\displaystyle \textbf{Reduce each of the following to the lowest terms:}

\displaystyle \textbf{Question 1: }\frac{x^2+x}{x^2-1}
\displaystyle \text{Answer:}
\displaystyle \frac{x^2+x}{x^2-1} =\frac{x(x+1)}{(x+1)(x-1)}
\displaystyle =\frac{x}{x-1},\qquad x\ne-1,\ 1.
\displaystyle \\

\displaystyle \textbf{Question 2: }\frac{x^2-16}{(x+4)^2}
\displaystyle \text{Answer:}
\displaystyle \frac{x^2-16}{(x+4)^2} =\frac{(x-4)(x+4)}{(x+4)(x+4)}
\displaystyle =\frac{x-4}{x+4},\qquad x\ne-4.
\displaystyle \\

\displaystyle \textbf{Question 3: }\frac{a^2-b^2}{a^2b-ab^2}
\displaystyle \text{Answer:}
\displaystyle \frac{a^2-b^2}{a^2b-ab^2} =\frac{(a-b)(a+b)}{ab(a-b)}
\displaystyle =\frac{a+b}{ab},\qquad a\ne0,\ b\ne0,\ a\ne b.
\displaystyle \\

\displaystyle \textbf{Question 4: }\frac{a^2+ab}{2a^3b-2ab^3}
\displaystyle \text{Answer:}
\displaystyle \frac{a^2+ab}{2a^3b-2ab^3} =\frac{a(a+b)}{2ab(a^2-b^2)}
\displaystyle =\frac{a(a+b)} {2ab(a-b)(a+b)}
\displaystyle =\frac{1}{2b(a-b)},\qquad a\ne0,\ b\ne0,\ a\ne b,\ a\ne-b.
\displaystyle \\

\displaystyle \textbf{Question 5: }\frac{x^2-9}{x^2+x-6}
\displaystyle \text{Answer:}
\displaystyle \frac{x^2-9}{x^2+x-6} =\frac{(x-3)(x+3)}{(x+3)(x-2)}
\displaystyle =\frac{x-3}{x-2},\qquad x\ne-3,\ 2.
\displaystyle \\

\displaystyle \textbf{Question 6: }\frac{x^2-2x}{x^2+3x-10}
\displaystyle \text{Answer:}
\displaystyle \frac{x^2-2x}{x^2+3x-10} =\frac{x(x-2)}{(x+5)(x-2)}
\displaystyle =\frac{x}{x+5},\qquad x\ne2,\ -5.
\displaystyle \\

\displaystyle \textbf{Question 7: }\frac{x+3}{x^2-x-12}
\displaystyle \text{Answer:}
\displaystyle \frac{x+3}{x^2-x-12} =\frac{x+3}{(x-4)(x+3)}
\displaystyle =\frac{1}{x-4},\qquad x\ne4,\ -3.
\displaystyle \\

\displaystyle \textbf{Question 8: }\frac{x^2+2x-15}{x^2+4x-21}
\displaystyle \text{Answer:}
\displaystyle \frac{x^2+2x-15}{x^2+4x-21} =\frac{(x+5)(x-3)}{(x+7)(x-3)}
\displaystyle =\frac{x+5}{x+7},\qquad x\ne3,\ -7.
\displaystyle \\

\displaystyle \textbf{Question 9: }\frac{x^2-5x+4}{x^2-3x-4}
\displaystyle \text{Answer:}
\displaystyle \frac{x^2-5x+4}{x^2-3x-4} =\frac{(x-1)(x-4)}{(x+1)(x-4)}
\displaystyle =\frac{x-1}{x+1},\qquad x\ne4,\ -1.
\displaystyle \\

\displaystyle \textbf{Question 10: } \frac{x^3-xy^2}{x^3+2x^2y+xy^2}
\displaystyle \text{Answer:}
\displaystyle \frac{x^3-xy^2}{x^3+2x^2y+xy^2} =\frac{x(x^2-y^2)}{x(x^2+2xy+y^2)}
\displaystyle =\frac{x(x-y)(x+y)}{x(x+y)^2}
\displaystyle =\frac{x-y}{x+y},\qquad x\ne0,\ x\ne-y.
\displaystyle \\

