3n+2 <= 14 and 6n/(n^2+5) > 1 Find possible values of N

3n+2 is less then or equaled to 14. 6n divided by N squared + 5 is less then 1. First, we are going to deal with the equation on the left. From this equation, takeaway 2 from both sides to get 3n<=12, divide both sides by 3 to get n is less then or equaled to 4. So we now know n<=4. Now for the second equation, multiply the right side of the equation by (n^2+5), you then get 6n>n^2+5. Takeaway 6n from both sides to get 0>n^2-6n+5. Factorise n^2-6n+5 to get 0>(n-5)(n-1). So we now know that n is between 5 and 1 for the right equation. To satisfy both conditions of the left equation and the right equation, we know that n is less then or equaled to 4, and n is between 5 and 1. This means then n is greater then 1 and less then equaled to 4. So the possible values must be 2 3 and 4.