Solve 12/((x + 1)(x + 5)) + 15/((x + 2)(x + 4)) = 2.

Step-by-step solution

1. 1

   Expand each denominator and spot the shared quadratic. Notice that the two pairs of factors have the same sum, $1+5=2+4=6$:

   $(x+1)(x+5)=x^{2}+6x+5,\qquad (x+2)(x+4)=x^{2}+6x+8$

   So with

   $u=x^{2}+6x$

   both denominators become linear in $u$. Clearing denominators directly would instead produce a quartic.

2. 2

   Rewrite in u.

   $\frac{12}{u+5}+\frac{15}{u+8}=2$

3. 3

   Clear the denominators and simplify.

   $12(u+8)+15(u+5)=2(u+5)(u+8)$

   $12u+96+15u+75=2\left(u^{2}+13u+40\right)$

   $27u+171=2u^{2}+26u+80\quad\Longrightarrow\quad 2u^{2}-u-91=0$

4. 4

   Solve for u.

   $u=\frac{1\pm\sqrt{1+728}}{4}=\frac{1\pm 27}{4}\quad\Longrightarrow\quad u=7\ \text{or}\ u=-\frac{13}{2}$

5. 5

   Unwind each value of u.

   $u=7$: $x^{2}+6x-7=0\Rightarrow(x+7)(x-1)=0\Rightarrow x=-7$ or $x=1$.

   $u=-\tfrac{13}{2}$: $2x^{2}+12x+13=0\Rightarrow x=\dfrac{-12\pm\sqrt{144-104}}{4}=\dfrac{-6\pm\sqrt{10}}{2}$, both real since $144-104=40>0$.

   $\boxed{x=1,\ -7,\ \frac{-6+\sqrt{10}}{2},\ \frac{-6-\sqrt{10}}{2}}$

6. 6

   Check the domain and one root. The excluded values are $x=-1,-2,-4,-5$; none of the four roots is among them ($\tfrac{-6\pm\sqrt{10}}{2}\approx -1.42$ and $-4.58$, both safely away). Checking $x=1$: $\tfrac{12}{2\cdot 6}+\tfrac{15}{3\cdot 5}=1+1=2$ ✓.