Решите уравнение:
1) $\frac{4x - 1}{12} - \frac{3x + 1}{8} = x + 1$;
2) $\frac{3x - 2}{9} - \frac{2x + 1}{6} = \frac{5 - x}{3}$.
$\frac{4x - 1}{12} - \frac{3x + 1}{8} = x + 1$
$\frac{2(4x - 1) - 3(3x + 1)}{24} = x + 1$
2(4x − 1) − 3(3x + 1) = 24(x + 1)
8x − 2 − 9x − 3 = 24x + 24
8x − 9x − 24x = 24 + 3 + 2
−25x = 29
$x = -\frac{29}{25} = -1\frac{4}{25}$
$\frac{3x - 2}{9} - \frac{2x + 1}{6} = \frac{5 - x}{3}$
$\frac{3x - 2}{9} - \frac{2x + 1}{6} - \frac{5 - x}{3} = 0$
$\frac{2(3x - 2) - 3(2x + 1) - 6(5 - x)}{18} = 0$
2(3x − 2) − 3(2x + 1) − 6(5 − x) = 0
6x − 4 − 6x − 3 − 30 + 6x = 0
6x − 6x + 6x = 4 + 3 + 30
6x = 37
$x = \frac{37}{6} = 6\frac{1}{6}$
Пожауйста, оцените решение