Add or subtract. Simplify if possible. __ 1. (3)/(m+5)+(8)/(m^2)-25 a. (11)/((m-5)(m+5)) (3m-7)/((m-5)(m+5)) b. (11)/(m^2)+m-20 d. (3m+23)/((m-5)(m+5)) __ 2 (b^2-2b-8)/(b^2)+b-2-(6)/(b-1) a. b-10 c. (b-4)/(b-1) b (b^2-2b-14)/(b^2)+b-2 d. (b-10)/(b-1) __ 3 (w^2+2w-24)/(w^2)+w-30+(8)/(w-5) a. (w-4)/(w-5) c. w+4 b (w^2+2w-16)/(w^2)+w-30 d. (w+4)/(w-5) __ 4. (3)/(d+3)+(1)/(d^2)-9 a. (4)/((d-3)(d+3)) c. (4)/(d^2)+d-6 b. (3d-8)/((d-3)(d+3)) d. (3d+10)/((d-3)(d+3))

1. To add the fractions $\frac {3}{m+5}+\frac {8}{m^{2}-25}$, we first need to find a common denominator. The denominator $m^{2}-25$ can be factored as $(m-5)(m+5)$. So, the common denominator is $(m+5)(m-5)$. <br /><br />Next, we rewrite each fraction with the common denominator and add the numerators:<br /><br />$\frac {3}{m+5} \cdot \frac {m-5}{m-5} + \frac {8}{m^{2}-25} \cdot \frac {m+5}{m+5} = \frac {3(m-5)}{(m+5)(m-5)} + \frac {8(m+5)}{(m+5)(m-5)} = \frac {3m-15+8m+40}{(m+5)(m-5)} = \frac {11m+25}{(m+5)(m-5)}$<br /><br />Therefore, the correct answer is d. $\frac {3m+23}{(m-5)(m+5)}$.<br /><br />2. To subtract the fractions $\frac {b^{2}-2b-8}{b^{2}+b-2}-\frac {6}{b-1}$, we first need to find a common denominator. The denominator $b^{2}+b-2$ can be factored as $(b-1)(b+2)$. So, the common denominator is $(b-1)(b+2)$. <br /><br />Next, we rewrite each fraction with the common denominator and subtract the numerators:<br /><br />$\frac {b^{2}-2b-8}{b^{2}+b-2} \cdot \frac {1}{1} - \frac {6}{b-1} \cdot \frac {b+2}{b+2} = \frac {b^{2}-2b-8}{(b-1)(b+2)} - \frac {6(b+2)}{(b-1)(b+2)} = \frac {b^{2}-2b-8-6(b+2)}{(b-1)(b+2)} = \frac {b^{2}-2b-8-6b-12}{(b-1)(b+2)} = \frac {b^{2}-8b-20}{(b-1)(b+2)}$<br /><br />Therefore, the correct answer is b. $\frac {b^{2}-2b-14}{b^{2}+b-2}$.<br /><br />3. To add the fractions $\frac {w^{2}+2w-24}{w^{2}+w-30}+\frac {8}{w-5}$, we first need to find a common denominator. The denominator $w^{2}+w-30$ can be factored as $(w-5)(w+6)$. So, the common denominator is $(w-5)(w+6)$. <br /><br />Next, we rewrite each fraction with the common denominator and add the numerators:<br /><br />$\frac {w^{2}+2w-24}{w^{2}+w-30} \cdot \frac {1}{1} + \frac {8}{w-5} \cdot \frac {w+6}{w+6} = \frac {w^{2}+2w-24}{(w-5)(w+6)} + \frac {8(w+6)}{(w-5)(w+6)} = \frac {w^{2}+2w-24+8(w+6)}{(w-5)(w+6)} = \frac {w^{2}+2w-24+8w+48}{(w-5)(w+6)} = \frac {w^{2}+10w+24}{(w-5)(w+6)}$<br /><br />Therefore, the correct answer is b. $\frac {w^{2}+2w-16}{w^{2}+w-30}$.<br /><br />4. To add the fractions $\frac {3}{d+3}+\frac {1}{d^{2}-9}$, we first need to find a common denominator. The denominator $d^{2}-9$ can be factored as $(d-3)(d+3)$. So, the common denominator is $(d+3)(d-3)$. <br /><br />Next, we rewrite each fraction with the common denominator and add the numerators:<br /><br />$\frac {3}{d+3} \cdot \frac {d-3}{d-3} + \frac {1}{d^{2}-9} \cdot \frac {1}{1} = \frac {3(d-3)}{(d+3)(d-3)} + \frac {1}{