Номер №149 — ГДЗ по алгебре за 7 класс, Макарычев

Перенесём слагаемые с переменной в левую часть, числа — в правую, приведём подобные и разделим на коэффициент.

а) \(3x - 8 = x + 6\), \(\;3x - x = 6 + 8\), \(\;2x = 14\), \(\;x = 7\).

б) \(7a - 10 = 2 - 4a\), \(\;7a + 4a = 2 + 10\), \(\;11a = 12\), \(\;a = \tfrac{12}{11} = 1\tfrac{1}{11}\).

в) \(\dfrac{1}{6}y - \dfrac{1}{2} = 3 - \dfrac{1}{2}y\), \(\;\dfrac{1}{6}y + \dfrac{1}{2}y = 3 + \dfrac{1}{2}\), \(\;\dfrac{2}{3}y = 3{,}5\), \(\;y = 3{,}5 \cdot \dfrac{3}{2} = 5{,}25\).

г) \(2{,}6 - 0{,}2b = 4{,}1 - 0{,}5b\), \(\;-0{,}2b + 0{,}5b = 4{,}1 - 2{,}6\), \(\;0{,}3b = 1{,}5\), \(\;b = 5\).

д) \(p - \dfrac{1}{4} = \dfrac{3}{8} + \dfrac{1}{2}p\), \(\;p - \dfrac{1}{2}p = \dfrac{3}{8} + \dfrac{1}{4}\), \(\;\dfrac{1}{2}p = \dfrac{5}{8}\), \(\;p = \dfrac{5}{4} = 1{,}25\).

е) \(0{,}8 - y = 3{,}2 + y\), \(\;-y - y = 3{,}2 - 0{,}8\), \(\;-2y = 2{,}4\), \(\;y = -1{,}2\).

ж) \(\dfrac{2}{7}x = \dfrac{1}{2}\), \(\;x = \dfrac{1}{2} : \dfrac{2}{7} = \dfrac{1}{2} \cdot \dfrac{7}{2} = \dfrac{7}{4} = 1{,}75\).

з) \(2x - 0{,}7x = 0\), \(\;1{,}3x = 0\), \(\;x = 0\).

Ответ: а) \(7\); б) \(1\tfrac{1}{11}\); в) \(5{,}25\); г) \(5\); д) \(1{,}25\); е) \(-1{,}2\); ж) \(1{,}75\); з) \(0\).

а) \(3x - 8 = x + 6\), \(\;2x = 14\), \(\;x = 7\). \(\quad\) б) \(7a - 10 = 2 - 4a\), \(\;11a = 12\), \(\;a = 1\tfrac{1}{11}\).

в) \(\tfrac{1}{6}y - \tfrac{1}{2} = 3 - \tfrac{1}{2}y\), \(\;\tfrac{2}{3}y = 3{,}5\), \(\;y = 5{,}25\). \(\quad\) г) \(2{,}6 - 0{,}2b = 4{,}1 - 0{,}5b\), \(\;0{,}3b = 1{,}5\), \(\;b = 5\).

д) \(p - \tfrac{1}{4} = \tfrac{3}{8} + \tfrac{1}{2}p\), \(\;\tfrac{1}{2}p = \tfrac{5}{8}\), \(\;p = 1{,}25\). \(\quad\) е) \(0{,}8 - y = 3{,}2 + y\), \(\;-2y = 2{,}4\), \(\;y = -1{,}2\).

ж) \(\tfrac{2}{7}x = \tfrac{1}{2}\), \(\;x = 1{,}75\). \(\quad\) з) \(2x - 0{,}7x = 0\), \(\;x = 0\).

Ответ: а) \(7\); б) \(1\tfrac{1}{11}\); в) \(5{,}25\); г) \(5\); д) \(1{,}25\); е) \(-1{,}2\); ж) \(1{,}75\); з) \(0\).