PLEASE HELP ME!!!!!!!!!!!!!
Write an inequality for the sum of twice a number and 5 is at most 15. Use y as the variable.

Answer:

The answer to the problem is 5

Step-by-step explanation:

The sum of twice a number y and 5 is at most 15" can be translated mathematically into the following inequality:

2y + 5 ≤ 15 since the sum, 2y + 5, is at most 15 but could be less than 15.

To solve this inequality for n, proceed as follows:

First, subtract 5 from both sides of the inequality as you would in solving an equation:

2y + 5 - 5 ≤ 15 - 5

2y + 0 ≤ 10

2y ≤ 10

Now, to finally solve the inequality for the variable y, divide both sides of the inequality by 2 as you would in solving an equation:

(2y)/2 ≤ 10/2

(2/2)y ≤ 10/2

(1)y ≤ 5

n ≤ 5 which is all real number less than or equal to 5.

Test Values (y = -1/2, 0, 3, 5, and n = 7):

For y = -1/2:

2y + 5 ≤ 15

2(-1/2) + 5 ≤ 15

-1 + 5 ≤ 15

-4 ≤ 15 (TRUE)

For y = 0:

2y + 5 ≤ 15

2(0) + 5 ≤ 15

0 + 5 ≤ 15

5 ≤ 15 (TRUE)

For y = 3:

2y + 5 ≤ 15

2(3) + 5 ≤ 15

6 + 5 ≤ 15

11 ≤ 15 (TRUE)

For y = 5:

2y + 5 ≤ 15

2(5) + 5 ≤ 15

10 + 5 ≤ 15

15 ≤ 15 (TRUE)

For y = 7:

2y + 5 ≤ 15

2(7) + 5 ≤ 15

14 + 5 ≤ 15

19 ≤ 15 (FALSE)

Therefore, the possible values for y which will make the relevant inequality, 2n + 5 ≤ 15, a true statement are:

{y | y is a real number and y ≤ 5}

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