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Qn. The number of integral solutions for the equation x + 2y = 2xy is p, then (Qn.no. 19. c)
x-1 = 1, -1 in the solution of the problem
) ·.· x + 2y = 2xy
=> x = 2y(x - 1)
Or 2y = x/(x-1) = 1+1/(x-1)
.·. x - 1 = 1, - 1 How do we get this step. Pls. explain
Or x = 2, 0
then 2y = 2, 0 => y = 1, 0
2y=x/(x-1)=(x-1+1)/(x-1)=1+1/(x-1)
Since y has to an integer, 2y is also integer. Therefore, 1+1/(x-1) is also an integer.
For the RHS to be an integer, 1/(x-1) has to be an integer which is possible only if (x-1)=-1 or 1
[For e.g for x-1=0, RHS is not defined and for x-1=2,3,4,5... RHS is not integral.]
Hence x=2 or 0
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