The given system of equations is
3(2x+y)=7xy (1)
3(x+3y)=11xy (2)
Observe that the given system is not linear because of the occurrence of xy term. Also, note that if x = 0, then y = 0 and vice versa. So, (0, 0) is a solution for the system and any other solution would have both x≠0 and y≠0.
Thus, we consider the case where x≠0,y≠0.
Dividing both sides of each equation by xy, we get
6y+3x=7,i.e.,3x+6y=7 (3)
and 9x+3y=11 (4)
Let a=1x and b=1y
Equations (3) and (4) become
3a+6b=7 (5)
9a+3b=11 (6)
which is a linear system in a and b.
To eliminate b, we have (6)×2→18a+6b=22 (7)
Subtracting (7) from (5) we get, -15a=15. That is, a=1.
Substituting a = 1 in (5) we get, b=23. Thus, a=1 and b=23.
When a=1, we have 1x=1, Thus x=1.
When a=23, we have 1y=23, Thus y=32.
Thus, the system has two solutions (1,32) and (0,0).