# Elements of a set. Find the elements of the set A A = {x is in Z: x=(6y-7)/(2y+1), y belongs to N}

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In order to determine the elements of the set A, we'll consider the given constraint of the set, that is that all the elements are integer numbers (x integer).

For x to be integer, the denominator (2y+1) has to be the divisor of the numerator (6y-7).

We'll write the division with reminder:

(6y-7)=3(2y+1)-10

We'll divide both sides by (2y+1) and we'l get:

(6y-7)/(2y+1) = 3-10/(2y+1)

But x = (6y-7)/(2y+1) and x integer, so, in order to obtain x integer, (2y+1) has to be a divisor of 10.

We'll write 10 divisors:

D10=+/-2;+/-5;+/-1;+/-10

(2y+1)=1

2y=0,

y=0

x=(6*0-7)/(2*0+1)

x = -7/1

x = -7

(2y+1)=-1

2y=-2

y=-1

x=(6*(-1)-7)/(2*(-1)+1)

x = 13

(2y+1)=2

2y=1

y=1/2 rejected (not integer)

(2y+1)=-2

2y=-3

y=-3/2 rejected (not integer)

(2y+1)=5

2y=4

y=2

x=(6*(2)-7)/(2*(2)+1)

x=5/5

x = 1

(2y+1)=-5

2y=-6

y=-3

x=(6*(-3)-7)/(2*(-3)+1)

x = 5

(2y+1)=10

2y=9

y=9/2 rejected (not integer)

(2y+1)=-10

2y=-11

y=-11/2 rejected (not integer)

A={-7,1,5,13}