Помогите, Раскройте скобки))

Решите задачу:

$12 * ( \frac{3}{4}x+ \frac{13}{18}y - \frac{1}{24}z) = \frac{12}{1} * \frac{3}{4} x + \frac{12}{1} * \frac{13}{18}y - \frac{12}{1} *\frac{1}{24}z = \\ \\ = \frac{12*3}{1*4} x + \frac{12*13}{1*18}y - \frac{12*1}{1*24} z= \frac{3*3}{1*1}x + \frac{2*13}{1*3}y- \frac{1*1}{1*2}z= \\ \\ =9x + \frac{26}{3}y - \frac{1}{2}z =9x + 8 \frac{2}{3}y - \frac{1}{2}z$

$1 \frac{1}{7} *(7p + \frac{21}{24}q - 1 \frac{3}{4} ) = \frac{8*7}{7*1}p + \frac{8*21}{7*24} q - \frac{8*7}{7*4} = \frac{8*1}{1*1} p + \frac{1*3}{1*3} q - \frac{2*1}{1*1} = \\ \\ =8p + 1q -2=8p + q-2 \\ \\$