The answer is y=x+1/2, since for any integer value of x, y becomes integer and a half: .

Answer: B.
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This question wants us to find a point (x,y) such that both x any y are not integers.

Evaluating answer options, we get

A. y=x: Here, if the x coordinate is an integer, so will the y coordinate be an integer
B. y=x+1/2: Whatever integer value we choose for x, y will not be an integer
C. Y=x+5: Here, if the x coordinate is an integer, so will the y coordinate be an integer
D. y=x*1/2: Here we have a case when x = 2, y = 1 and the point will have both coordinates as integers.
E. y=x/2 +5: Here, if x=2, the value of y is 6. Again both the coordinates are integers.

Therefore, Option B(y=x+1/2) is the only line which cannot have both coordinates as integers
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Bunuel what does this question want? I cant understand ...

Hey dave13 ,

Question is saying if you draw the graph of the equation given in the options, which of the graph will not have both x and y coordinates as integers.

Does that make sense?
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dave13 , I'm just expanding a little on the two excellent answers above, because you mentioned that you knew this slope-intercept equation for a line.

y's value depends on x. For every x, there is only one y.

So you can plug in an integer for x, as Bunuel and pushpitkc noted, and get a y value.*

The result for each option, as abhimahna noted, will give you a point on the line as if it were graphed. As he also noted, one of the answers' coordinates, if graphed, will NOT be (x,y) = (integer, integer)

For options B, D, and E, which have x/2, try x = 0 and x = 2 (or any even integer).
Is y an integer?

For A and C, you can use x = 0 (or just note that y = b). Is y an integer?

Answer B will never yield (x,y) = (integer, integer).

Answer B) y = x + 1/2
x = 0, y = 1/2 (y = b)
x = 1, y = 1 1/2
x = 2, y = 2 1/2
As Bunuel and pushpitkc explain, adding 1/2 to integer x will never yield y = integer.

Answer B

* In all the equations, some of the x values will not be integers. Example: y = x. If x = 1/2, y = 1/2. But if x = 0,1,2,3 etc., y = 0,1,2,3. Those coordinates ARE (integer, integer). Not what the prompt wants.
In one of the equations, however, when x IS an integer, y can never be an integer. In all the other equations, you can find an x value whose y value IS an integer. If you can find a y value that is an integer, that answer is disqualified.
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Ok, I chose B only. But I don’t understand what’s the problem with D?
If x is even, y is an integer.
Likewise if x is odd, y is not an integer...

Can someone elaborate on this?

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