1. Find 2 consecutive even numbers such that 3 times the larger number exceeds 2 times the smaller number by 22.
2. Jacky is y years old. His age is 2 times his son's age plus 1,and is 3 times his daughter's age plus 2. If the sum of their age is (98-y),find their ages.
*要用方程式
[快呀] 2條F1 maths
2008-03-29 7:47 pm
回答 (2)
2008-03-29 8:04 pm
✔ 最佳答案
1.Assume small number is X, the large number is X + 2,
then:
3X - 2(X + 2) = 22
3X - 2X - 4 = 22
X - 4 = 22
X = 26
the small number is 26, and the large number is 28
2. Jacky is y year., assume son is x years and daughter is z years.
y = 2x + 1, then x = ( y - 1 ) / 2
y = 3 z + 2, then z = (y - 2 ) / 3
then
y + x + z = 98 - y
2y + x + z = 98
2 y +(y -1) / 2 + (y - 2) / 3 = 98
12y + 3 y - 3 + 2y - 4 = 588
17y = 595
y = 35#
x = ( 35 - 1 ) / 2
x = 17#
z = ( 35 - 2 ) / 3
z = 11#
Jacky is 35 years old, son is 17 years old, daughter is 11 years old.
2008-03-29 7:56 pm
1) Let x be the smaller number, then x+2 is the larger number
therefore -> 3(x+2)-2x = 22
3x+6 -2x=22
x=16
therefore-> smaller number is 16, larger is 18
2) because -> jacky = y
therefore -> son = (y-1)/2
daugther = (y-2)/3
y+(y-1)/2 + (y-2)/3 = 98-y
y+y/2+y/3+y = 98+1/2+2/3
y+5/6y = 99+1/6
y = 54+1/11
therefore-> jacky = 54 and 1/11
son -> 26 and 6/11
daughter -> 17 and 4/11
therefore -> 3(x+2)-2x = 22
3x+6 -2x=22
x=16
therefore-> smaller number is 16, larger is 18
2) because -> jacky = y
therefore -> son = (y-1)/2
daugther = (y-2)/3
y+(y-1)/2 + (y-2)/3 = 98-y
y+y/2+y/3+y = 98+1/2+2/3
y+5/6y = 99+1/6
y = 54+1/11
therefore-> jacky = 54 and 1/11
son -> 26 and 6/11
daughter -> 17 and 4/11
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