Here looking at the values,

Only option D , which is \(y= x+\frac{1}{2}\) is not possible because for any integer value of x, y will be fraction
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How is option E not right?

I would also like to know because I did this with 0 and I remembered you can't divide by 0

The question or OA is flawed..
A.\(y=4\).... x could be any integer...both integer possible

B. \(y=\frac{1}{2x}\).......2xy=1 or xy=\(\frac{1}{2}\)..
If both x and y are integers, xy should be an integer... so both x and y cannot be integers

C.\(y=x+3\)... Possiible

D. \(y= x+\frac{1}{2}\).......\(y- x=\frac{1}{2}\)..
The difference of two integers cannot be a fraction... so both x and y cannot be integers

E. \(y=\frac{1}{2x} + 3....y-3=\frac{1}{2x}\)........If x is an integer, \(\frac{1}{2x}\) will be a fraction, then y-3 cannot be a fraction

so answer = B, D, and E

Note - editing the answer accordingly
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First, know that we have to get an integer for the x coordinate, and an integer for the y coordinate.

for Y = 4, and Y=x+3, we know the obvious answer,

for choice b

y=1/2x ... lets think..

the smallest integer (absolute value-wise) is 0, lets start with that

0=1/2x... nope
y=1/[2(0)].... nope these answers are impossible.

lets try 1

1=1/(2x)

2x=1
then x = 1/2... so that doesn't work

what about..
y=1/2(1)

y=1/2... so no, that wont work.

try a bigger number, like 100

100=1/(2x)
200x=1
x=1/200....

try for x

y=1/[2(100)]
y=1/200.... nope still a fraction and integer pair. no matter what number you put, there will never be integers in both x and y coordinates.

For choice D
plug in any integer

1+1/2= 3/2
1000+ 1/2 = 2001/2.. no matter what you plug in, the number will always end in 1/2. so there will be no integer pairs for the x and y coordinates.

for E

thats just a continuation of B which was 1/(2x)
so, 1/(2x)+3 will always be a fraction, plus the number 3.

we choose B.D,E

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