Calculus with Applications (10th Edition)

The solution is $b=2$
$\dfrac{5}{b+5}-\dfrac{4}{b^{2}+2b}=\dfrac{6}{b^{2}+7b+10}$ Factor all rational expressions completely, if possible: $\dfrac{5}{b+5}-\dfrac{4}{b(b+2)}=\dfrac{6}{(b+5)(b+2)}$ Multiply the whole equation by $b(b+5)(b+2)$: $b(b+5)(b+2)\Big[\dfrac{5}{b+5}-\dfrac{4}{b(b+2)}=\dfrac{6}{(b+5)(b+2)}\Big]$ $5b(b+2)-4(b+5)=6b$ $5b^{2}+10b-4b-20=6b$ Take $6b$ to the left side and simplify: $5b^{2}+10b-4b-20-6b=0$ $5b^{2}-20=0$ Take $20$ to the right side: $5b^{2}=20$ Take $5$ to divide the right side: $b^{2}=\dfrac{20}{5}$ $b^{2}=4$ Take the square root of both sides: $\sqrt{b^{2}}=\pm\sqrt{4}$ $b=\pm2$ The original equation is undefined for $b=-2$, so the solution is only $b=2$