas Posts: 2,640 Thanks: 1319  Quote:
$x + \dfrac{a}{x}  b + d = c$ is this what you meant to post? if not, use some parentheses to clarify numerator & denominator for the fraction.  
January 17th, 2017, 10:31 AM  #3 
Senior Member Joined: May 2016 From: USA Posts: 802 Thanks: 318 
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, 10:36 AM  #4 
Newbie Joined: Jan 2017 From: Britain Posts: 18 Thanks: 0  apologies
Sorry haven't learnt how to use maths notation code yetshould be: (x+a)/xb+d = c apologies! 
January 17th, 2017, 10:37 AM  #5 
Newbie Joined: Jan 2017 From: Britain Posts: 18 Thanks: 0  apologies again!
(x+a)/(xb+d) =c Thanks 
January 17th, 2017, 10:56 AM  #6 
Math Team Joined: Jul 2011 From: Texas Posts: 2,640 Thanks: 1319  
January 17th, 2017, 11:07 AM  #7 
Newbie Joined: Jan 2017 From: Britain Posts: 18 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: xxc becomes x(1c) Could you (or anyone else with the time) explain how xxc goes to x(1c) I would have gone: xx times c = just c! I'm obviously missing something 
January 17th, 2017, 11:22 AM  #8 
Math Team Joined: Jul 2011 From: Texas Posts: 2,640 Thanks: 1319 
$x  xc = x(1)  x(c)$ common factor for both terms is $x$, so pull out the common factor from both terms (just doing the distributive property backwards) ... $x(1)  x(c) = x(1  c)$ 
February 10th, 2017, 10:56 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 10:59 AM. 

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