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

Solve for y:11/6-(y+2/3) = 3/2Put 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) = 3/2Put 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 = 3/211/6+(2 (-2-3 y))/6 = (11+2 (-2-3 y))/6:(11+2 (-2-3 y))/6 = 3/22 (-2-3 y) = -6 y-4:(-4-6 y+11)/6 = 3/2Grouping like terms, -(6 y)-4+11 = (11-4)-6 y:((11-4)-6 y)/6 = 3/211-4 = 7:(7-6 y)/6 = 3/2Multiply both sides of (7-6 y)/6 = 3/2 by 6:(6 (7-6 y))/6 = 6×3/26×3/2 = (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)/26/2 = (2×3)/2 = 3:7-6 y = 3×33×3 = 9:7-6 y = 9Subtract 7 from both sides:(7-7)-6 y = 9-77-7 = 0:-6 y = 9-79-7 = 2:-6 y = 2Divide 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/-3Простите за английский!От вас "лучший ответ"!