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Решите задачу:

$3 \frac{1}{8} -x=2 \frac{1}{8}$
$\frac{25}{8} -x= \frac{17}{8}$
$x= \frac{25}{8} - \frac{17}{8}$
$x= \frac{8}{8} =1$

$12 \frac{1}{15} -y=1 \frac{1}{5}$
$\frac{181}{15} -y= \frac{6}{5}$
$y= \frac{181}{15} - \frac{6}{5}$
$y= \frac{181-18}{15}$
$y= \frac{163}{15} =10 \frac{13}{15}$

$x+4 \frac{1}{3} =7 \frac{3}{9}$
$x+ \frac{13}{3} = \frac{22}{3}$
$x= \frac{22}{3} - \frac{13}{3}$
$x= \frac{9}{9} =1$

$4 \frac{1}{2} -y=2 \frac{1}{8}$
$\frac{9}{2} -y= \frac{17}{8}$
$y= \frac{9}{2} - \frac{17}{8}$
$y= \frac{36-17}{8}$
$y= \frac{19}{8} =2 \frac{3}{8}$

$y-5 \frac{3}{7} = \frac{1}{14}$
$y- \frac{38}{7} = \frac{1}{14}$
$y= \frac{38}{7} + \frac{1}{14}$
$y= \frac{76+1}{14}$
$y= \frac{11}{2} =5 \frac{1}{2}$

$y+13 \frac{1}{4} =2 \frac{4}{8}$
$y+ \frac{53}{4} =2 \frac{2}{4}$
$y+ \frac{53}{4} = \frac{10}{4}$
$y= \frac{10}{4} - \frac{53}{4}$
$y=- \frac{43}{4} =-10 \frac{3}4}$