Question

3/x+1/y=7;5/x-4/y=6 [x≠0,y≠0]

Answer 1

Let 1/x=p and 1/y=q
3/x+1/y=7
or,3p+q=7
or,q=7-3y......................(1)
5/x -4/y=6
or,5p-4q=6
putting the value of q  as 7-3y from (1) we have,
5p-(28-12p)=6
or,17p=6+28=34
or, p=34/17=2............(2)
q=7-3y
Putting the value of y as 2 from (2) we have,
q=7-6=1
So,x=1/1=1 and y=1/2=1/2
Ans: y=1/2
x=1

Answer 2

Let 1/x = a and 1/y = b
Rewrite the equations now. 
3a + b = 7  ......(1) ;   5a - 4b = 6 ........(2)

Multiply eq (1) by 4
12a + 4b = 28 ......(3)
Adding equations (2) and (3)

5a - 4b = 6
12a + 4b = 28
-------------------
17a = 34
 a = 2

Substituting a = 2 in eq(1)
3(2) + b = 7
b = 1

a = 2 ⇒  x = 1/2
b = 1  ⇒  y = 1