Решебник по математике 6 класс Виленкин Часть 1, 2 Часть 2. Параграф 5. Решение уравнений Задание 101

\[\boxed{\text{5.\ 101}}\]

\[\textbf{а)}\ \frac{x - 4}{8} = \frac{7}{4}\]

\[4 \cdot (x - 4) = 7 \cdot 8\]

\[4x - 16 = 56\]

\[4x = 56 + 16\]

\[4x = 72\]

\[x = 72\ :4\]

\[x = 18.\]

\[\textbf{б)}\ \frac{5}{3x + 2} = \frac{2,5}{27,5}\]

\[(3x + 2) \cdot 2,5 = 5 \cdot 27,5\]

\[3x + 2 = \frac{5 \cdot 27,5}{2,5}\]

\[3x + 2 = 20 \cdot 27,5\]

\[3x + 2 = 55\]

\[3x = 55 - 2\]

\[3x = 53\]

\[x = \frac{53}{3}\]

\[x = 17\frac{2}{3}.\]

\[\textbf{в)}\ \frac{x + 6}{4} = \frac{2x - 15}{7}\]

\[7 \cdot (x + 6) = 4 \cdot (2x - 15)\]

\[7x + 42 = 8x - 60\]

\[7x - 8x = - 60 - 42\]

\[- x = - 102\]

\[x = 102.\]

\[\textbf{г)}\ \frac{0,3}{x + 5} = \frac{0,8}{x - 9}\ \ \ \ \ \ \ | \cdot 10\]

\[\frac{3}{x + 5} = \frac{8}{x - 9}\]

\[8 \cdot (x + 5) = 3 \cdot (x - 9)\]

\[8x + 40 = 3x - 27\]

\[8x - 3x = - 27 - 40\]

\[5x = - 67\]

\[x = - 67\ :5\]

\[x = - 13,4.\]