Question

How many distinct natural numbers are in the solution set of {x:|2x+6|≥4} and (intersection) {x:|3x+20|≤29} ? Please explain.

Answer 1

I'm not sure about this notation. Are those absolute value signs?

Answer 2

yes

Answer 3

Start with |2x + 6| >= 4 is 2 sets, the first of which, for 2x + 6 greater than 0 -> 2x +6 >= 4 and the second, for 2x + 6 less than 0, 2x + 6 <= -4.

Answer 4

yeah i know that the first equation splits in x≥-1 or x≤-5 and the second equation splits into x≤3 and x≥-49/3
i just don't know how to find how many natural numbers there are in the solution

Answer 5

For |3x + 20| <= 29, you have -29 <= 3x + 20 <= 29

Answer 6

I'd say it's infinity

Answer 7

You can list each of the "halves" with ellipses notation (...) after solving and then work the intersection.

Answer 8

just kidding

Answer 9

okay so do you know the solution set of the two equations?

Answer 10

So, working with the second set, or -29 <= 3x + 20 <= 29 we get -49 <= 3x <= 9 or (-49)/3 <= x <= 3. So this will give the explicit values for x for the SECOND set, once you take into account only natural numbers. Remember your definition of natural numbers.

Answer 11

yes you do. now the natural numbers are from 1 on to infinity (0 is disputed) so you can discard all the negative numbers in the solution sets. the intersection of the positive solution sets is you answer

Answer 12

can you help me find the intersection of the positive solution sets?

Answer 13

Remeber that natural numbers are 1,2,3,... That is the "counting" numbers. Starting with "1" and no negatives. "0" is not disputed in the set of natural numbers. If you include "0", you get the "whole" numbers.

Answer 14

wait is the answer 3?

Answer 15

one of the solutions sets is -1to infinity and the other one is x less than equal to 3. yes

Answer 16

The solution set to to the second inequality is [1,2,3]. Now, we have to look again at the first inequality.

Answer 17

it's 3

Answer 18

We do not want negative numbers, so for the first inequality, we just have to look at 2x +6 >= 4. That set will be all natural numbers. So the intersection set is {1,2,3}. Three numbers. Done.