\displaystyle \textbf{Question 11: } \frac{2x^2+7x+3}{3x^2+10x+3}
\displaystyle \text{Answer:}
\displaystyle \frac{2x^2+7x+3}{3x^2+10x+3} =\frac{2x^2+x+6x+3}{3x^2+9x+x+3}
\displaystyle =\frac{x(2x+1)+3(2x+1)}{3x(x+3)+(x+3)}
\displaystyle =\frac{(2x+1)(x+3)}{(3x+1)(x+3)}
\displaystyle =\frac{2x+1}{3x+1},\qquad x\ne-3,\ -\frac{1}{3}.
\displaystyle \\

\displaystyle \textbf{Question 12: } \frac{2x^2+x-3}{3x^2+x-4}
\displaystyle \text{Answer:}
\displaystyle \frac{2x^2+x-3}{3x^2+x-4} =\frac{2x^2-2x+3x-3}{3x^2-3x+4x-4}
\displaystyle =\frac{2x(x-1)+3(x-1)}{3x(x-1)+4(x-1)}
\displaystyle =\frac{(2x+3)(x-1)}{(3x+4)(x-1)}
\displaystyle =\frac{2x+3}{3x+4},\qquad x\ne1,\ -\frac{4}{3}.
\displaystyle \\

\displaystyle \textbf{Simplify:}

\displaystyle \textbf{Question 13: } \frac{x^2-16}{x^2-9}\times\frac{x+3}{x+4}
\displaystyle \text{Answer:}
\displaystyle \frac{x^2-16}{x^2-9}\times\frac{x+3}{x+4}
\displaystyle =\frac{(x-4)(x+4)}{(x-3)(x+3)} \times\frac{x+3}{x+4}
\displaystyle =\frac{x-4}{x-3},\qquad x\ne3,\ -3,\ -4.
\displaystyle \\

\displaystyle \textbf{Question 14: } \frac{a^2-9b^2}{a^2-16b^2}\times\frac{a+4b}{a-3b}
\displaystyle \text{Answer:}
\displaystyle \frac{a^2-9b^2}{a^2-16b^2}\times\frac{a+4b}{a-3b}
\displaystyle =\frac{(a-3b)(a+3b)}{(a-4b)(a+4b)} \times\frac{a+4b}{a-3b}
\displaystyle =\frac{a+3b}{a-4b},\qquad a\ne3b,\ 4b,\ -4b.
\displaystyle \\

\displaystyle \textbf{Question 15: } \frac{x^2+6x+5}{x^2+5x}\times\frac{x^3-x}{x^2-1}
\displaystyle \text{Answer:}
\displaystyle \frac{x^2+6x+5}{x^2+5x}\times\frac{x^3-x}{x^2-1}
\displaystyle =\frac{(x+1)(x+5)}{x(x+5)} \times\frac{x(x-1)(x+1)}{(x-1)(x+1)}
\displaystyle =x+1,\qquad x\ne0,\ -5,\ 1,\ -1.
\displaystyle \\

\displaystyle \textbf{Question 16: } \frac{x+y}{x^2-xy}\div\frac{x^2+xy}{x-y}
\displaystyle \text{Answer:}
\displaystyle \frac{x+y}{x^2-xy}\div\frac{x^2+xy}{x-y}
\displaystyle =\frac{x+y}{x(x-y)}\times\frac{x-y}{x(x+y)}
\displaystyle =\frac{1}{x^2},\qquad x\ne0,\ x\ne y,\ x\ne-y.
\displaystyle \\

