2.252 Решите уравнение: a) $$1-y=\frac{7}{24}+\frac{1}{4}$$; б) $$1 + m = \frac{3}{5} + \frac{6}{15}$$; в) $$l+3\frac{5}{6}=7\frac{1}{6}-2\frac{2}{3}$$.

a) $$1-y=\frac{7}{24}+\frac{1}{4}$$

$$1-y=\frac{7}{24}+\frac{6}{24}$$

$$1-y=\frac{13}{24}$$

$$y=1-\frac{13}{24}$$

$$y=\frac{24}{24}-\frac{13}{24}$$

$$y=\frac{11}{24}$$

б) $$1 + m = \frac{3}{5} + \frac{6}{15}$$

$$1 + m = \frac{9}{15} + \frac{6}{15}$$

$$1 + m = \frac{15}{15}$$

$$1 + m = 1$$

$$m = 1 - 1$$

$$m = 0$$

в) $$l+3\frac{5}{6}=7\frac{1}{6}-2\frac{2}{3}$$

$$l+3\frac{5}{6}=7\frac{1}{6}-2\frac{4}{6}$$

$$l+3\frac{5}{6}=4\frac{3}{6}$$

$$l=4\frac{3}{6}-3\frac{5}{6}$$

$$l=3\frac{9}{6}-3\frac{5}{6}$$

$$l=\frac{4}{6}=\frac{2}{3}$$