1) Using $z=x+y$, solve $$\frac{dy}{dx}=\frac{x+y+2}{x+y+5}$$
My attempt,so $$\frac{dy}{dx}=\frac{z+2}{z+5}$$
How to integrate then to become y?
2)Using $v=2x-y$, solve $$\frac{dy}{dx}=\frac{2x-y-1}{2x-y+3}$$
so, $$\frac{dy}{dx}=\frac{v-1}{v+3}$$
How to proceed then?
Hint: $$z=x+y\implies {dz\over dx}=1+{dy\over dx}$$ Then $$\frac{dz}{dx}=\frac{z+2}{z+5}+1\implies \frac{dz}{dx}=\frac{2z+7}{z+5}$$ is a separable equation. $$\frac{z+5}{2z+7} dz=dx$$ integrate both sides of the latter.