How do I find...

If x and y are nonzero integers, which of the following must be an integer?

A) x+ y/ x

B) x+y^2/ x

C) x^2 + xy/ x

D) x^2 + y^2/ x

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- May 30th 2010, 02:24 PMdnntauI don't really even understand what they are asking.
How do I find...

If x and y are nonzero integers, which of the following must be an integer?

A) x+ y/ x

B) x+y^2/ x

C) x^2 + xy/ x

D) x^2 + y^2/ x - May 30th 2010, 02:33 PMe^(i*pi)
- May 30th 2010, 02:33 PMilovemath98Plug in Numbers
A good idea is to plug in numbers for x and y. Pick nonzero integers as the problem says. For example, x = 3 and y = 2. Plug these values in for A, B, C, and D. See which one will give you an integer!

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How do I find...

If x and y are nonzero integers, which of the following must be an integer?

A) x+ y/ x

B) x+y^2/ x

C) x^2 + xy/ x

D) x^2 + y^2/ x - May 30th 2010, 02:33 PMpickslides
- May 30th 2010, 02:35 PMSpringFan25
A nonzero integer is a whole number other than zero (1,2,3,4,5.....,-1,-2,-3,-4,-5......).

The question gives you two whole numbers (x,y). It wants you to work out if the formulae given will also be whole numbers or not.

Some basic rules to remember are:

Rule1: A whole number**plus**a whole number is a whole number

Rule2: A whole number**minus**a whole number is a whole number

Rule3: A whole number**times**a whole number is a whole number

Rule4: A whole number**divided by**a whole number is a**not always**whole number

I'll do two of the options for you, see if you can do the other two yourself:

(A)

x + (y/x)

y/x is not necessarily a whole number (rule 4)

So we have (whole number) + (maybe not a whole number)

which is**not**always a whole number

(C)

$\displaystyle x^2 + xy/x$

simplify this first

$\displaystyle x \times x + y$

Now $\displaystyle x \times x$ is a whoole number (rule 3)

and $\displaystyle y$ is a whole number (from the question)

So we have (whole number) + (whole number) which gives a whole number

Although you have the answer (C), check you can show B & D are not the answer - May 30th 2010, 02:36 PMdnntau
Thanks.