\displaystyle \textbf{Question 17: } \frac{x^2-5x}{2x-3y}\div\frac{x^2-25}{4x^2-9y^2}
\displaystyle \text{Answer:}
\displaystyle \frac{x^2-5x}{2x-3y}\div\frac{x^2-25}{4x^2-9y^2}
\displaystyle =\frac{x(x-5)}{2x-3y}\times \frac{(2x-3y)(2x+3y)}{(x-5)(x+5)}
\displaystyle =\frac{x(2x+3y)}{x+5},\qquad 2x\ne3y,\ 2x\ne-3y,\ x\ne5,\ x\ne-5.
\displaystyle \\

\displaystyle \textbf{Question 18: } \frac{x^2+x-6}{x^2+2x-3}\div \frac{x^2+5x-14}{x^2+4x-5}
\displaystyle \text{Answer:}
\displaystyle \frac{x^2+x-6}{x^2+2x-3}\div \frac{x^2+5x-14}{x^2+4x-5}
\displaystyle =\frac{(x+3)(x-2)}{(x+3)(x-1)} \times\frac{(x+5)(x-1)}{(x+7)(x-2)}
\displaystyle =\frac{x+5}{x+7},\qquad x\ne-3,\ 1,\ -5,\ 2,\ -7.
\displaystyle \\

\displaystyle \textbf{Question 19: } \frac{3x^2-x-2}{x^2+x-2}\div \frac{3x^2-7x-6}{x^2-x-6}
\displaystyle \text{Answer:}
\displaystyle \frac{3x^2-x-2}{x^2+x-2}\div \frac{3x^2-7x-6}{x^2-x-6}
\displaystyle =\frac{(3x+2)(x-1)}{(x+2)(x-1)} \times\frac{(x-3)(x+2)}{(3x+2)(x-3)}
\displaystyle =1,\qquad x\ne1,\ -2,\ 3,\ -\frac{2}{3}.
\displaystyle \\

\displaystyle \textbf{Question 20: } \frac{2x^2-3x-2}{x^2+7x+12}\times \frac{x^2+x-12}{x^2+3x-10}
\displaystyle \div \frac{2x^2-5x-3}{x^2+8x+15}
\displaystyle \text{Answer:}
\displaystyle \frac{2x^2-3x-2}{x^2+7x+12} =\frac{(2x+1)(x-2)}{(x+4)(x+3)}
\displaystyle \frac{x^2+x-12}{x^2+3x-10} =\frac{(x+4)(x-3)}{(x+5)(x-2)}
\displaystyle \frac{2x^2-5x-3}{x^2+8x+15} =\frac{(2x+1)(x-3)}{(x+3)(x+5)}
\displaystyle \therefore \frac{(2x+1)(x-2)}{(x+4)(x+3)} \times\frac{(x+4)(x-3)}{(x+5)(x-2)}
\displaystyle \times\frac{(x+3)(x+5)}{(2x+1)(x-3)}=1.
\displaystyle x\ne-4,\ -3,\ -5,\ 2,\ 3,\ -\frac{1}{2}.
\displaystyle \\

\displaystyle \textbf{Simplify:}

\displaystyle \textbf{Question 21: } \frac{2x+3}{3}-\frac{2x-4}{4}
\displaystyle \text{Answer:}
\displaystyle \frac{2x+3}{3}-\frac{2x-4}{4}
\displaystyle =\frac{4(2x+3)-3(2x-4)}{12}
\displaystyle =\frac{8x+12-6x+12}{12}
\displaystyle =\frac{2x+24}{12}=\frac{x+12}{6}.
\displaystyle \\

\displaystyle \textbf{Question 22: } \frac{3}{x-1}-\frac{3}{x+1}
\displaystyle \text{Answer:}
\displaystyle \frac{3}{x-1}-\frac{3}{x+1}
\displaystyle =\frac{3(x+1)-3(x-1)}{(x-1)(x+1)}
\displaystyle =\frac{3x+3-3x+3}{(x-1)(x+1)}
\displaystyle =\frac{6}{(x-1)(x+1)} =\frac{6}{x^2-1},\qquad x\ne1,\ -1.
\displaystyle \\

