Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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

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

Grade:11

1 Answers

Nehal Wani
21 Points
11 years ago

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

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free