Algebra II double question homework help?

Question

Write and solve an equality to find three consecutive whole numbers with a sum between 13 and 16.

The length of a sidee of any triangle is less than the sum of the elngths of the other two sides. In (Triangle) PQR, PR = RQ+4 and RQ<11. Write and solve an inequality for PQ.

Help how do i do these?

Answer 1

Call the 3 consecutive whole numbers a,b,c then:
b = a+1
c = b+1 = (a+1)+1=a+2

13 < a + b +c < 16
substitute
13 < a + a + 1 + a +2 < 16
group like terms
13 < 3a+3 < 16
subtract 3
10 < 3a < 13
divide by 3
10/3 < a < 13/3
the only such whole number is a = 4
thus (a,b,c)=(a,a+1,a+2)=(4,5,6)

If RQ < 11 then
PR = RQ +4 < 11+4 = 15

The identity states
PQ < PR + RQ
since RQ < 11 and PR < 15
PQ < 11 + 15 = 26

Note: Previous 2 responses use <= and >= where < and > are more appropriate to the exact problem as stated.

Answer 2

Let x be the first number, then the two consecutive numbers are x+1 and x+2. The "equality", then, is

13 <= x + x + 1 + x + 2 <= 16

13 <= 3x + 3 <= 16

10 <= 3x <= 14

3.3_ <= x <= 4.6_

Since x must be an integer, the only possible solution is x = 4, making the numbers 4, 5 and 6. The sum is 14, which fits.

Check: use 3, 4 and 5: the sum is 12, which doesn't work.
Use 5, 6 and 7: the sum is 18, which also doesn't work, so our solution is verified.

Edit: Andy, honey? If you tell me you need between 10 and 20 dollars and I give you 20, are you going to object? Generally speaking, between a and b includes both a and b. Besides - the answer turns out the same either way. Don't be quite such a prig, shall we?

Answer 3

Rq<11
Pr= RQ+4
that means that
PR< or = to 15 (11+4)
since PQ must be less than PR+ RQ, 11+15= 26
PQ must be less than 26