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

Question

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

Nehal Wani · Answer

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

1 Answers