$\frac{1}{2a - 2} + \frac{2}{4a - 4} = \frac{1}{2(a - 1)} + \frac{2}{4(a - 1)} = \frac{2 + 2}{4(a - 1)} = \frac{4}{4(a - 1)} = \frac{1}{a - 1}$

$\frac{7a}{3x + 3} - \frac{a}{6x + 6} = \frac{7a}{3(x + 1)} - \frac{a}{6(x + 1)} = \frac{14a - a}{6(x + 1)} = \frac{13a}{6(x + 1)}$

$\frac{2m}{4m + 4n} + \frac{4n}{8m + 8n} = \frac{2m}{4(m + n)} + \frac{4n}{8(m + n)} = \frac{4m + 4n}{8(m + n)} = \frac{4(m + n)}{8(m + n)} = \frac{1}{2}$

$\frac{2p}{10p - 10q} - \frac{3q}{15p - 15q} = \frac{2p}{10(p - q)} - \frac{3q}{15(p - q)} = \frac{6p - 6q}{30(p - q)} = \frac{6(p - q)}{30(p - q)} = \frac{1}{5}$

$\frac{2x}{ax + bx} + \frac{3y}{ay + by} = \frac{2x}{x(a + b)} + \frac{3y}{y(a + b)} = \frac{2xy + 3xy}{xy(a + b)} = \frac{5xy}{xy(a + b)} = \frac{5}{a + b}$

$\frac{y}{ax - bx} - \frac{x}{ay - by} = \frac{y}{x(a - b)} - \frac{x}{y(a - b)} = \frac{y^2 - x^2}{xy(a - b)}$

$\frac{1}{2x^2y - xy} + \frac{2}{y - 2xy} = \frac{1}{xy(2x - 1} + \frac{2}{y(1 - 2x)} = \frac{1}{xy(2x - 1} - \frac{2}{y(2x - 1)} = \frac{1 - 2x}{xy(2x - 1} = -\frac{2x - 1}{xy(2x - 1} = -\frac{1}{xy}$

$\frac{3}{3m^2n - 6mn^2} - \frac{2}{4mn - 2m^2} = \frac{3}{3mn(m - 2n)} - \frac{2}{2m(2n - m)} = \frac{3}{3mn(m - 2n)} + \frac{2}{2m(m - 2n)} = \frac{6 + 6n}{6mn(m - 2n)} = \frac{6(1 + n)}{6mn(m - 2n)} = \frac{1 + n}{mn(m - 2n)}$

$\frac{15}{10p^3q - 15p^2q^2} - \frac{6q}{9pq^3 - 6p^2q^2} = \frac{15}{5p^2q(2p - 3q)} - \frac{6q}{3pq^2(3q - 2p)} = \frac{15}{5p^2q(2p - 3q)} + \frac{6q}{3pq^2(2p - 3q)} = \frac{45q + 30pq}{15p^2q^2(2p - 3q)} = \frac{15q(3 + 2p)}{15p^2q^2(2p - 3q)} = \frac{3 + 2p}{p^2q(2p - 3q)}$

$\frac{3b}{2a^3b - 8a^2b^2} - \frac{5a}{12a^3b - 3a^4} = \frac{3b}{2a^2b(a - 4b)} - \frac{5a}{3a^3(4b - a)} = \frac{3b}{2a^2b(a - 4b)} + \frac{5a}{3a^3(a - 4b)} = \frac{9ab + 10ab}{6a^3b(a - 4b)} = \frac{19ab}{6a^3b(a - 4b)} = \frac{19}{6a^2(a - 4b)}$