Algebra quandary?

Question

So, x and y are integers, x > y > 0 and x^2 - y^2 = 12.
Which can be the value of x - y and why?
Those are: 1 or 2 or 4

Answer 1

x^2 - y^2 = (x+y)(x-y) = 12.

x and y are positive, so x+y > x-y.

the ways to multiply 2 numbers together and get 12 are:

12*1
6*2
4*3

so x-y can only be 1, 2, or 3.

for each possibility, you get two equations in two variables, e.g.
x+y = 12
x-y = 1
you can solve the system, and see if the solution has integer values.

out of the three possibilities, the only one that has integer solutions is
x+y = 6
x-y = 2.

Answer 2

First, make use of the fact that x^2 - y^2 = (x+y)*(x-y).
From that, you get (x+y)*(x-y) = 12
Since x and y are integers, (x+y) and (x-y) must be integers.

How can you factor a 12?
4*3
6*2
12*1
We can disregard any factors with two negative terms, since (x+y) must be positive.

For 4*3: if x+y = 4 and x-y is 3, then x=3.5 and y=0.5.
For 6*2: if x+y = 6 and x-y is 2, then x=4 and y=2.
For 12*1: if x+y = 12 and x-y is 1, then x=x=6.5 and y=5.5

Since x and y must be integers, the only one of these that works is the middle one: x=4 and y=2.
In that case, x-y = 4-2 = 2, so the only possible answer is 2.

Hope that helps!

Answer 3

x-y = 2 where x = 4 and y =2 is the only combination that satisfies x>y>0,
x-y =2, and
x^2-y^2 = 12.

If x-y = 1, then (x-y)(x+y)=12 = x+y This would require 2x =13 or x = 6.5 and y= 5.5 These values obviously do not satisfy x^2 + y^2 = 12.

Similarly if x-y = 4, then (x-y)(x+y) = 12 so (x+y) = 3. This wo
uld require 2x = 7 or x = 3.5 and y = 8.5. These two numbers do not satisfy x^2+y^2 = 12.

So x-y = 2 is the solution.

Answer 4

x=4,y=2

Answer 5

x-y = 2.

x² - y² = (x+y)(x-y) = 12

there are only a few possible integer factor pairs for 12:

12•1
6•2
4•3

if you add (x+y) and (x-y) you get 2x -- the sum of the factors must be even. That's only true for 6•2, so x+y = 6 and x-y = 2. x = 4, y = 2.

Answer 6

x^2 - y^2 = 12
(x+y)(x-y)=12

if x-y=1 then x+y =12 which means 2x=13 and that's not possible

if x-y = 2, then x+y = 6 which means 2x = 8, x=4 and y = 2 (so this is correct)

if x-y = 4 then x+y=3 which means 2x=7 and that again is not possible