Question

34. Find four consecutive odd integers such that five times the sum of the first and third integers exceeds four times the sum of the second and last integers by 14. Explain.

Answer 1

treat the first integer as x, the next as x + 2, and the third as x + 4

Answer 2

x
x +2
x+4
whre x is odd
5(x + x+4) = 4(x+2 +x+4) +14

Answer 3

10x +20 = 8x + 12 +14
2x = 6
x= 3
5
7

Answer 4

@P0sitr0n should have a '>' sign?

Answer 5

read till the end.

Answer 6

i read the word 'exceeds'

Answer 7

by 14, so its 14 plus?

Answer 8

okay

Answer 9

3, 5, and 7 do not seem to work?

Answer 10

18, 20, 22 is the answer, but how...

Answer 11

these are not odd integers...?

Answer 12

lol true

Answer 13

@P0sitr0n, seems to be an arithmetic error above

Answer 14

5(x + x+4) = 4(x+2 +x+4) +14 does not equal 10x +20 = 8x + 12 +14

Answer 15

Oh wait sorry, the answer is 13, 15, 17, 19

Answer 16

And it's 4 consecutive odd integers.

Answer 17

i get 11, 13, 15

Answer 18

the right hand value of the above equation should be
10x +20 = 8x + 24 +14

Answer 19

solve that and x will equal 9, but since you need to exceed, you can use 9, you have to move up one odd integer.

Answer 20

you CAN'T use 9, sorry

Answer 21

Alright, can you list the steps from the beginning?

Answer 22

@Albert0898, do you see how we arrived at the equation?

Answer 23

No, this one is kind of confusing to me.

Answer 24

okay, I will try to explain. you need three consecutive odd integers.

start with the first one, we called that x. the next would be 'x + 2', and the the last one would be 'x + 4'

still with me?

Answer 25

Yes

Answer 26

okay, then the problem calls for the FIVE TIMES THE SUM of the first and third numbers, which are x and 'x + 4' to EXCEED FOUR TIMES THE SUM of the second and last integers ('x + 2' and 'x + 4') by 14.

Answer 27

we arrive at this inequality:\[5(x + x + 4) > 4(x + 2 + x + 4) + 14\]

Answer 28

\[5x + 5x + 20 > (4x + 8 + 4x + 16) + 14\]

Answer 29

AHHHHH, I see now!! I wasn't sure if I was allowed to put a greater than sign because I couldn't figure out how I'd get the equivalent to 'x'. But please do continue

Answer 30

\[10x + 20 > 8x + 38\]\[2x > 18\]\[x > 9\]

Answer 31

glad to know you see it. wonderful!

since x has to be GREATER than 9, we have to move to 11

Answer 32

So x has to be greater than 9? There's an infinite amount of choices for that.

Answer 33

is there?

Answer 34

The answer is 13, 15, 17, 19 ~ I sort of get it now

Answer 35

good. i guess there are an infinite amount. you only need three integers and 11 is the first you can start with

Answer 36

and thanks for the medal. good luck

Answer 37

You're Welcome and Thank You!