\displaystyle \textbf{Question 23: } \frac{3x-1}{4x}+\frac{3-5x}{12x}
\displaystyle \text{Answer:}
\displaystyle \frac{3x-1}{4x}+\frac{3-5x}{12x}
\displaystyle =\frac{3(3x-1)+(3-5x)}{12x}
\displaystyle =\frac{9x-3+3-5x}{12x}
\displaystyle =\frac{4x}{12x}=\frac{1}{3},\qquad x\ne0.
\displaystyle \\

\displaystyle \textbf{Question 24: } \frac{x}{x-1}-\frac{x^2}{x^2-1}
\displaystyle \text{Answer:}
\displaystyle \frac{x}{x-1}-\frac{x^2}{x^2-1}
\displaystyle =\frac{x}{x-1}-\frac{x^2}{(x-1)(x+1)}
\displaystyle =\frac{x(x+1)-x^2}{(x-1)(x+1)}
\displaystyle =\frac{x^2+x-x^2}{(x-1)(x+1)}
\displaystyle =\frac{x}{(x-1)(x+1)} =\frac{x}{x^2-1},\qquad x\ne1,\ -1.
\displaystyle \\

\displaystyle \textbf{Question 25: } \frac{6}{a+b}-\frac{5}{a-b}+\frac{11b}{a^2-b^2}
\displaystyle \text{Answer:}
\displaystyle \frac{6}{a+b}-\frac{5}{a-b}+\frac{11b}{a^2-b^2}
\displaystyle =\frac{6(a-b)-5(a+b)+11b}{(a-b)(a+b)}
\displaystyle =\frac{6a-6b-5a-5b+11b}{(a-b)(a+b)}
\displaystyle =\frac{a}{(a-b)(a+b)} =\frac{a}{a^2-b^2},\qquad a\ne b,\ a\ne-b.
\displaystyle \\

\displaystyle \textbf{Question 26: } \frac{6}{x-2}+\frac{8}{2x-4}
\displaystyle \text{Answer:}
\displaystyle \frac{6}{x-2}+\frac{8}{2x-4}
\displaystyle =\frac{6}{x-2}+\frac{8}{2(x-2)}
\displaystyle =\frac{6}{x-2}+\frac{4}{x-2}
\displaystyle =\frac{10}{x-2},\qquad x\ne2.
\displaystyle \\

\displaystyle \textbf{Question 27: } \frac{1}{x^2-11x+30}+\frac{1}{x^2-9x+20}
\displaystyle \text{Answer:}
\displaystyle \frac{1}{x^2-11x+30}+\frac{1}{x^2-9x+20}
\displaystyle =\frac{1}{(x-6)(x-5)}+\frac{1}{(x-5)(x-4)}
\displaystyle =\frac{(x-4)+(x-6)}{(x-6)(x-5)(x-4)}
\displaystyle =\frac{2x-10}{(x-6)(x-5)(x-4)}
\displaystyle =\frac{2(x-5)}{(x-6)(x-5)(x-4)}
\displaystyle =\frac{2}{(x-6)(x-4)},\qquad x\ne4,\ 5,\ 6.
\displaystyle \\

\displaystyle \textbf{Question 28: } \frac{1}{x^2-8x+15}+\frac{1}{x^2-4x+3} -\frac{2}{x^2-6x+5}
\displaystyle \text{Answer:}
\displaystyle \frac{1}{x^2-8x+15}+\frac{1}{x^2-4x+3} -\frac{2}{x^2-6x+5}
\displaystyle =\frac{1}{(x-3)(x-5)}+\frac{1}{(x-3)(x-1)} -\frac{2}{(x-1)(x-5)}
\displaystyle =\frac{(x-1)+(x-5)-2(x-3)} {(x-1)(x-3)(x-5)}
\displaystyle =\frac{x-1+x-5-2x+6} {(x-1)(x-3)(x-5)}
\displaystyle =\frac{0}{(x-1)(x-3)(x-5)}=0,\qquad x\ne1,\ 3,\ 5.
\displaystyle \\

