# y′(x)=x+y/z z′(x)=x−y/y

I have a system of differential equations which I need to solve and obtain y(x) and z(x). I tried elimination method and got to a point but I don't know what to do after here. Any help would be appreciated Question:
${y}^{\prime }\left(x\right)=\frac{x+y}{z}$
${z}^{\prime }\left(x\right)=\frac{x-y}{y}$
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Bestvinajw
${y}^{\prime }\left(x\right)=\frac{x+y}{z}\phantom{\rule{0ex}{0ex}}{z}^{\prime }\left(x\right)=\frac{x-y}{y}$
${y}^{\prime }\left(x\right)z=x+y\phantom{\rule{0ex}{0ex}}{z}^{\prime }\left(x\right)y=x-y$
Sum both differential equations:
${y}^{\prime }z+{z}^{\prime }y=2x$
$\left(yz{\right)}^{\prime }=2x$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}yz={x}^{2}+C$
The second DE is:
${z}^{\prime }\left(x\right)y=x-y$
${z}^{\prime }=\frac{x}{y}-1$
${z}^{\prime }=\frac{zx}{{x}^{2}+C}-1$
That you can solve.

Some details
${z}^{\prime }=\frac{zx}{{x}^{2}+C}-1$
You can't integrate both sides the way you did because there is the z function on RHS:
${z}^{\prime }-\frac{zx}{{x}^{2}+C}=-1$
$\sqrt{{x}^{2}+C}{z}^{\prime }-\frac{zx}{\sqrt{{x}^{2}+C}}=-\sqrt{{x}^{2}+C}$
${\left(\frac{z}{\sqrt{{x}^{2}+C}}\right)}^{\prime }=-\frac{1}{\sqrt{{x}^{2}+C}}$
Now you can integrate both sides.
$\frac{z}{\sqrt{{x}^{2}+C}}=-\int \frac{dx}{\sqrt{{x}^{2}+C}}+{C}_{2}$
$z\left(x\right)=\sqrt{{x}^{2}+C}\left({C}_{2}-\mathrm{arctan}\frac{x}{\sqrt{{x}^{2}+C}}\right)$