# 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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### 1 Answer

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}