how do you solve 5 <= x + 2 <= 11?

5 ≤ x + 2 ≤ 11

3 ≤ x ≤ 9

5 ≤ x + 2 ≤ 11
5 - 2 ≤ x ≤ 11 - 2
3 ≤ x ≤ 9

the answer is 3<x and x<9

11-x greater advantageous than/equivalent to 2(x+3) <---------- If thats the luxurious version of the equation this is your answer: 11 - x greater advantageous than/equivalent to 2x + 6 5 - x greater advantageous than/equivalent to 2x 5 greater advantageous than/equivalent to 3x x greater advantageous than/equivalent to 5/3

3 <= x <= 9

subtract 2 everywhere then x - 2 = x + 2 - 2 = x + 0 = x and you're done

3 ≤ x ≤ 9

x is all the numbers greater than or equal to 3 and less than or equal to 9

5 <= 2 + x <= 11
3<=x<=9

5-2 <= x <= 11-2
3<= x <=9

5 -2<= x <= 11-2
3<=x<=9
x lies between [3,9]