If x and y are integers such that x < y < 0, what is x – y?

[furtadovinod]

Hi,

A minor issue in the answer explanation to this problem. I am not copying out the entire explanation.

If x and y are integers such that x < y < 0, what is x - y?

(1) (x + y)(x - y) = 5
(2) xy = 6

Part of the explanation given is

As you can see, the difference between squares grows as the squares themselves get larger. The only difference between two squares that equals 5 is the difference between 4 and 9. Since x and y are both negative, this tells us that x = -3 and y = -2; therefore, x - y = -5.

The bolded portion should be
x - y = (-3) - (-2) = (-3) + 2 = -1

[www.mbachase.com]

[Attention mbachase.com: we appreciate your desire to help GMAT students. Unfortunately, we don't host posts from competitors on our forums. If you'd like to offer free services to your students (and I think that's a great idea!), then you might consider starting your own forums. Good luck! Stacey]

[Ben Ku]

Hi furtadovinod,

Thanks for the correction. I'll be sure to forward this to the right people so that it gets corrected on our explanations.

[heydharini]

Hi,

I came across the same question in my Manhattan test today and I felt that the answer could be D


If x and y are integers such that x < y < 0, what is x - y?

(1) (x + y)(x - y) = 5
(2) xy = 6

Considering B alone here, [according to this question A is the answer and the explanation is provided - only correction needed x - y = -1]

xy = 6
The options here are -3 and -2
Since we know x<y<0 , x= -3 and y=-2

x - y = -3 - (-2) = -1

which means D should be the correct answer. Manhattan Staff, please let me know if am right?

Thanks,

[mkorrapati]

Hey heydharini,

What about x = -6 and y = -1?

This makes xy = 6 as well.

Now you have two different solutions for x-y. So B and D can not be the answers.

[RonPurewal]

Hey heydharini,

What about x = -6 and y = -1?

This makes xy = 6 as well.

Now you have two different solutions for x-y. So B and D can not be the answers.

correct.

[lancequas]

I got the answer right by accident I guess; I surmized that x^2-y^2 =5 thus x-y= +/- sq. root 5. Knowing x and y are both negative and x is farther away from zero x-y must = neg. sq. root 5. I'm still not seeing why this is wrong. Please correct me and explain the answer

[adiagr]

The term is [(x^2) - (y^2)] and not (x-y)^2.

You cannot take square root for the left hand side i.e. for

[(x^2) - (y^2)].

I hope that helps.

Aditya.

[RonPurewal]

The term is [(x^2) - (y^2)] and not (x-y)^2.

You cannot take square root for the left hand side i.e. for

[(x^2) - (y^2)].

I hope that helps.

Aditya.

yes.

this confusion is going to be a very bad thing on the test, if it persists. therefore, you should probably pick up an algebra book -- or surf the internet -- and look up "difference of squares" and "perfect square trinomials"; these are the two factoring patterns that you are confusing.

you may also want to check out the uses of the distributive property; you are mistakenly applying it to square roots, when in fact it only applies to multiplication and division.

[yousuf_azim]

What is the ans?

BR

[jnelson0612]

What is the ans?

BR

The answer is A.

[HIK]

Hi Guys,

Is there any way of finding out a solution algebrically?

During the exam testing numbers, always makes one wonder if there may be more solutions?

I understand that x and y are integers so the options for testing numbers is less, but still if there is an algebric solution, it'll be great.

Thanks,
Hassan

[tim]

Hassan, if you are interested in a solution that uses more algebra than what you've already seen in the posts above, that is probably not a good idea. The fact that x and y are integers means that you should factor that into your solution method in order to make things easier. A purely algebraic solution that only restricts x and y to integers at the end would be much less efficient..

[chughbrajesh]

(x-y)(x+y)=5

We need to figure out the value for x-y. Couldn't x-y= 1 or 5? So we are not sure about the (x-y) value?

Also if x is smaller than y and y is smaller than 0 then it means that
x must be -4 and y must be -1 to make (x-y)=-5. The question says "x < y < 0".
Then (x+y) will not equal -1 to make a positive 5. This will not happen because one -4 and other is -1. no way, it will equal -1.

[RonPurewal]

(x-y)(x+y)=5

We need to figure out the value for x-y. Couldn't x-y= 1 or 5? So we are not sure about the (x-y) value?

Also if x is smaller than y and y is smaller than 0 then it means that
x must be -4 and y must be -1 to make (x-y)=-5. The question says "x < y < 0".
Then (x+y) will not equal -1 to make a positive 5. This will not happen because one -4 and other is -1. no way, it will equal -1.

You're correct that x - y doesn't equal -5. Both numbers are negative, so x + y will be more negative (= farther to the left on the number line) than x - y.

Since everything is in terms of integers, that leaves only one possibility: x - y = -1 and x + y = -5. So, sufficient.

(You'll get x = -3 and y = -2, although, as long as you know it guarantees a unique answer, it's unnecessary to solve the system once you've found it).