Solve 1/((x + 1)(x + 3)) + 9/((x − 1)(x + 5)) = −1.

Step-by-step solution

1. 1

   Expand both denominators and look for the common quadratic part. The factor pairs have the same sum, $1+3=(-1)+5=4$:

   $(x+1)(x+3)=x^{2}+4x+3,\qquad (x-1)(x+5)=x^{2}+4x-5$

   Setting $v=x^{2}+4x$ makes both denominators linear in $v$.

2. 2

   Rewrite in v.

   $\frac{1}{v+3}+\frac{9}{v-5}=-1$

3. 3

   Clear the denominators.

   $(v-5)+9(v+3)=-(v+3)(v-5)$

   $10v+22=-\left(v^{2}-2v-15\right)=-v^{2}+2v+15$

   $v^{2}+8v+7=0$

4. 4

   Solve for v.

   $(v+1)(v+7)=0\quad\Longrightarrow\quad v=-1\ \text{or}\ v=-7$

5. 5

   Unwind each value, discarding the impossible one.

   $v=-1$: $x^{2}+4x+1=0\Rightarrow x=\dfrac{-4\pm\sqrt{12}}{2}=-2\pm\sqrt3$, both real.

   $v=-7$: $x^{2}+4x+7=0$ has discriminant $16-28=-12<0$, so it gives no real roots.

   $\boxed{x=-2+\sqrt3\ \text{or}\ x=-2-\sqrt3}$

6. 6

   Check the domain and verify. The excluded values are $x=-1,-3,1,-5$; the roots $-2\pm\sqrt3\approx -0.268$ and $-3.732$ avoid all of them ✓. Checking $x=-2+\sqrt3$: here $v=x^{2}+4x=-1$, so the left side is $\tfrac{1}{2}+\tfrac{9}{-6}=0.5-1.5=-1$ ✓.