ГДЗ по алгебре 7 класс Макарычев Задание 638

\[\boxed{\text{638\ (638).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \frac{6y + 7}{4} + \frac{8 - 5y}{3} = 5\ \ \ \ | \cdot 12\]

\[3 \cdot (6y + 7) + 4 \cdot (8 - 5y) = 60\]

\[18y + 21 + 32 - 20y = 60\]

\[- 2y = 60 - 53\]

\[- 2y = 7\]

\[y = - 3,5\]

\[Ответ:y = - 3,5.\]

\[\textbf{б)}\ \frac{5a - 1}{3} = \frac{2a - 3}{5} - 1\ \ \ \ \ | \cdot 15\]

\[5 \cdot (5a - 1) = 3 \cdot (2a - 3) - 15\]

\[25a - 5 = 6a - 9 - 15\]

\[25a - 6a = - 24 + 5\]

\[19a = - 19\]

\[a = - 1\]

\[Ответ:a = - 1.\]

\[\textbf{в)}\ \frac{11x - 4}{7} - \frac{x - 9}{2} = 5\ \ \ \ | \cdot 14\]

\[2 \cdot (11x - 4) - 7 \cdot (x - 9) = 70\]

\[22x - 8 - 7x + 63 = 70\]

\[15x = 70 - 55\]

\[15x = 15\]

\[x = 1\]

\[Ответ:x = 1.\]

\[\textbf{г)}\ \frac{2c - 1}{9} + \frac{c}{4} = \frac{c + 3}{6}\ \ \ \ | \cdot 36\]

\[4 \cdot (2c - 1) + 9c = 6 \cdot (c + 3)\]

\[8c - 4 + 9c = 6c + 18\]

\[17c - 6c = 18 + 4\]

\[11c = 22\]

\[c = 2\]

\[Ответ:c = 2.\]

\[3 \cdot (3p - 1) - 2 \cdot (2p + 6) - 72 =\]

\[= 0\]

\[9p - 3 - 4p - 12 = 72\]

\[5p = 72 + 15\]

\[5p = 87\]

\[p = \frac{87}{5} = 17\frac{2}{5}\]

\[p = 17,4\]

\[Ответ:p = 17,4.\]

\[60 - 3 \cdot (1 - 2x) =\]

\[= 2 \cdot (3x + 20) + 4x\]

\[60 - 3 + 6x = 6x + 40 + 4x\]

\[6x - 10x = 40 - 57\]

\[- 4x = - 17\]

\[4x = 17\]

\[x = 17\ :4 = 4\frac{1}{4}\]

\[x = 4,25\]

\[Ответ:x = 4,25.\]