Question

   Solve the following equations by substitution method 3/x+1/y=7;5/x-4/y=6 [x≠0,y≠0]

Answer 1

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

Answer 2

3/x + 1/y = 7    -----(1)
3*(1/y) + 1/y = 7  -----(2)
5/x  -  4/y  =  6    -----(3)
5*(1/x) - 4*(1/y)  = 6 ----(4)
let 1/y = p and 1/y = q ---(5)
(5)in (2)and(4)=>
3p+q=7 ----(6)
5p-4q=6-----(7)
from (6)
q=7-3p -------(8)
(8)in(7)=>
5p - 4(7-3p) = 6
5p - 28 + 12p = 6
17p               = 6+28
                17p=34
                 p=34/17
                   =2  -----(9)
(9) in(8)=> q=7 - 3*2=7-6=1
we have p=1/x
              2=1/x
              2x=1
                x=1/2
we have q=1/y
               1=1/y
               y=1
solution x=1/2
            y= 1