Find the solutions of
\begin{align*}
2x-y+1&=0 \\
x+y&=0.
\end{align*}
This happens to be $x=-1/3, \; y=1/3$. So we employ the substitution $u=x+1/3$ and $v=y-1/3$. This changes the DE to
$$(2u-v) \, du+(u+v) \, dv=0,$$
which is homogeneous. The substitution $vt=u$ with $v \, dt+t \, dv=du$ transforms the DE into the separable DE
$$v(2t-1) \, dt+(2t^2+1) \, dv=0,$$
which has the solution
$$\ln|v|=\frac{\arctan\left(\sqrt{2} \, t \right)}{\sqrt{2}}+\ln\left(\frac{1}{\sqrt{2t^2+1}} \right)+C.$$
We then back substitute what $t$ was, followed by what $u$ and $v$ were, to obtain
$$\ln\left|y-\frac13\right|=\frac{\arctan\left(\sqrt{2} \, \frac{x+1/3}{y-1/3} \right)}{\sqrt{2}}+\ln\left(\frac{1}{\sqrt{2\left(\frac{x+1/3}{y-1/3}\right)^{\!2}+1}} \right)+C.$$
Now then, for the extra credit, we follow the same basic pattern. We solve the system
\begin{align*}
ax+by+c&=0 \\
fx+gy+h&=0
\end{align*}
to obtain
$$x=\frac{cg-bh}{bf-ag}=:x_0, \qquad y=\frac{ah-cf}{bf-ag}=:y_0.$$
We set $u=x-x_0$ and $v=y-y_0$. Then $du=dx$ and $dv=dy$, and the DE becomes
$$(a(u+x_0)+b(v+y_0)+c) \, du+(f(u+x_0)+g(v+y_0)+h) \, dv=0,$$
or
$$(au+bv) \, du+(fu+gv) \, dv=0,$$
which, as before, is homogeneous. Let $vt=u$ as before, and the DE becomes
$$(at+b)(v \, dt+t \, dv)+(ft+g) \, dv=0,$$
which is separable. We get
\begin{align*}
v(at+b) \, dt+(at^2+(b+f)t+g) \, dv&=0 \\
-\frac{(at+b) \, dt}{at^2+(b+f)t+g} &=\frac{dv}{v}. \qquad \qquad (1)
\end{align*}
Now subsequent steps depend on the roots of the quadratic $at^2+(b+f)t+g$. If the discriminant $(b+f)^2-4ag>0$, then we obtain
$$\ln|v|=-\frac{(b-f) \arctan\left(\frac{2 a t+b+f}{\sqrt{4 a g-b^2-2 b
f-f^2}}\right)}{\sqrt{4 a g-b^2-2 b f-f^2}}-\frac{1}{2} \ln |t (a t+b+f)+g|.$$
Back-substituting yields
$$\ln|v|=-\frac{(b-f) \arctan\left(\frac{2 a (u/v)+b+f}{\sqrt{4 a g-b^2-2 b
f-f^2}}\right)}{\sqrt{4 a g-b^2-2 b f-f^2}}-\frac{1}{2} \ln |(u/v) (a (u/v)+b+f)+g|,$$
and I will leave the last substitution up to the reader.
If the discriminant $(b+f)^2-4ag=0$, then there are two repeated real roots, and the LHS of (1) above integrates to
$$-4 a \left(\frac{f-b}{4 a (2 a t+b+f)}+\frac{\ln (2 a t+b+f)}{4 a}\right),$$
which you can simplify using the usual rules.
Finally, if the discriminant $(b+f)^2-4ag<0$, then there are complex conjugate roots. In this case, we complete the square to obtain
$$at^2+(b+f)t+g=a\left(t+\frac{b+f}{2a}\right)^{\!2}+\alpha^2,$$
where
$$\alpha^2=\frac{4ag-(b+f)^2}{4a}.$$
Now the LHS of (1) integrates to
$$\frac{\left(4 a^2 \alpha^2-4 a b+b+f\right) \ln \left(4 a^2
\left(\alpha^2+t\right)+b+f\right)}{4 a}-a t.$$
And again, I invite the reader to finish substituting.