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Question

3(2x+y)=7xy;3(x+3y)=11xy using elimination method.

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Solution

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 x0 and y0.
Thus, we consider the case where x0,y0.
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)×218a+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).

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