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
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