system of equations with 3 variables in denominator

Question

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.

Answer 1

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 $$(x,y,z)=(6,-3,2).$$

Answer 2

$$\frac{3}{x+y}+\frac{2}{x-z}=\frac{3}{2}$$ $$\frac{1}{x+y}-\frac{10}{y-z}=\frac{7}{3}$$ $$\frac{3}{x-z}+\frac{5}{\color{red}{x-z}}=-\frac{1}{4}$$

I assume that the part in red should be $y-z$; if not, please clarify.

Hint. Letting: $$a=\frac{1}{x+y} \;,\; b=\frac{1}{x-z} \;,\;c=\frac{1}{y-z}$$ turn the system into: $$\left\{\begin{array}{rcr} 3a&+&2b&&&=&\tfrac{3}{2}\\ a&&&-&10c&=&\tfrac{7}{3}\\ &&3b&+&5c&=&-\tfrac{1}{4} \end{array}\right.$$ Perhaps this looks easier?