F1 Maths ~~~ Inequalities

2014-10-10 5:47 am
1) x + 3 > 10

2) 8x < 48

3) 11 + 8x > 32

4) 2(x-1) > 3

5) Find the two smallest consecutive integers whose sum is greater than 35.

回答 (2)

2014-10-10 7:44 am
✔ 最佳答案
1)
x + 3 > 10
x + 3 -3 > 10 - 3
x > 7

2)
8x < 48
8x/8< 48/8
x < 6

3)
11 + 8x > 32
11 + 8x - 11> 32 - 11
8x > 21
8x/8 > 21/8
x > 21/8

4)
2(x - 1) > 3
2(x - 1)/2 > 3/2
x - 1 > 3/2
x - 1 + 1 > (3/2) + 1
x > 5/2

5)
Let n and (n + 1) be the two consecutive integers.

n + (n + 1) > 35
2n + 1 > 35
2n + 1 - 1 > 35 - 1
2n > 34
n > 17
smallest n = 18
smallest (n + 1) = 19

The two smallest consecutive integers are 18 and 19.
2014-10-10 6:20 am
1)x+3>10
=x>7
2)8x<48
=6<x
3)11+8x>32
=8x>32-11
=8x>21
=x>21/8
4)2(x-1)>3
=2x-2>3
=2x>5
=x=2.5
5)18+19
=37

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參考: me


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