First Order ODE/(x + y + 4) over (x + y - 6)

Theorem
The first order ODE:
 * $(1): \quad \dfrac {\d y} {\d x} = \dfrac {x + y + 4} {x + y - 6}$

has the solution:
 * $y - x = 5 \ln \paren {x + y - 1} + C$

Proof
We note that $(1)$ is in the form:
 * $\dfrac {\d y} {\d x} = F \paren {\dfrac {a x + b y + c} {d x + e y + f} }$

where:
 * $a e = b d = 1$

Hence we can use First Order ODE in form $y' = F \paren {\dfrac{a x + b y + c} {d x + e y + f} }$ where $a e = b d$.

Let:
 * $z = x + y$

to obtain:
 * $\dfrac {\d z} {\d x} = b F \paren {\dfrac {a z + a c} {d z + f} } + a$

where:
 * $a = 1$
 * $c = 4$
 * $d = 1$
 * $f = -6$

which gives:

Substitute $v = z - 1$ which gives:
 * $\dfrac {\d z} {\d v} = 1$

and thence:
 * $v = x + y - 1$:

Substituting $x + y - 1$ for $v$: