Solve the given simultaneous equations

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The discussion focuses on solving the simultaneous equations given by the expressions involving x and y. One participant employs a brute force method to derive the solutions, finding x and y values of (4, 1) and (2.5, -0.5). Another contributor suggests a more efficient approach using symmetry and substitution, leading to two systems of equations. The solutions derived from these systems are x + 2y = 6 and x + y = 2, with constraints on x and y values to avoid undefined terms. The conversation highlights the potential for different methods to arrive at the same solutions in solving simultaneous equations.
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Homework Statement
##\dfrac{x+2}{y-4} +\dfrac{2(y-4)}{x+2} + 3=0##

##x-y=3##
Relevant Equations
knowledge of simultaneous equations
##\dfrac{x+2}{y-4} +\dfrac{2(y-4)}{x+2} + 3=0##
##x-y=3##

My approach, will call it brute force, i have

##y=x-3##
then,

##\dfrac{(x+2)^2+2(y-4)^2}{(x+2)(y-4)}=-3##
##(x+2)^2+2(y-4)^2=-3(x+2)(y-4)##
##(x+2)^2+2(x-7)^2=-3(x+2)(x-7)##
##x^2+4x+4+2(x^2-14x+49)+3(x^2-5x-14)=0##
##6x^2-39x+60=0##
##x_1=4, y_1=1## and ##x_2=2.5, y_2=-0.5##

there could be a much better and simpler approach that prompted my post.

cheers man!
 
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That's how most people would solve it.
 
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I don't know if it's really easier or much different, but you can use the symmetry and a substitution
##u = \frac{(x+2)}{(y-4)}## and solve ##u^2 + 3u +2 =0## with ##u= -1, -2##
switching back to x, y gives ##x = 6 - 2y## or ##x = 2-y ##
This can then be combined with the given constraint ##x-y=3## to get the two solutions.

PS: Yes, I left out some intermediate steps.
 
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## \frac{x+2}{y-4}+\frac{2(y-4)}{x+2}+3=0 ##
## \frac{x+2+2(y-4)}{y-4}+\frac{2(y-4)+x+2}{x+2}=0 \Rightarrow (x+2y-6)(x+y-2)=0 ##

and there are two systems of equations with two different solutions.
The first one is
## x+2y=6 ##
## x-y=3 ##
and the second one is
## x+y=2 ##
## x-y=3 ##.
 
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It should be stated implicitly, because of the first equation, that ##y \ne 4## and ##x \ne -2##. Then, due to the second equation, which is equivalent to y = x -3, we have to disallow x = 7 and y = -5.
 
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You cna simplify the algebra by setting a = x + 2, b = y - 4 so that \begin{split}<br /> \frac{a}{b} + 2 \frac{b}{a} + 3 &amp;= 0 \\<br /> a - b &amp;= 9 \end{split} Then multiplying the first by ab gives <br /> a^2 + 3ab + 2b^2 = (a + b)(a + 2b) = 0 and thus we have two solutions, <br /> \left.\begin{array}{rcl}<br /> a + b &amp;=&amp; 0 \\<br /> a - b &amp;=&amp; 9 \end{array} \right\}\quad \mbox{and}\quad \left\{\begin{array}{rcl}<br /> a + 2b &amp;=&amp; 0 \\<br /> a - b &amp;=&amp; 9 \end{array}\right..
 
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