 # system of equations with 3 variables in denominator I'm having problems with this system of equations: 3/(x+y)+2/(x-z)=3/2 1/(x+y)-(10)/(y-z)=(7)/(3) 3/(x-z)+5/(y-z)=-1/4 I've tried substitution method but that took me nowhere, I'm not sure I can solve it and would appreciate if anyone showed me the proper method to solve this problem. drobtinicnu 2022-09-13 Answered
system of equations with 3 variables in denominator
I'm having problems with this system of equations:
$\frac{3}{x+y}+\frac{2}{x-z}=\frac{3}{2}$
$\frac{1}{x+y}-\frac{10}{y-z}=\frac{7}{3}$
$\frac{3}{x-z}+\frac{5}{y-z}=-\frac{1}{4}$
I've tried substitution method but that took me nowhere, I'm not sure I can solve it and would appreciate if anyone showed me the proper method to solve this problem.
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I hope you mean the following.
$\frac{3}{x+y}+\frac{2}{x-z}=\frac{3}{2},$
$\frac{1}{x+y}-\frac{10}{y-z}=\frac{7}{3},$
$\frac{3}{x-z}+\frac{5}{y-z}=-\frac{1}{4}.$
If so, let $\frac{1}{x+y}=c$, $\frac{1}{x-z}=b$ and $\frac{1}{y-z}=a$
Thus,
$3c+2b=\frac{3}{2},$
$c-10a=\frac{7}{3}$
and
$3b+5a=-\frac{1}{4}.$
The last two equations give
$6b+c=\frac{11}{6},$
which with first gives
$c=\frac{1}{3},$
$b=\frac{1}{4}$
and from here we'll get
$a=-\frac{1}{5}.$
Thus,
$x+y=3,$
$x-z=4$
and
$y-z=-5,$
which gives
$\left(x,y,z\right)=\left(6,-3,2\right).$

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