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

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

has the solution:
 * $y - x = 5 \ln \left({x + y - 1}\right) + C$

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

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

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

Let:
 * $z = x + y$

to obtain:
 * $\dfrac {\mathrm d z} {\mathrm d x} = b F \left({\dfrac {a z + a c} {d z + f} }\right) + a$

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

which gives:
 * $\dfrac {\mathrm d z} {\mathrm d x} = \dfrac {z + 4} {z - 6} + 1$

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

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

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