急問方程式(F.1) (20分)

1)
8y - 2 = 2y + 34
8y - 2y = 34 + 2
6y = 36
y = 6

2)
y+ (3/4)y = 28
(7/4)y = 28
7y = 28 x 4
7y = 112
y = 16

3)
Let A be the smallest number of the three consecutives,
A + (A+ 1) + (A + 2) = 48
3A + 3 = 48
3A = 45
A = 15

Thus, the three consecutives are 15 , 16 and 17.

4)
Let B be the smallest number of the two consecutive even numbers,
B + (B + 2) = 86
2B + 2 = 86
2B = 84
B = 42

Thus, the two consecutives even numbers are 42 and 44.

1) 8y-2=2y+34
6y=36
y=6
2) y+3/4y=28
7/4y=28
y=28*4/7
y=16
3) The sum of three consecutive(連續) numbers is 48.
48/3=16
the numbers are 15,16,17
4) The sum of two consecutive even numbers is 86.
86/2=43
there are N possible answer
like42,44
40,46
38,48
36,50
and so on...........

2009-07-13 12:53:49 補充：
there just one
no N possible answer
is42,44

1) 8y-2=2y+34
8y-2y=34+2
6y=36
y=6

2)y+3/4y=28
4y+3y=84
7y=84
y=12

3) let the smallest number be x.
x+x+1+x+1+1=48
3x+3=48
3x=45
x=15
∴The numbers are 15, 16 and 17.

4)let the smaller number be x.
x+x+2=86
2x+2=86
2x=84
x=42
∴the numbers are 42 and 44.

8y-2=2y+34
8y-2y=34+2
6y=36
y=6

y+3/4y=28
7/4y=28
y=16

x+(x+1)+(x+2)=48
3x+3=48
3x=45
x=15
Therefore, the numbers are 15,16,17

x+(x+2)=86
2x+2=86
2x=84
x=42
Therefore,the numbers are 42 and 44