Решить уравнение: 1 целая 5\6-(у+2/3)=1 целая 1\2

Solve for y: 11/6-(y+2/3) = 3/2 Put each term in y+2/3 over the common denominator 3: y+2/3 = (3 y)/3+2/3: 11/6-(3 y)/3+2/3 = 3/2 (3 y)/3+2/3 = (3 y+2)/3: 11/6-1/3 (3 y+2) = 3/2 Put each term in 11/6-(3 y+2)/3 over the common denominator 6: 11/6-(3 y+2)/3 = 11/6+(2 (-2-3 y))/6: 11/6+(2 (-2-3 y))/6 = 3/2 11/6+(2 (-2-3 y))/6 = (11+2 (-2-3 y))/6: (11+2 (-2-3 y))/6 = 3/2 2 (-2-3 y) = -6 y-4: (-4-6 y+11)/6 = 3/2 Grouping like terms, -(6 y)-4+11 = (11-4)-6 y: ((11-4)-6 y)/6 = 3/2 11-4 = 7: (7-6 y)/6 = 3/2 Multiply both sides of (7-6 y)/6 = 3/2 by 6: (6 (7-6 y))/6 = 6×3/2 6×3/2 = (6×3)/2: (6 (7-6 y))/6 = (6×3)/2 (6 (7-6 y))/6 = 6/6×(7-6 y) = 7-6 y: 7-6 y = (6×3)/2 6/2 = (2×3)/2 = 3: 7-6 y = 3×3 3×3 = 9: 7-6 y = 9 Subtract 7 from both sides: (7-7)-6 y = 9-7 7-7 = 0: -6 y = 9-7 9-7 = 2: -6 y = 2 Divide both sides of -6 y = 2 by -6: (-6 y)/(-6) = 2/(-6) (-6)/(-6) = 1: y = 2/(-6) The gcd of 2 and -6 is 2, so 2/(-6) = (2×1)/(2 (-3)) = 2/2×1/(-3) = 1/(-3): Answer: | | y = 1/-3 Простите за английский!От вас "лучший ответ"!