中二數學的問題

method of substitution :

6/(x+y) - 5/(x-y) = 7
4/(x+y) + 7/(x-y)=15

Let a be 1/(x+y) ,b be 1/(x-y) :

6a - 5b = 7.............(1)
4a + 7b = 15..........(2)

By(1),a=(5b+7)/6,sub it to (2):

4(5b+7)/6 + 7b =15
20b+28 + 42b =15*6
62b+28 = 90
62b=62
b = 1,so a=(5+7)/6
a = 2

a=1/(x+y)=2
b=1/(x-y)=1

2x+2y=1........(1)
x-y=1.............(2)

by(2),x=y+1,sub to (1):

2(y+1)+2y=1
4y=-1
y=-1/4, x=-1/4+1
x=3/4
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method of elimination :

6/(x+y) - 5/(x-y) = 7...........(1)
4/(x+y) + 7/(x-y)=15..........(2)

(1)*7+(2)*5 :
42/(x+y) - 35/(x-y) + 20/(x+y) + 35/(x-y) = 7*7+15*5
62/(x+y) = 124

x+y=1/2,sub to (1): 6/(1/2) - 5/(x-y) = 7
x-y= 1

so x+y + x-y = 1/2 + 1
2x=1 + 1/2
x = 3/4
y = 3/4 -1 = -1/4