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1. Let the four odd numbers be x, x+2, x+4 and x+6
x+x+2+x+4+x+6 = 120
4x + 12 = 120
4x = 108
x = 27
The four odd numbers are 27, 29, 31, 33
2. 3(x-1)-2(x+6)=x-32
3x-3 -2x-12 = x - 32
x - 15 = x - 32
-15 = -32
Contradiction exists.
No solution

1. Let the smallest odd number Be x

x + (x+2) + (x+2+2) + (x+2+2+2) = 120

4x+12 = 120

4x = 108

x = 27

Therefore,the four consecutive odd numbers are 27, (27+2)=29 , (29+2) = 31 and (31+2) = 33

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2. 3(x-1)-2(x+6) = x-32

3x-3-2x-6 = x-32

x-9 = x-32

xxxxxx 9 = 32 xxxxxxx

NO ANSWER ~!!

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1.
(2y+1)+(2y+3)+(2y+5)+(2y+7)=120
8y+16=120
8y=104
y=13

First number is 2y+1
=2(13)+1
=27

Second would be 29
Third would be 31
Fouth would be 33

27+29+31+33=120

2.
3(x-1)-2(x+6)=x-32
3x-3-2x-12=x-32
x-15=x-32
x-x=-17

No answer

2007-11-11 14:16:55 補充：
2y= Surely even number2y 1= Surely odd number