\displaystyle \textbf{Question 29: } \frac{x-3}{x^2-7x+12}+\frac{2(x-1)}{x^2-4x+3} -\frac{3(x-4)}{x^2-5x+4}
\displaystyle \text{Answer:}
\displaystyle \frac{x-3}{x^2-7x+12}+\frac{2(x-1)}{x^2-4x+3} -\frac{3(x-4)}{x^2-5x+4}
\displaystyle =\frac{x-3}{(x-3)(x-4)} +\frac{2(x-1)}{(x-3)(x-1)} -\frac{3(x-4)}{(x-4)(x-1)}
\displaystyle =\frac{1}{x-4}+\frac{2}{x-3}-\frac{3}{x-1}
\displaystyle =\frac{(x-3)(x-1)+2(x-4)(x-1)-3(x-4)(x-3)} {(x-1)(x-3)(x-4)}
\displaystyle =\frac{x^2-4x+3+2x^2-10x+8-3x^2+21x-36} {(x-1)(x-3)(x-4)}
\displaystyle =\frac{7x-25}{(x-1)(x-3)(x-4)},\qquad x\ne1,\ 3,\ 4.
\displaystyle \\

\displaystyle \textbf{Question 30: } \frac{x^2-x-6}{x^2-9}+\frac{x^2+2x-24}{x^2-x-12}
\displaystyle \text{Answer:}
\displaystyle \frac{x^2-x-6}{x^2-9}+\frac{x^2+2x-24}{x^2-x-12}
\displaystyle =\frac{(x-3)(x+2)}{(x-3)(x+3)} +\frac{(x+6)(x-4)}{(x-4)(x+3)}
\displaystyle =\frac{x+2}{x+3}+\frac{x+6}{x+3}
\displaystyle =\frac{2x+8}{x+3} =\frac{2(x+4)}{x+3},\qquad x\ne-3,\ 3,\ 4.
\displaystyle \\

\displaystyle \textbf{Question 31: } \left(\frac{2x}{4x+1}-\frac{x}{2x+3}\right) \div\frac{10x+5}{2x+3}
\displaystyle \text{Answer:}
\displaystyle \left(\frac{2x}{4x+1}-\frac{x}{2x+3}\right) \div\frac{10x+5}{2x+3}
\displaystyle =\frac{2x(2x+3)-x(4x+1)} {(4x+1)(2x+3)} \times\frac{2x+3}{10x+5}
\displaystyle =\frac{4x^2+6x-4x^2-x} {(4x+1)(2x+3)} \times\frac{2x+3}{5(2x+1)}
\displaystyle =\frac{5x}{(4x+1)(2x+3)} \times\frac{2x+3}{5(2x+1)}
\displaystyle =\frac{x}{(4x+1)(2x+1)},\qquad x\ne-\frac14,\ -\frac32,\ -\frac12.
\displaystyle \\

\displaystyle \textbf{Question 32: } \left(\frac{a^2+a-12}{a^2+6a+8} +\frac{a^2-6a+5}{a^2-3a-10}\right) \text{ of }\frac{a^2+8a+12}{a^2+4a-12}
\displaystyle \text{Answer:}
\displaystyle \left(\frac{a^2+a-12}{a^2+6a+8} +\frac{a^2-6a+5}{a^2-3a-10}\right) \times\frac{a^2+8a+12}{a^2+4a-12}
\displaystyle =\left(\frac{(a+4)(a-3)}{(a+4)(a+2)} +\frac{(a-5)(a-1)}{(a-5)(a+2)}\right) \times\frac{(a+2)(a+6)}{(a+6)(a-2)}
\displaystyle =\left(\frac{a-3}{a+2}+\frac{a-1}{a+2}\right) \times\frac{a+2}{a-2}
\displaystyle =\frac{2a-4}{a+2}\times\frac{a+2}{a-2}
\displaystyle =\frac{2(a-2)}{a+2}\times\frac{a+2}{a-2}=2.
\displaystyle a\ne-4,\ -2,\ 5,\ -6,\ 2.
\displaystyle \\

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