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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```
7 years ago

Nehal Wani
21 Points
```										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
```
7 years ago
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