For what integer values of x, will 10/(2x-3) be integer.?

|2x-3|<=10

=>

x<=13/2=6.5

or

2x-3>=-10

=>

x>=-3.5=-3

Thus

-3=<x<=6

Taking

x=-1, 1, 2, 4=>

10/(2x-3)=

-2, -10, 10, 2

correspondingly.

=>

Ans. x=-1,1,2,4.

for 2x - 3 = ±1, ±2, ±5, ±10, 10/(2x - 3) is an integer

integer solutions: x = -1, 1, 2, 4

10/(2x-3) ... factors of 10 = +/- 10, 5. 2.1

2x - 3 = 10 no integer solution .. same for -10

2x - 3 = 5 ... x = 4

2x - 3 = -5 ... x = -1

2x - 3 = 1 ... x = 2

2x - 3 = -1 ... x = 1

x = 4, -1, 2, 1 <<< answer

If 10/(2x-3) is integer, then 2x-3 can only take the values +-1,+-2, and +-5, +-10

So suppose 2x-3 = +-1, then 2x = 3+-1 or 2x = 4 or 2 so x=1 or 2

Suppose 2x-3 =+-2 then 2x=3+-2 or 2x = 5 or 1 and x = 5/2 or 1/2 and these values are not integer

Suppose 2x-3 =+-5, then 2x = 3+-5, or 2x = 8 or -2 and x= 4 or -1

If 2x-3 = +-10 then x values are non-integer

So the permissible values of x are 1, -1, 2, 4

You need 2x -3 to be equal to 2 or 5 or 10. Also -2, -5, and -10. I'm hoping you can solve for that!