Maths [20點]

(1)Let x be the age of Edmund now and y be the age of his sister now.
x=y+7
x-9=2(y-9)


(2) Let x be the age of the father now and y be the age of the his son now .
x+y=50
x+5=3(y+5)

Edmund is 7 years older than his sister. 9 years ago, he was twice as old as his sister. How old is Edmund now?

~~Method 1~~
Let Edmund is x years old now.
　 x-9=2(x-7)
　 x-9=2x-14
2x-x=9+14
　 x=23

~~Method 2~~
Let Edmund is x years old and his sister is y years old now.
x-7=y---------------(1)
x-9=2(y-9)---------(2)
From (1), x-7=y
　　　　 x=y+7--------(3)
Sub. (3) into (2),
　 x-9=2(y-9)
(y+7)-9=2(y-9)
y+7-9=2y-18
2y-y=7-9+18
　　 y=16-------------(4)

Sub. (4) into (3),
x=y+7
x=16+7
x=23

Therefore, Edmund is 23 years old now.



The sum of the ages of a father and his son two years ago was 50. Five years later from now, the father will be three times as old as his son.

Let father is x years old and his son is y years old now.
(x-2+5)+(y-2+5)=50---------(1)
　　　　　 x=3y---------(2)
Sub. (2) into (1),
(x-2+5)+(y-2+5)=50
(3y-2+5)+(y-2+5)=50
3y-2+5+y-2+5=50
　　　　　　 4y=44
　　　　　　　y=11------------(3)

Sub. (3) into (2),
x=3y
x=3(11)
x=33

Therefore, father is 33 years old and his son is 11 years old now.

Edmund is 7 years older than his sister. 9 years ago, he was twice as old as his sister. How old is Edmund now?

Let n be the age of Edmund now, hence his sister's age is n-7
9 years ago,
n-9 = 2 x (n-7-9)
n-9 = 2n -32
n=23

Hence Edmund is 23 years old now.


The sum of the ages of a father and his son two years ago was 50. Five years later from now, the father will be three times as old as his son.

Let n be the age of father 2 years ago, hence his son's age was 50-n at that time.
Now father is n+2, his son is 50-n+2 years old.

Hence,
n+2+5 = 3 x (50-n+2+5)
n+7 = 171 - 3n
4n = 164
n = 41

Hence, father is 41+2 = 43 years old now, while his son is 11 years old now.